W7. Eigenvalues, Eigenvectors, Diagonalization of Matrices

1. Summary

1.1 Introduction to Eigenvalues and Eigenvectors

In linear algebra, eigenvalues and eigenvectors represent one of the most powerful concepts for understanding the structure and behavior of matrices. At their core, eigenvalues and eigenvectors answer a fundamental question: Are there any special directions in space where a linear transformation behaves like simple scaling?

When we apply a matrix \(A\) to a vector \(\mathbf{v}\), we typically get a vector that points in a completely different direction and has a different magnitude. However, for certain special vectors—called eigenvectors—the matrix transformation only scales the vector without changing its direction. The scaling factor is the eigenvalue.

Definition (Eigenvector and Eigenvalue): Let \(A\) be an \(n \times n\) matrix. A nonzero vector \(\mathbf{v}\) is called an eigenvector of \(A\) if there exists a scalar \(\lambda\) such that:

\[A\mathbf{v} = \lambda\mathbf{v}\]

The scalar \(\lambda\) is called the eigenvalue corresponding to \(\mathbf{v}\).

This equation is the defining property: when we multiply \(A\) by the eigenvector \(\mathbf{v}\), we get the same vector multiplied by a constant. The vector “stays on its line” through the origin—its direction is unchanged, only its magnitude (and possibly sign) changes.

Geometric Intuition: Imagine a matrix as a transformation that stretches, rotates, and shears space. An eigenvector points in a direction that is special for this transformation: applying the transformation along this direction is equivalent to scaling. This makes eigenvectors the “natural axes” or “principal directions” of the transformation.

1.1.1 Simple Examples

Let’s verify with a concrete example. Consider the matrix \(A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\) and the vector \(\mathbf{v} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\).

Compute \(A\mathbf{v}\):

\[A\mathbf{v} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 1 \end{bmatrix}\]

Since \(A\mathbf{v} = 3\mathbf{v}\), the vector \(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\) is an eigenvector with eigenvalue \(\lambda = 3\).

1.2 Finding Eigenvalues: The Characteristic Equation

To find eigenvalues systematically, we use the characteristic equation. Starting from the definition \(A\mathbf{v} = \lambda\mathbf{v}\), we can rewrite this as:

\[A\mathbf{v} - \lambda\mathbf{v} = \mathbf{0}\]

\[(A - \lambda I)\mathbf{v} = \mathbf{0}\]

where \(I\) is the identity matrix. This is a homogeneous system of linear equations. For a nonzero solution \(\mathbf{v}\) to exist, the matrix \((A - \lambda I)\) must be singular (non-invertible), which happens if and only if:

\[\det(A - \lambda I) = 0\]

This equation is called the characteristic equation of \(A\). The left side, \(\det(A - \lambda I)\), is called the characteristic polynomial and is a polynomial of degree \(n\) in the variable \(\lambda\).

Algorithm to Find Eigenvalues:

1. Compute the characteristic polynomial: Calculate \(\det(A - \lambda I)\).
2. Set equal to zero: Solve \(\det(A - \lambda I) = 0\).
3. Find the roots: The roots of the characteristic polynomial are the eigenvalues of \(A\).

Important Note: The Fundamental Theorem of Algebra guarantees that over the complex numbers \(\mathbb{C}\), every \(n \times n\) matrix has at least one eigenvalue (and at most \(n\) distinct eigenvalues). However, a real matrix may have complex eigenvalues.

Special Case—Triangular Matrices: If \(A\) is upper or lower triangular, the characteristic polynomial becomes:

\[\det(A - \lambda I) = \prod_{i=1}^n (a_{ii} - \lambda)\]

Therefore, the eigenvalues of a triangular matrix are exactly its diagonal entries.

1.3 Finding Eigenvectors

Once we have found an eigenvalue \(\lambda\), we find the corresponding eigenvectors by solving the homogeneous system:

\[(A - \lambda I)\mathbf{v} = \mathbf{0}\]

The set of all solutions to this equation forms the eigenspace for eigenvalue \(\lambda\). This is a subspace of \(\mathbb{R}^n\) and is always nonempty (it contains at least the zero vector). From this eigenspace, we select nonzero vectors to be our eigenvectors.

Algorithm to Find Eigenvectors:

1. Form the matrix \((A - \lambda I)\) by subtracting \(\lambda\) from the diagonal entries of \(A\).
2. Row reduce \((A - \lambda I)\) to find its null space.
3. Express solutions parametrically in terms of free variables.
4. Identify basis vectors for the null space—these are the eigenvectors.

1.4 Trace and Determinant Relationships

Two important relationships connect eigenvalues to properties we can compute directly from the matrix:

Trace Property: The sum of all eigenvalues (counted with algebraic multiplicity) equals the trace of \(A\):

\[\lambda_1 + \lambda_2 + \cdots + \lambda_n = \text{tr}(A) = a_{11} + a_{22} + \cdots + a_{nn}\]

Determinant Property: The product of all eigenvalues (counted with algebraic multiplicity) equals the determinant of \(A\):

\[\lambda_1 \cdot \lambda_2 \cdots \lambda_n = \det(A)\]

These properties provide useful checks when computing eigenvalues by hand. If your computed eigenvalues don’t satisfy these relationships, you’ve made an error.

1.5 Linear Independence of Eigenvectors

A crucial property that guarantees nice structure is:

Theorem: If \(\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\) are eigenvectors corresponding to distinct eigenvalues \(\lambda_1, \lambda_2, \ldots, \lambda_k\), then these vectors are linearly independent.

This means that eigenvectors corresponding to different eigenvalues can never be linearly dependent. This property is fundamental to diagonalization.

1.6 Introduction to Diagonalization

The ultimate goal of eigenvalue analysis is often diagonalization—rewriting the matrix in a simpler form using its eigenvalues and eigenvectors.

An \(n \times n\) matrix \(A\) is diagonalizable if we can find an invertible matrix \(P\) and a diagonal matrix \(D\) such that:

\[A = PDP^{-1}\]

Equivalently, \(D = P^{-1}AP\).

What are \(P\) and \(D\)? - The diagonal matrix \(D\) has the eigenvalues of \(A\) on its diagonal: \(D = \text{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n)\). - The matrix \(P\) has the corresponding eigenvectors as its columns: \(P = [\mathbf{v}_1 \ \mathbf{v}_2 \ \ldots \ \mathbf{v}_n]\).

When Can We Diagonalize? The key requirement is that we need \(n\) linearly independent eigenvectors. This is automatically satisfied if:

• Case 1: \(A\) has \(n\) distinct eigenvalues. In this case, the corresponding \(n\) eigenvectors are automatically linearly independent.
• Case 2: \(A\) has repeated eigenvalues. We must check that for each repeated eigenvalue \(\lambda\), the number of linearly independent eigenvectors equals its multiplicity in the characteristic polynomial. This is checked by examining the dimension of the eigenspace.

A matrix that has repeated eigenvalues but fewer than \(n\) linearly independent eigenvectors is called defective or not diagonalizable.

1.6.1 Why Diagonalization is Useful

Diagonalization is powerful because it simplifies matrix computations:

1. Powers of matrices: If \(A = PDP^{-1}\), then \(A^k = PD^kP^{-1}\). Since \(D\) is diagonal, \(D^k\) is easy to compute: \(D^k = \text{diag}(\lambda_1^k, \lambda_2^k, \ldots, \lambda_n^k)\).
2. Solving differential equations: Systems of differential equations \(\frac{d\mathbf{x}}{dt} = A\mathbf{x}\) are easily solved when \(A\) is diagonalized.
3. Understanding transformations: Diagonalization reveals the “principal axes” along which the transformation acts simply as scaling.

1.7 Spectral Properties

Several fundamental theorems describe special properties related to eigenvalues:

Theorem (Cayley-Hamilton): Every square matrix satisfies its own characteristic equation. If \(p(\lambda) = \det(A - \lambda I)\) is the characteristic polynomial, then \(p(A) = 0\) (the zero matrix).

Theorem (Spectral Mapping): If \(A\) has eigenvalues \(\lambda_1, \ldots, \lambda_n\), and \(p(x)\) is any polynomial, then \(p(A)\) has eigenvalues \(p(\lambda_1), \ldots, p(\lambda_n)\).

Theorem (Symmetric Matrices): If \(A\) is a real symmetric matrix (i.e., \(A = A^T\)), then:

• All eigenvalues are real numbers
• Eigenvectors corresponding to distinct eigenvalues are orthogonal
• \(A\) is always diagonalizable by an orthogonal matrix \(Q\): \(A = Q\Lambda Q^T\)

1.8 Special Cases and Applications

Understanding eigenvalues is essential across mathematics and science:

• Physics: Eigenvalues represent natural frequencies of vibrating systems, principal stresses in materials, and energy levels in quantum mechanics.
• Computer Science: PageRank (Google’s algorithm) is based on eigenvalues, as is facial recognition and image compression.
• Data Science: Principal Component Analysis (PCA), a fundamental technique in machine learning, uses eigenvalues to reduce dimensionality.
• Economics: Eigenvalue analysis models input-output systems and portfolio optimization.

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2. Definitions

• Eigenvalue: A scalar \(\lambda\) such that \(A\mathbf{v} = \lambda\mathbf{v}\) for some nonzero vector \(\mathbf{v}\) of matrix \(A\).
• Eigenvector: A nonzero vector \(\mathbf{v}\) such that \(A\mathbf{v} = \lambda\mathbf{v}\) for some scalar \(\lambda\) (eigenvalue) of matrix \(A\).
• Characteristic Polynomial: The polynomial \(p(\lambda) = \det(A - \lambda I)\) for an \(n \times n\) matrix \(A\).
• Characteristic Equation: The equation \(\det(A - \lambda I) = 0\), whose roots are the eigenvalues of \(A\).
• Eigenspace: The set of all eigenvectors (plus the zero vector) corresponding to a given eigenvalue \(\lambda\); equivalently, the null space of \((A - \lambda I)\).
• Algebraic Multiplicity: The multiplicity of an eigenvalue \(\lambda\) as a root of the characteristic polynomial.
• Geometric Multiplicity: The dimension of the eigenspace of eigenvalue \(\lambda\) (i.e., the number of linearly independent eigenvectors for that eigenvalue).
• Diagonalizable Matrix: A square matrix \(A\) for which there exists an invertible matrix \(P\) and diagonal matrix \(D\) such that \(A = PDP^{-1}\).
• Defective Matrix: A square matrix that has repeated eigenvalues but is not diagonalizable (geometric multiplicity < algebraic multiplicity for some eigenvalue).
• Trace of a Matrix: The sum of the diagonal entries: \(\text{tr}(A) = \sum_{i=1}^n a_{ii}\). This equals the sum of eigenvalues.
• Similar Matrices: Two matrices \(A\) and \(B\) are similar if there exists an invertible matrix \(P\) such that \(B = P^{-1}AP\). Similar matrices have the same eigenvalues.
• Orthogonal Matrix: A square matrix \(Q\) such that \(Q^T Q = QQ^T = I\), i.e., \(Q^{-1} = Q^T\).
• Spectral Theorem: The theorem stating that every real symmetric matrix is diagonalizable by an orthogonal matrix.

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3. Formulas

• Eigenvalue-Eigenvector Relationship: \(A\mathbf{v} = \lambda\mathbf{v}\) (defining equation)
• Characteristic Equation: \(\det(A - \lambda I) = 0\)
• Finding Eigenvectors: \((A - \lambda I)\mathbf{v} = \mathbf{0}\) (solve this homogeneous system)
• Trace and Sum of Eigenvalues: \(\text{tr}(A) = \lambda_1 + \lambda_2 + \cdots + \lambda_n\)
• Determinant and Product of Eigenvalues: \(\det(A) = \lambda_1 \cdot \lambda_2 \cdots \lambda_n\)
• Diagonalization: \(A = PDP^{-1}\) where \(P = [\mathbf{v}_1 \ \mathbf{v}_2 \ \ldots \ \mathbf{v}_n]\) and \(D = \text{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n)\)
• Powers via Diagonalization: \(A^k = PD^kP^{-1}\) where \(D^k = \text{diag}(\lambda_1^k, \lambda_2^k, \ldots, \lambda_n^k)\)
• Inverse via Diagonalization: \(A^{-1} = PD^{-1}P^{-1}\) where \(D^{-1} = \text{diag}(1/\lambda_1, 1/\lambda_2, \ldots, 1/\lambda_n)\) (if \(A\) is invertible)
• Spectral Decomposition (Symmetric Matrices): \(A = Q\Lambda Q^T\) where \(Q\) is orthogonal and \(\Lambda\) is diagonal
• Eigenvalue of Matrix Power: If \(\lambda\) is an eigenvalue of \(A\), then \(\lambda^k\) is an eigenvalue of \(A^k\)
• Eigenvalue of Inverse: If \(\lambda\) is an eigenvalue of an invertible matrix \(A\), then \(1/\lambda\) is an eigenvalue of \(A^{-1}\)
• Cayley-Hamilton Theorem: \(p(A) = 0\) where \(p(\lambda) = \det(A - \lambda I)\) is the characteristic polynomial

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4. Examples

4.1. Eigenvalue Properties (Lab 6, Task 1)

Find the eigenvalues of the matrix \(A = \begin{bmatrix} 1 & -1 \\ 2 & 4 \end{bmatrix}\). Verify that the trace equals the sum of eigenvalues and the determinant equals their product.

Click to see the solution

Key Concept: The characteristic equation \(\det(A - \lambda I) = 0\) yields the eigenvalues. These must satisfy the trace and determinant relationships.

1. Set up the characteristic equation: \[\det(A - \lambda I) = \det \begin{bmatrix} 1 - \lambda & -1 \\ 2 & 4 - \lambda \end{bmatrix} = 0\]
2. Expand the determinant: \[(1 - \lambda)(4 - \lambda) - (-1)(2) = 0\] \[\lambda^2 - 5\lambda + 4 + 2 = 0\] \[\lambda^2 - 5\lambda + 6 = 0\]
3. Solve: \[(\lambda - 2)(\lambda - 3) = 0 \implies \lambda_1 = 2,\quad \lambda_2 = 3\]
4. Verify trace property: \[\text{tr}(A) = 1 + 4 = 5, \quad \lambda_1 + \lambda_2 = 2 + 3 = 5 \quad \checkmark\]
5. Verify determinant property: \[\det(A) = 1 \cdot 4 - (-1)(2) = 6, \quad \lambda_1 \cdot \lambda_2 = 2 \cdot 3 = 6 \quad \checkmark\]

Answer: \(\lambda_1 = 2\), \(\lambda_2 = 3\). Both trace and determinant relations are verified.

4.2. Eigenvector Verification (Lab 6, Task 2)

Part A. For matrix \(A = \begin{bmatrix} 2 & 2 \\ -4 & 8 \end{bmatrix}\) and vector \(\mathbf{v} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\): is \(\mathbf{v}\) an eigenvector of \(A\)? If yes, what is \(\lambda\)?

Part B. For matrix \(B = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}\) and vector \(\mathbf{w} = \begin{bmatrix} -3 \\ 1 \end{bmatrix}\): is \(\mathbf{w}\) an eigenvector of \(B\)? If yes, what is \(\lambda\)?

Click to see the solution

Key Concept: A vector \(\mathbf{v}\) is an eigenvector of \(A\) if \(A\mathbf{v} = \lambda\mathbf{v}\) for some scalar \(\lambda\). Compute \(A\mathbf{v}\) and check if the result is a scalar multiple of \(\mathbf{v}\).

Part A:

1. Compute \(A\mathbf{v}\): \[A\mathbf{v} = \begin{bmatrix} 2 & 2 \\ -4 & 8 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 4 \\ 4 \end{bmatrix} = 4\begin{bmatrix} 1 \\ 1 \end{bmatrix}\]
2. Conclusion: Yes, \(\mathbf{v}\) is an eigenvector with \(\lambda = 4\).

Part B:

1. Compute \(B\mathbf{w}\): \[B\mathbf{w} = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix} \begin{bmatrix} -3 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} = 0\begin{bmatrix} -3 \\ 1 \end{bmatrix}\]
2. Conclusion: Yes, \(\mathbf{w}\) is an eigenvector with \(\lambda = 0\).

Answer: (A) \(\mathbf{v}\) is an eigenvector of \(A\) with \(\lambda = 4\). (B) \(\mathbf{w}\) is an eigenvector of \(B\) with \(\lambda = 0\).

4.3. Construct Matrix from Eigenvalues (Lab 6, Task 3)

Give three examples of \(2 \times 2\) matrices with eigenvalues \(\lambda_1 = 4\), \(\lambda_2 = 5\).

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Key Concept: The eigenvalues of a triangular (or diagonal) matrix are its diagonal entries. Any matrix similar to a diagonal matrix with entries 4 and 5 also has those eigenvalues, giving infinitely many solutions.

1. Diagonal matrix: \[A_1 = \begin{bmatrix} 4 & 0 \\ 0 & 5 \end{bmatrix}\]
2. Upper triangular matrix: \[A_2 = \begin{bmatrix} 4 & 1 \\ 0 & 5 \end{bmatrix}\]
3. Lower triangular matrix: \[A_3 = \begin{bmatrix} 4 & 0 \\ 3 & 5 \end{bmatrix}\]

Answer: Three valid examples are \(\begin{bmatrix} 4 & 0 \\ 0 & 5 \end{bmatrix}\), \(\begin{bmatrix} 4 & 1 \\ 0 & 5 \end{bmatrix}\), \(\begin{bmatrix} 4 & 0 \\ 3 & 5 \end{bmatrix}\). (Many other answers are possible.)

4.4. 3×3 Eigenvalues and Eigenvectors (Lab 6, Task 4)

Find the eigenvalues and eigenvectors of: \[A = \begin{bmatrix} 0 & 6 & 8 \\ 0.5 & 0 & 0 \\ 0 & 0.5 & 0 \end{bmatrix}\]

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Key Concept: For a 3×3 matrix, compute the characteristic polynomial by cofactor expansion, factor to find eigenvalues, then solve \((A - \lambda I)\mathbf{v} = \mathbf{0}\) for each eigenvalue.

Step 1: Compute the characteristic polynomial

\[\det(A - \lambda I) = \det \begin{bmatrix} -\lambda & 6 & 8 \\ 0.5 & -\lambda & 0 \\ 0 & 0.5 & -\lambda \end{bmatrix}\]

Expanding along the first column: \[= -\lambda(\lambda^2) - 0.5(-6\lambda - 4) = -\lambda^3 + 3\lambda + 2\]

Step 2: Factor and find eigenvalues

\[-(\lambda - 2)(\lambda + 1)^2 = 0\]

Therefore: \(\lambda_1 = 2\) (multiplicity 1) and \(\lambda_2 = -1\) (multiplicity 2).

Step 3: Eigenvectors for \(\lambda_1 = 2\)

Row-reduce \((A - 2I)\): \[\begin{bmatrix} -2 & 6 & 8 \\ 0.5 & -2 & 0 \\ 0 & 0.5 & -2 \end{bmatrix} \to \begin{bmatrix} 1 & 0 & -16 \\ 0 & 1 & -4 \\ 0 & 0 & 0 \end{bmatrix}\]

Free variable \(z = t\): \(x = 16t\), \(y = 4t\).

\[\mathbf{v}_1 = \begin{bmatrix} 16 \\ 4 \\ 1 \end{bmatrix}\]

Step 4: Eigenvectors for \(\lambda_2 = -1\)

Row-reduce \((A + I)\): \[\begin{bmatrix} 1 & 6 & 8 \\ 0.5 & 1 & 0 \\ 0 & 0.5 & 1 \end{bmatrix} \to \begin{bmatrix} 1 & 0 & -4 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix}\]

Free variable \(z = t\): \(x = 4t\), \(y = -2t\).

\[\mathbf{v}_2 = \begin{bmatrix} 4 \\ -2 \\ 1 \end{bmatrix}\]

Answer: \(\lambda_1 = 2\) with \(\mathbf{v}_1 = \begin{bmatrix} 16 \\ 4 \\ 1 \end{bmatrix}\); \(\lambda_2 = -1\) (multiplicity 2) with \(\mathbf{v}_2 = \begin{bmatrix} 4 \\ -2 \\ 1 \end{bmatrix}\).

4.5. Diagonalization (Lab 6, Task 5)

Diagonalize the matrix \(A = \begin{bmatrix} \frac{1}{2} & \frac{3}{2} \\ \frac{3}{2} & \frac{1}{2} \end{bmatrix}\) (find \(C\) and \(D\) such that \(A = CDC^{-1}\)).

Click to see the solution

Key Concept: To diagonalize \(A\), find its eigenvalues and eigenvectors, form \(C\) with eigenvectors as columns and \(D\) with eigenvalues on the diagonal, then \(A = CDC^{-1}\).

Step 1: Find eigenvalues

\[\det(A - \lambda I) = \left(\tfrac{1}{2} - \lambda\right)^2 - \tfrac{9}{4} = \lambda^2 - \lambda - 2 = (\lambda + 1)(\lambda - 2) = 0\]

Therefore: \(\lambda_1 = -1\), \(\lambda_2 = 2\).

Step 2: Eigenvector for \(\lambda_1 = -1\)

\[(A + I)\mathbf{v} = \mathbf{0}: \quad \begin{bmatrix} \frac{3}{2} & \frac{3}{2} \\ \frac{3}{2} & \frac{3}{2} \end{bmatrix} \mathbf{v} = \mathbf{0} \implies x = -y\]

\[\mathbf{v}_1 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}\]

Step 3: Eigenvector for \(\lambda_2 = 2\)

\[(A - 2I)\mathbf{v} = \mathbf{0}: \quad \begin{bmatrix} -\frac{3}{2} & \frac{3}{2} \\ \frac{3}{2} & -\frac{3}{2} \end{bmatrix} \mathbf{v} = \mathbf{0} \implies x = y\]

\[\mathbf{v}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\]

Step 4: Form \(C\), \(D\), and \(C^{-1}\)

\[C = \begin{bmatrix} -1 & 1 \\ 1 & 1 \end{bmatrix}, \quad D = \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}\]

\[\det(C) = -2, \quad C^{-1} = \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix}\]

Answer: \[A = \begin{bmatrix} -1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix}\]

4.6. Matrix Power via Diagonalization (Lab 6, Task 6)

For \(A = \begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix}\), compute \(A^{100}\) using diagonalization.

Click to see the solution

Key Concept: If \(A = CDC^{-1}\), then \(A^{100} = CD^{100}C^{-1}\), and \(D^{100}\) is trivial to compute since \(D\) is diagonal.

Step 1: Find eigenvalues

\[\text{tr}(A) = 6, \quad \det(A) = 5 \implies \lambda^2 - 6\lambda + 5 = (\lambda - 1)(\lambda - 5) = 0\]

Therefore: \(\lambda_1 = 5\), \(\lambda_2 = 1\).

Step 2: Find eigenvectors

For \(\lambda_1 = 5\): \((A - 5I)\mathbf{v} = \mathbf{0}\): \(x = 3y \implies \mathbf{v}_1 = \begin{bmatrix} 3 \\ 1 \end{bmatrix}\)

For \(\lambda_2 = 1\): \((A - I)\mathbf{v} = \mathbf{0}\): \(x = -y \implies \mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\)

Step 3: Form the diagonalization

\[C = \begin{bmatrix} 3 & 1 \\ 1 & -1 \end{bmatrix}, \quad D = \begin{bmatrix} 5 & 0 \\ 0 & 1 \end{bmatrix}\]

\[\det(C) = -4, \quad C^{-1} = \frac{1}{4}\begin{bmatrix} 1 & 1 \\ 1 & -3 \end{bmatrix}\]

Step 4: Compute \(A^{100}\)

\[D^{100} = \begin{bmatrix} 5^{100} & 0 \\ 0 & 1 \end{bmatrix}\]

\[A^{100} = C D^{100} C^{-1} = \frac{1}{4}\begin{bmatrix} 3 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 5^{100} & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -3 \end{bmatrix}\]

\[= \frac{1}{4}\begin{bmatrix} 3 \cdot 5^{100} + 1 & 3 \cdot 5^{100} - 3 \\ 5^{100} - 1 & 5^{100} + 3 \end{bmatrix}\]

Answer: \[A^{100} = \frac{1}{4}\begin{bmatrix} 3 \cdot 5^{100} + 1 & 3(5^{100} - 1) \\ 5^{100} - 1 & 5^{100} + 3 \end{bmatrix}\]

4.7. Find Eigenvalues of Triangular Matrix (Assignment 6, Task 1)

Find the eigenvalues and corresponding eigenvectors of the matrix: \[A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}\]

Click to see the solution

Key Concept: For a 2×2 matrix, compute the characteristic polynomial, solve for eigenvalues, then find eigenvectors for each eigenvalue.

Step 1: Find eigenvalues

\[\det(A - \lambda I) = \det\begin{pmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{pmatrix} = 0\]

\[(4 - \lambda)(3 - \lambda) - 2 = 0\] \[12 - 7\lambda + \lambda^2 - 2 = 0\] \[\lambda^2 - 7\lambda + 10 = 0\] \[(\lambda - 2)(\lambda - 5) = 0\]

Therefore: \(\lambda_1 = 2\) and \(\lambda_2 = 5\)

Step 2: Find eigenvector for \(\lambda_1 = 2\)

Solve \((A - 2I)\mathbf{v} = \mathbf{0}\):

\[\begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\]

This gives: \(2x + y = 0 \Rightarrow y = -2x\)

\[\mathbf{v}_1 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}\]

Step 3: Find eigenvector for \(\lambda_2 = 5\)

Solve \((A - 5I)\mathbf{v} = \mathbf{0}\):

\[\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\]

This gives: \(-x + y = 0 \Rightarrow y = x\)

\[\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\]

Answer: Eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = 5\). Corresponding eigenvectors are \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}\) and \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\).

4.8. Eigenspace and Algebraic/Geometric Multiplicity (Assignment 6, Task 2)

For the matrix \(C = \begin{pmatrix} 2 & 0 & 0 \\ 1 & 2 & 1 \\ -1 & 0 & 1 \end{pmatrix}\):

1. Find the characteristic polynomial of \(C\).
2. Find all eigenvalues of \(C\).
3. For each eigenvalue, find a basis for the corresponding eigenspace.

Click to see the solution

Key Concept: The characteristic polynomial is found via determinant. Eigenvalues are its roots. For each eigenvalue, the eigenspace is the null space of \((A - \lambda I)\).

Step 1: Find the characteristic polynomial

\[\det(C - \lambda I) = \det\begin{pmatrix} 2 - \lambda & 0 & 0 \\ 1 & 2 - \lambda & 1 \\ -1 & 0 & 1 - \lambda \end{pmatrix}\]

Since the first column has two zeros, expand along it: \[(2 - \lambda) \det\begin{pmatrix} 2 - \lambda & 1 \\ 0 & 1 - \lambda \end{pmatrix} = (2 - \lambda)[(2 - \lambda)(1 - \lambda) - 0]\]

\[= (2 - \lambda)^2(1 - \lambda)\]

Step 2: Find eigenvalues

From \((2 - \lambda)^2(1 - \lambda) = 0\): \[\lambda_1 = 2 \text{ (multiplicity 2)}, \quad \lambda_2 = 1 \text{ (multiplicity 1)}\]

Step 3: Find eigenspace for \(\lambda = 2\)

Solve \((C - 2I)\mathbf{v} = \mathbf{0}\):

\[\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 1 \\ -1 & 0 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\]

From the second row: \(x + z = 0 \Rightarrow z = -x\). The variable \(y\) is free.

\[\begin{pmatrix} x \\ y \\ z \end{pmatrix} = x\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} + y\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\]

Basis for eigenspace: \(\left\{ \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \right\}\) (geometric multiplicity = 2)

Step 4: Find eigenspace for \(\lambda = 1\)

Solve \((C - I)\mathbf{v} = \mathbf{0}\):

\[\begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ -1 & 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\]

From the first row: \(x = 0\). From the second: \(y + z = 0 \Rightarrow z = -y\).

\[\begin{pmatrix} x \\ y \\ z \end{pmatrix} = y\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}\]

Basis for eigenspace: \(\left\{ \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} \right\}\) (geometric multiplicity = 1)

Answer: (a) Characteristic polynomial: \((2 - \lambda)^2(1 - \lambda)\) (b) Eigenvalues: \(\lambda = 2\) (multiplicity 2) and \(\lambda = 1\) (multiplicity 1) (c) Eigenspace bases:

• For \(\lambda = 2\): \(\text{span}\left\{ \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \right\}\)
• For \(\lambda = 1\): \(\text{span}\left\{ \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} \right\}\)

4.9. Construct Matrix from Eigenvectors and Eigenvalues (Assignment 6, Task 3)

Let \(A\) be a \(2 \times 2\) matrix with eigenvalues \(\lambda_1 = 3\) and \(\lambda_2 = -2\), and corresponding eigenvectors \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\) and \(\mathbf{v}_2 = \begin{pmatrix} -1 \\ 1 \end{pmatrix}\). Find the matrix \(A\).

Click to see the solution

Key Concept: Use the defining equations \(A\mathbf{v}_i = \lambda_i\mathbf{v}_i\) to set up a system of linear equations for the matrix entries.

1. Write out the eigen-equations: Let \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\). Then: \[A\mathbf{v}_1 = \lambda_1\mathbf{v}_1 \Rightarrow \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = 3 \begin{pmatrix} 1 \\ 2 \end{pmatrix}\]

   \[A\mathbf{v}_2 = \lambda_2\mathbf{v}_2 \Rightarrow \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} -1 \\ 1 \end{pmatrix} = -2 \begin{pmatrix} -1 \\ 1 \end{pmatrix}\]

2. Expand first equation: \[\begin{pmatrix} a + 2b \\ c + 2d \end{pmatrix} = \begin{pmatrix} 3 \\ 6 \end{pmatrix}\]

   This gives: \(a + 2b = 3\) and \(c + 2d = 6\)

3. Expand second equation: \[\begin{pmatrix} -a + b \\ -c + d \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \end{pmatrix}\]

   This gives: \(-a + b = 2\) and \(-c + d = -2\)

4. Solve the system: From equations 1 and 3:

   • Adding: \(3b = 5 \Rightarrow b = \frac{5}{3}\)
   • Then: \(a = 3 - 2 \cdot \frac{5}{3} = 3 - \frac{10}{3} = -\frac{1}{3}\)

   From equations 2 and 4:

   • Adding: \(3d = 4 \Rightarrow d = \frac{4}{3}\)
   • Then: \(c = 6 - 2 \cdot \frac{4}{3} = 6 - \frac{8}{3} = \frac{10}{3}\)

5. Form the matrix: \[A = \begin{pmatrix} -\frac{1}{3} & \frac{5}{3} \\ \frac{10}{3} & \frac{4}{3} \end{pmatrix}\]

Answer: \(A = \begin{pmatrix} -\frac{1}{3} & \frac{5}{3} \\ \frac{10}{3} & \frac{4}{3} \end{pmatrix}\)

4.10. Verify Eigenvalue-Eigenvector Pair (Assignment 6, Task 4)

For matrix \(D = \begin{pmatrix} 5 & -4 & 2 \\ -4 & 5 & 2 \\ 2 & 2 & 2 \end{pmatrix}\), verify that \(\lambda = 1\) is an eigenvalue with eigenvector \(\mathbf{v} = \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}\).

Click to see the solution

Key Concept: To verify an eigenvalue-eigenvector pair, compute \(A\mathbf{v}\) and check if it equals \(\lambda\mathbf{v}\).

1. Compute \(D\mathbf{v}\): \[D\mathbf{v} = \begin{pmatrix} 5 & -4 & 2 \\ -4 & 5 & 2 \\ 2 & 2 & 2 \end{pmatrix} \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}\]
2. Calculate each component:
   • First row: \(5(2) + (-4)(2) + 2(-1) = 10 - 8 - 2 = 0\)
   • Second row: \((-4)(2) + 5(2) + 2(-1) = -8 + 10 - 2 = 0\)
   • Third row: \(2(2) + 2(2) + 2(-1) = 4 + 4 - 2 = 6\)
3. Result: \[D\mathbf{v} = \begin{pmatrix} 0 \\ 0 \\ 6 \end{pmatrix}\]
4. Check against \(\lambda\mathbf{v}\): \[\lambda\mathbf{v} = 1 \cdot \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}\]
5. Compare: \[\begin{pmatrix} 0 \\ 0 \\ 6 \end{pmatrix} \neq \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}\]

Answer: The pair \(\lambda = 1, \mathbf{v} = \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}\) does NOT satisfy \(D\mathbf{v} = \lambda\mathbf{v}\). There is an error in the original problem statement.

4.11. Prove Eigenvalue Property for Invertible Matrices (Assignment 6, Task 5)

Prove that if \(\lambda\) is an eigenvalue of an invertible matrix \(A\), then \(1/\lambda\) is an eigenvalue of \(A^{-1}\).

Click to see the solution

Key Concept: Use the definition of eigenvalue and properties of matrix inverses to transform the equation for \(A\) into one for \(A^{-1}\).

Proof:

1. Start with the given condition: Suppose \(\lambda\) is an eigenvalue of invertible matrix \(A\) with eigenvector \(\mathbf{v}\). Then: \[A\mathbf{v} = \lambda\mathbf{v}\]
2. Note that \(\lambda \neq 0\): Since \(A\) is invertible, \(\det(A) \neq 0\). But \(\det(A) = \prod \lambda_i\), so no eigenvalue can be zero.
3. Multiply both sides by \(A^{-1}\): \[A^{-1}(A\mathbf{v}) = A^{-1}(\lambda\mathbf{v})\] \[\mathbf{v} = \lambda A^{-1}\mathbf{v}\]
4. Divide both sides by \(\lambda\) (nonzero): \[\frac{1}{\lambda}\mathbf{v} = A^{-1}\mathbf{v}\]
5. Rearrange to standard form: \[A^{-1}\mathbf{v} = \frac{1}{\lambda}\mathbf{v}\]
6. Conclusion: This shows that \(\mathbf{v}\) is an eigenvector of \(A^{-1}\) with eigenvalue \(1/\lambda\).

Answer: If \(\lambda\) is an eigenvalue of invertible \(A\), then \(1/\lambda\) is an eigenvalue of \(A^{-1}\), with the same eigenvector.

4.12. True or False: Eigenvalue Properties (Assignment 6, Task 6)

Determine whether each statement is true or false. Justify your answer.

(a) If 0 is an eigenvalue of \(A\), then \(A\) is singular (non-invertible).

(b) If \(\lambda\) is an eigenvalue of \(A\), then \(\lambda^2\) is an eigenvalue of \(A^2\).

(c) A matrix can have different eigenvalues with the same eigenvector.

(d) If \(A\) and \(B\) have the same eigenvalues, then \(A = B\).

Click to see the solution

Key Concept: Reason through each statement using the definitions and properties of eigenvalues and eigenvectors.

(a) TRUE:

If 0 is an eigenvalue of \(A\), then there exists a nonzero vector \(\mathbf{v}\) such that: \[A\mathbf{v} = 0\mathbf{v} = \mathbf{0}\]

This means \(\mathbf{v}\) is in the null space of \(A\), so \(A\) has a non-trivial null space. Therefore, \(A\) is not invertible (singular).

(b) TRUE:

If \(A\mathbf{v} = \lambda\mathbf{v}\), compute \(A^2\mathbf{v}\): \[A^2\mathbf{v} = A(A\mathbf{v}) = A(\lambda\mathbf{v}) = \lambda A\mathbf{v} = \lambda(\lambda\mathbf{v}) = \lambda^2\mathbf{v}\]

So \(\mathbf{v}\) is an eigenvector of \(A^2\) with eigenvalue \(\lambda^2\).

(c) FALSE:

Suppose \(\mathbf{v}\) is an eigenvector with two different eigenvalues \(\lambda_1\) and \(\lambda_2\): \[A\mathbf{v} = \lambda_1\mathbf{v} \quad \text{and} \quad A\mathbf{v} = \lambda_2\mathbf{v}\]

Then: \[\lambda_1\mathbf{v} = \lambda_2\mathbf{v}\] \[(\lambda_1 - \lambda_2)\mathbf{v} = \mathbf{0}\]

Since \(\mathbf{v} \neq \mathbf{0}\), we must have \(\lambda_1 = \lambda_2\). So each eigenvector corresponds to exactly one eigenvalue.

(d) FALSE:

Counter-example: Consider \[A = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}\]

\(A\) is diagonal, so its eigenvalues are 1 and 2.

For \(B\), it’s upper triangular, so its eigenvalues are also the diagonal entries: 1 and 2.

However, \(A \neq B\) (their off-diagonal entries differ).

Answer: (a) TRUE, (b) TRUE, (c) FALSE, (d) FALSE.

4.13. Determine Diagonalizability (Assignment 6, Task 7)

Describe all matrices \(S\) that diagonalize the matrix: \[A = \begin{bmatrix} 4 & 0 \\ 1 & 2 \end{bmatrix}\]

Click to see the solution

Key Concept: A diagonalizing matrix \(S\) has columns that are eigenvectors of \(A\). Different choices of eigenvectors (and their ordering) give different matrices \(S\), all valid diagonalizers.

Step 1: Find eigenvalues

Since \(A\) is upper triangular, the eigenvalues are the diagonal entries: \(\lambda_1 = 4\) and \(\lambda_2 = 2\).

Step 2: Find eigenvector for \(\lambda_1 = 4\)

Solve \((A - 4I)\mathbf{v} = \mathbf{0}\):

\[\begin{bmatrix} 0 & 0 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\]

From the second row: \(x - 2y = 0 \Rightarrow x = 2y\)

\[\mathbf{v}_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix}\]

Step 3: Find eigenvector for \(\lambda_2 = 2\)

Solve \((A - 2I)\mathbf{v} = \mathbf{0}\):

\[\begin{bmatrix} 2 & 0 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\]

From the first row: \(2x = 0 \Rightarrow x = 0\). The variable \(y\) is free.

\[\mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\]

Step 4: Characterize all diagonalizing matrices

Since any nonzero scalar multiple of an eigenvector is still an eigenvector (for the same eigenvalue), the diagonalizing matrices are:

\[S = \begin{bmatrix} 2a & 0 \\ a & b \end{bmatrix} \quad \text{where } a \neq 0 \text{ and } b \neq 0\]

The first column is \(a\) times \(\mathbf{v}_1\), and the second column is \(b\) times \(\mathbf{v}_2\).

Answer: All matrices of the form \(S = \begin{bmatrix} 2a & 0 \\ a & b \end{bmatrix}\) with \(a \neq 0, b \neq 0\) diagonalize \(A\).

4.14. Identify Non-Diagonalizable Matrix (Assignment 6, Task 8)

Which of these matrices cannot be diagonalized?

\[A_1 = \begin{bmatrix} 2 & -2 \\ 2 & -2 \end{bmatrix}, \quad A_2 = \begin{bmatrix} 2 & 0 \\ 2 & -2 \end{bmatrix}, \quad A_3 = \begin{bmatrix} 2 & 0 \\ 2 & 2 \end{bmatrix}\]

Click to see the solution

Key Concept: A matrix is diagonalizable if it has \(n\) linearly independent eigenvectors. If it has repeated eigenvalues, we must check the dimension of the eigenspace.

Analysis of \(A_1 = \begin{bmatrix} 2 & -2 \\ 2 & -2 \end{bmatrix}\):

Find eigenvalues: \[\det(A_1 - \lambda I) = \det\begin{bmatrix} 2 - \lambda & -2 \\ 2 & -2 - \lambda \end{bmatrix} = (2-\lambda)(-2-\lambda) + 4\] \[= -4 - 2\lambda + 2\lambda + \lambda^2 + 4 = \lambda^2\]

So \(\lambda = 0\) with multiplicity 2.

Find eigenvectors: Solve \(A_1\mathbf{v} = \mathbf{0}\): \[\begin{bmatrix} 2 & -2 \\ 2 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \mathbf{0} \Rightarrow 2x - 2y = 0 \Rightarrow x = y\]

\[\mathbf{v} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\]

There is only one linearly independent eigenvector, but we need two. Thus \(A_1\) is NOT diagonalizable.

Analysis of \(A_2 = \begin{bmatrix} 2 & 0 \\ 2 & -2 \end{bmatrix}\):

Eigenvalues (diagonal entries, since upper triangular): \(\lambda_1 = 2, \lambda_2 = -2\) (distinct).

Since eigenvalues are distinct, the corresponding eigenvectors are automatically linearly independent. Thus \(A_2\) IS diagonalizable.

Analysis of \(A_3 = \begin{bmatrix} 2 & 0 \\ 2 & 2 \end{bmatrix}\):

Eigenvalues (diagonal entries, since upper triangular): \(\lambda = 2\) with multiplicity 2.

Find eigenvectors: Solve \((A_3 - 2I)\mathbf{v} = \mathbf{0}\): \[\begin{bmatrix} 0 & 0 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \mathbf{0}\]

From the second row: \(2x = 0 \Rightarrow x = 0\). The variable \(y\) is free.

\[\mathbf{v} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\]

There is only one linearly independent eigenvector for the repeated eigenvalue. Thus \(A_3\) is NOT diagonalizable.

Answer: \(A_1\) and \(A_3\) cannot be diagonalized. Only \(A_2\) is diagonalizable.

4.15. Verify Eigenvalue and Find Eigenvalues (Lecture 6, Example 1)

Find the eigenvalues of the matrix \(\mathbf{A} = \begin{bmatrix} 1 & -1 \\ 2 & 4 \end{bmatrix}\). Verify that the trace equals the sum of the eigenvalues and that the determinant equals their product.

Click to see the solution

Key Concept: The characteristic equation \(\det(A - \lambda I) = 0\) provides the eigenvalues. These satisfy the trace and determinant relationships.

1. Set up the characteristic equation: \[\det(\mathbf{A} - \lambda\mathbf{I}) = \det \begin{bmatrix} 1 - \lambda & -1 \\ 2 & 4 - \lambda \end{bmatrix} = 0\]

2. Calculate the determinant: \[(1 - \lambda)(4 - \lambda) - (-1)(2) = 0\] \[4 - \lambda - 4\lambda + \lambda^2 + 2 = 0\] \[\lambda^2 - 5\lambda + 6 = 0\]

3. Use the quadratic formula: \[\lambda = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2}\]

   Therefore: \(\lambda_1 = 3\) and \(\lambda_2 = 2\)

4. Verify trace property: \[\text{tr}(\mathbf{A}) = 1 + 4 = 5\] \[\lambda_1 + \lambda_2 = 3 + 2 = 5 \quad \checkmark\]

5. Verify determinant property: \[\det(\mathbf{A}) = 1 \cdot 4 - (-1) \cdot 2 = 4 + 2 = 6\] \[\lambda_1 \cdot \lambda_2 = 3 \cdot 2 = 6 \quad \checkmark\]

Answer: Eigenvalues are \(\lambda_1 = 3\) and \(\lambda_2 = 2\). Both trace and determinant relationships are verified.

4.16. Determine if a Vector is an Eigenvector (Lecture 6, Example 2)

Consider the matrix \(\mathbf{A} = \begin{bmatrix} 2 & 2 \\ -4 & 8 \end{bmatrix}\) and vector \(\mathbf{v} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\). Is \(\mathbf{v}\) an eigenvector of \(\mathbf{A}\)? If so, what is the eigenvalue?

Also consider the matrix \(\mathbf{B} = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}\) and vector \(\mathbf{w} = \begin{bmatrix} -3 \\ 1 \end{bmatrix}\). Is \(\mathbf{w}\) an eigenvector of \(\mathbf{B}\)? If so, what is the eigenvalue?

Click to see the solution

Key Concept: A vector is an eigenvector if \(A\mathbf{v} = \lambda\mathbf{v}\) for some scalar \(\lambda\). Compute \(A\mathbf{v}\) and check if it’s a scalar multiple of \(\mathbf{v}\).

(a) For matrix \(\mathbf{A}\) and vector \(\mathbf{v}\):

1. Compute \(\mathbf{A}\mathbf{v}\): \[\mathbf{A}\mathbf{v} = \begin{bmatrix} 2 & 2 \\ -4 & 8 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 + 2 \\ -4 + 8 \end{bmatrix} = \begin{bmatrix} 4 \\ 4 \end{bmatrix}\]

2. Check if this is a scalar multiple of \(\mathbf{v}\): \[\begin{bmatrix} 4 \\ 4 \end{bmatrix} = 4 \begin{bmatrix} 1 \\ 1 \end{bmatrix}\]

   Yes! This means \(\mathbf{A}\mathbf{v} = 4\mathbf{v}\).

(b) For matrix \(\mathbf{B}\) and vector \(\mathbf{w}\):

1. Compute \(\mathbf{B}\mathbf{w}\): \[\mathbf{B}\mathbf{w} = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix} \begin{bmatrix} -3 \\ 1 \end{bmatrix} = \begin{bmatrix} -3 + 3 \\ -6 + 6 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\]

2. Check if this is a scalar multiple of \(\mathbf{w}\): \[\begin{bmatrix} 0 \\ 0 \end{bmatrix} = 0 \begin{bmatrix} -3 \\ 1 \end{bmatrix}\]

   Yes! This means \(\mathbf{B}\mathbf{w} = 0\mathbf{w}\).

Answer:

• Yes, \(\mathbf{v}\) is an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda = 4\).
• Yes, \(\mathbf{w}\) is an eigenvector of \(\mathbf{B}\) with eigenvalue \(\lambda = 0\).

4.17. Find Three Matrices with Specified Eigenvalues (Lecture 6, Example 3)

Find three \(2 \times 2\) matrices that have eigenvalues \(\lambda_1 = 4\) and \(\lambda_2 = 5\).

Click to see the solution

Key Concept: There are infinitely many matrices with the same eigenvalues. We can construct them using triangular or diagonal matrices, where eigenvalues appear on the diagonal.

This problem has infinitely many solutions because the eigenvalues alone don’t determine the matrix uniquely. The easiest examples use triangular or diagonal matrices:

1. Diagonal matrix: \[\mathbf{D} = \begin{bmatrix} 4 & 0 \\ 0 & 5 \end{bmatrix}\] Eigenvalues are clearly 4 and 5 (diagonal entries).
2. Upper triangular matrix: \[\mathbf{U} = \begin{bmatrix} 4 & x \\ 0 & 5 \end{bmatrix}\] where \(x\) can be any nonzero number (e.g., \(x = 1, 2, 3, \ldots\)). Eigenvalues are 4 and 5 (diagonal entries).
3. Lower triangular matrix: \[\mathbf{L} = \begin{bmatrix} 4 & 0 \\ y & 5 \end{bmatrix}\] where \(y\) can be any nonzero number. Eigenvalues are 4 and 5 (diagonal entries).

Answer: Three possible matrices are: \[\begin{bmatrix} 4 & 0 \\ 0 & 5 \end{bmatrix}, \quad \begin{bmatrix} 4 & 1 \\ 0 & 5 \end{bmatrix}, \quad \begin{bmatrix} 4 & 0 \\ 2 & 5 \end{bmatrix}\]

(Many other answers are possible.)

4.18. Find Eigenvalues and Eigenvectors of a 3×3 Matrix (Lecture 6, Example 4)

Find the eigenvalues and eigenvectors of the matrix: \[\mathbf{A} = \begin{bmatrix} 0 & 6 & 8 \\ 0.5 & 0 & 0 \\ 0 & 0.5 & 0 \end{bmatrix}\]

Click to see the solution

Key Concept: For a 3×3 matrix, we compute the characteristic polynomial using cofactor expansion, find the roots (eigenvalues), then solve the system \((A - \lambda I)\mathbf{v} = \mathbf{0}\) for each eigenvalue.

Step 1: Calculate the characteristic polynomial

\[f(\lambda) = \det(\mathbf{A} - \lambda\mathbf{I}) = \det \begin{bmatrix} -\lambda & 6 & 8 \\ 0.5 & -\lambda & 0 \\ 0 & 0.5 & -\lambda \end{bmatrix}\]

Expand along the first column: \[f(\lambda) = -\lambda \det \begin{bmatrix} -\lambda & 0 \\ 0.5 & -\lambda \end{bmatrix} - 0.5 \det \begin{bmatrix} 6 & 8 \\ 0.5 & -\lambda \end{bmatrix} + 0\]

\[= -\lambda(\lambda^2) - 0.5(-6\lambda - 4)\] \[= -\lambda^3 + 3\lambda + 2\]

Step 2: Factor and find eigenvalues

\[f(\lambda) = -(\lambda - 2)(\lambda + 1)^2 = 0\]

Therefore: \(\lambda_1 = 2\) (with multiplicity 1) and \(\lambda_2 = -1\) (with multiplicity 2)

Step 3: Find eigenvectors for \(\lambda_1 = 2\)

Solve \((\mathbf{A} - 2\mathbf{I})\mathbf{v} = \mathbf{0}\):

\[\begin{bmatrix} -2 & 6 & 8 \\ 0.5 & -2 & 0 \\ 0 & 0.5 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\]

Row reduce to RREF: \[\begin{bmatrix} 1 & 0 & -16 \\ 0 & 1 & -4 \\ 0 & 0 & 0 \end{bmatrix}\]

In parametric form: \(x = 16t, y = 4t, z = t\)

\[\mathbf{v}_1 = t \begin{bmatrix} 16 \\ 4 \\ 1 \end{bmatrix}\]

Step 4: Find eigenvectors for \(\lambda_2 = -1\)

Solve \((\mathbf{A} + \mathbf{I})\mathbf{v} = \mathbf{0}\):

\[\begin{bmatrix} 1 & 6 & 8 \\ 0.5 & 1 & 0 \\ 0 & 0.5 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\]

Row reduce to RREF: \[\begin{bmatrix} 1 & 0 & -4 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix}\]

In parametric form: \(x = 4t, y = -2t, z = t\)

\[\mathbf{v}_2 = t \begin{bmatrix} 4 \\ -2 \\ 1 \end{bmatrix}\]

Answer: Eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = -1\) (multiplicity 2). Corresponding eigenvectors are \(\mathbf{v}_1 = \begin{bmatrix} 16 \\ 4 \\ 1 \end{bmatrix}\) and \(\mathbf{v}_2 = \begin{bmatrix} 4 \\ -2 \\ 1 \end{bmatrix}\) (each up to scalar multiples).

4.19. Diagonalize a 2×2 Matrix (Lecture 6, Example 5)

Diagonalize the matrix: \[\mathbf{A} = \begin{bmatrix} \frac{1}{2} & \frac{3}{2} \\ \frac{3}{2} & \frac{1}{2} \end{bmatrix}\]

Click to see the solution

Key Concept: Diagonalization requires finding eigenvalues and eigenvectors, then forming matrices \(\mathbf{C}\) (eigenvectors as columns) and \(\mathbf{D}\) (eigenvalues on diagonal) such that \(A = CDC^{-1}\).

Step 1: Find eigenvalues

\[\det(\mathbf{A} - \lambda\mathbf{I}) = \det \begin{bmatrix} \frac{1}{2} - \lambda & \frac{3}{2} \\ \frac{3}{2} & \frac{1}{2} - \lambda \end{bmatrix} = 0\]

\[\left(\frac{1}{2} - \lambda\right)^2 - \frac{9}{4} = 0\]

\[\lambda^2 - \lambda + \frac{1}{4} - \frac{9}{4} = 0\]

\[\lambda^2 - \lambda - 2 = 0\]

\[(\lambda + 1)(\lambda - 2) = 0\]

Therefore: \(\lambda_1 = -1\) and \(\lambda_2 = 2\)

Step 2: Find eigenvectors for \(\lambda_1 = -1\)

Solve \((\mathbf{A} + \mathbf{I})\mathbf{v} = \mathbf{0}\):

\[\begin{bmatrix} \frac{3}{2} & \frac{3}{2} \\ \frac{3}{2} & \frac{3}{2} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\]

This gives: \(\frac{3}{2}x + \frac{3}{2}y = 0 \Rightarrow x = -y\)

\[\mathbf{v}_1 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}\]

Step 3: Find eigenvectors for \(\lambda_2 = 2\)

Solve \((\mathbf{A} - 2\mathbf{I})\mathbf{v} = \mathbf{0}\):

\[\begin{bmatrix} -\frac{3}{2} & \frac{3}{2} \\ \frac{3}{2} & -\frac{3}{2} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\]

This gives: \(-\frac{3}{2}x + \frac{3}{2}y = 0 \Rightarrow x = y\)

\[\mathbf{v}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\]

Step 4: Form the diagonalization

\[\mathbf{C} = \begin{bmatrix} -1 & 1 \\ 1 & 1 \end{bmatrix}, \quad \mathbf{D} = \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}\]

To find \(\mathbf{C}^{-1}\): \[\det(\mathbf{C}) = -1 - 1 = -2\]

\[\mathbf{C}^{-1} = \frac{1}{-2} \begin{bmatrix} 1 & -1 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix}\]

Answer: \[\mathbf{A} = \begin{bmatrix} -1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix}\]

4.20. Use Diagonalization to Compute Matrix Powers (Lecture 6, Example 6)

For the matrix \(\mathbf{A} = \begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix}\), compute \(\mathbf{A}^{100}\).

Click to see the solution

Key Concept: Using diagonalization, \(A^{100} = CD^{100}C^{-1}\), where \(D^{100}\) is easy to compute since \(D\) is diagonal.

Step 1: Find eigenvalues

Using trace and determinant: \[\text{tr}(\mathbf{A}) = 4 + 2 = 6\] \[\det(\mathbf{A}) = 4 \cdot 2 - 3 \cdot 1 = 5\]

So \(\lambda_1 + \lambda_2 = 6\) and \(\lambda_1 \cdot \lambda_2 = 5\).

Solving \(\lambda^2 - 6\lambda + 5 = 0\): \((\lambda - 1)(\lambda - 5) = 0\)

Therefore: \(\lambda_1 = 5\) and \(\lambda_2 = 1\)

Step 2: Find eigenvectors for \(\lambda_1 = 5\)

Solve \((\mathbf{A} - 5\mathbf{I})\mathbf{v} = \mathbf{0}\):

\[\begin{bmatrix} -1 & 3 \\ 1 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \mathbf{0}\]

This gives: \(x = 3y\)

\[\mathbf{v}_1 = \begin{bmatrix} 3 \\ 1 \end{bmatrix}\]

Step 3: Find eigenvectors for \(\lambda_2 = 1\)

Solve \((\mathbf{A} - \mathbf{I})\mathbf{v} = \mathbf{0}\):

\[\begin{bmatrix} 3 & 3 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \mathbf{0}\]

This gives: \(x = -y\)

\[\mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\]

Step 4: Form the diagonalization

\[\mathbf{C} = \begin{bmatrix} 3 & 1 \\ 1 & -1 \end{bmatrix}, \quad \mathbf{D} = \begin{bmatrix} 5 & 0 \\ 0 & 1 \end{bmatrix}\]

Step 5: Compute \(\mathbf{A}^{100}\)

\[\mathbf{A}^{100} = \mathbf{C} \mathbf{D}^{100} \mathbf{C}^{-1}\]

\[\mathbf{D}^{100} = \begin{bmatrix} 5^{100} & 0 \\ 0 & 1 \end{bmatrix}\]

Therefore: \[\mathbf{A}^{100} = \begin{bmatrix} 3 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 5^{100} & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ 1 & -1 \end{bmatrix}^{-1}\]

(The numerical computation of \(5^{100}\) is impractical without a computer, but the formula shows the approach.)

Answer: \(\mathbf{A}^{100} = \mathbf{C} \begin{bmatrix} 5^{100} & 0 \\ 0 & 1 \end{bmatrix} \mathbf{C}^{-1}\), which can be computed once \(\mathbf{C}^{-1}\) is found.

4.21. Eigenvalues of a Triangular Matrix — Trace and Determinant Check (Tutorial 6, Task 1)

Find the eigenvalues of the matrix: \[A = \begin{bmatrix} 3 & 4 & 2 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix}\] Verify that \(\lambda_1 + \lambda_2 + \lambda_3 = \text{tr}(A)\) and \(\lambda_1 \lambda_2 \lambda_3 = \det(A)\).

Click to see the solution

Key Concept: For a triangular matrix, the eigenvalues are the diagonal entries directly.

1. Read off eigenvalues from the diagonal: \[\lambda_1 = 3, \quad \lambda_2 = 1, \quad \lambda_3 = 0\]

2. Verify trace: \[\text{tr}(A) = 3 + 1 + 0 = 4, \qquad \lambda_1 + \lambda_2 + \lambda_3 = 3 + 1 + 0 = 4 \quad \checkmark\]

3. Verify determinant: \[\det(A) = 3 \cdot 1 \cdot 0 = 0, \qquad \lambda_1 \lambda_2 \lambda_3 = 3 \cdot 1 \cdot 0 = 0 \quad \checkmark\]

Answer: \(\lambda_1 = 3\), \(\lambda_2 = 1\), \(\lambda_3 = 0\). Both trace and determinant relations hold.

4.22. Eigenvalues and Eigenvectors of a Symmetric 3×3 Matrix (Tutorial 6, Task 2)

Find the eigenvalues and eigenvectors of: \[B = \begin{bmatrix} 0 & 0 & 2 \\ 0 & 2 & 0 \\ 2 & 0 & 0 \end{bmatrix}\]

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Key Concept: Compute the characteristic polynomial \(\det(B - \lambda I) = 0\), then find eigenvectors for each eigenvalue.

1. Characteristic polynomial: \[\det(B - \lambda I) = \det \begin{bmatrix} -\lambda & 0 & 2 \\ 0 & 2-\lambda & 0 \\ 2 & 0 & -\lambda \end{bmatrix}\] Expanding along the second row: \((2 - \lambda)[(-\lambda)(-\lambda) - 4] = (2-\lambda)(\lambda^2 - 4) = (2-\lambda)(2-\lambda)(2+\lambda)\) \[= -({\lambda - 2})^2(\lambda + 2) = 0\]

2. Eigenvalues: \(\lambda_1 = 2\) (algebraic multiplicity 2), \(\lambda_2 = -2\) (multiplicity 1).

3. Eigenvectors for \(\lambda_1 = 2\): Solve \((B - 2I)\mathbf{v} = \mathbf{0}\): \[\begin{bmatrix} -2 & 0 & 2 \\ 0 & 0 & 0 \\ 2 & 0 & -2 \end{bmatrix} \to x_1 = x_3, \; x_2 \text{ free}\] Two independent eigenvectors: \(\mathbf{v}_1 = \begin{bmatrix}1\\0\\1\end{bmatrix}\), \(\mathbf{v}_2 = \begin{bmatrix}0\\1\\0\end{bmatrix}\).

4. Eigenvectors for \(\lambda_2 = -2\): Solve \((B + 2I)\mathbf{v} = \mathbf{0}\): \[\begin{bmatrix} 2 & 0 & 2 \\ 0 & 4 & 0 \\ 2 & 0 & 2 \end{bmatrix} \to x_1 = -x_3, \; x_2 = 0\] Eigenvector: \(\mathbf{v}_3 = \begin{bmatrix}1\\0\\-1\end{bmatrix}\).

Answer: \(\lambda_1 = \lambda_2 = 2\) with eigenvectors \(\begin{bmatrix}1\\0\\1\end{bmatrix}\), \(\begin{bmatrix}0\\1\\0\end{bmatrix}\); \(\lambda_3 = -2\) with eigenvector \(\begin{bmatrix}1\\0\\-1\end{bmatrix}\).

4.23. Characteristic Polynomial Matching (Tutorial 6, Task 3)

Find the values of \(a\), \(b\), and \(c\) in the third row of: \[A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ a & b & c \end{bmatrix}\] so that the characteristic polynomial \(|A - \lambda I|\) equals \(-\lambda^3 + 4\lambda^2 + 5\lambda + 6\).

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Key Concept: This is a companion matrix. For such a matrix, the characteristic polynomial is directly \(-\lambda^3 + c\lambda^2 + b\lambda + a\).

1. Compute \(\det(A - \lambda I)\) by cofactor expansion along the first column: \[\det \begin{bmatrix} -\lambda & 1 & 0 \\ 0 & -\lambda & 1 \\ a & b & c-\lambda \end{bmatrix} = -\lambda^3 + c\lambda^2 + b\lambda + a\]

2. Match coefficients with \(-\lambda^3 + 4\lambda^2 + 5\lambda + 6\): \[c = 4, \quad b = 5, \quad a = 6\]

Answer: \(a = 6\), \(b = 5\), \(c = 4\).

4.24. Eigenvectors vs. Eigenvalues — Fill in the Blanks (Tutorial 6, Task 4)

Fill in the blanks:

(a) If you know that \(\mathbf{x}\) is an eigenvector of \(A\), to find \(\lambda\) you need to _______.

(b) If you know that \(\lambda\) is an eigenvalue of \(A\), to find \(\mathbf{x}\) you need to _______.

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Key Concept: The eigenvector equation \(A\mathbf{x} = \lambda\mathbf{x}\) can be used in both directions.

(a) Compute \(A\mathbf{x}\) and then find the scalar \(\lambda\) such that \(A\mathbf{x} = \lambda\mathbf{x}\). Since \(\mathbf{x} \neq \mathbf{0}\), any component where \(x_i \neq 0\) gives \(\lambda = (A\mathbf{x})_i / x_i\).

(b) Solve the homogeneous linear system \((A - \lambda I)\mathbf{x} = \mathbf{0}\). The eigenvectors are the nonzero vectors in the null space of \(A - \lambda I\).

Answer: (a) Compute \(A\mathbf{x}\) and read off \(\lambda = (A\mathbf{x})_i/x_i\) for any nonzero component. (b) Find the null space of \((A - \lambda I)\) by row reduction.

4.25. All Eigenvalues and Two Diagonalizing Matrices (Tutorial 6, Task 5)

Find all eigenvalues and eigenvectors of: \[A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}\] and write two different diagonalizing matrices \(S\).

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Key Concept: This rank-1 matrix has eigenvalues 0 (with geometric multiplicity 2) and \(\text{tr}(A) = 3\).

1. Eigenvalues: \(\text{tr}(A) = 3\), \(\det(A) = 0\) (rank 1). Characteristic polynomial: \(\lambda^2(\lambda - 3) = 0\), so \(\lambda_1 = \lambda_2 = 0\), \(\lambda_3 = 3\).

2. Eigenvectors for \(\lambda = 0\): Null space of \(A\): \[A\mathbf{x} = \mathbf{0} \Rightarrow x_1 + x_2 + x_3 = 0\] Two free variables; two independent eigenvectors, e.g.: \(\mathbf{v}_1 = \begin{bmatrix}1\\-1\\0\end{bmatrix}\), \(\mathbf{v}_2 = \begin{bmatrix}1\\0\\-1\end{bmatrix}\).

3. Eigenvectors for \(\lambda = 3\): \((A - 3I)\mathbf{x} = \mathbf{0}\) gives \(x_1 = x_2 = x_3\), so \(\mathbf{v}_3 = \begin{bmatrix}1\\1\\1\end{bmatrix}\).

4. Two diagonalizing matrices (any choice of two independent eigenvectors for \(\lambda = 0\)): \[S_1 = \begin{bmatrix}1&1&1\\-1&0&1\\0&-1&1\end{bmatrix}, \quad S_2 = \begin{bmatrix}1&-1&1\\-1&-1&1\\0&2&1\end{bmatrix}\]

Answer: \(\lambda = 0\) (multiplicity 2), \(\lambda = 3\) (multiplicity 1). Any matrix whose columns are two independent vectors from \(N(A)\) and the vector \(\begin{bmatrix}1\\1\\1\end{bmatrix}\) is a valid \(S\).

4.26. Prove a Matrix Power Formula via Diagonalization (Tutorial 6, Task 6)

Diagonalize \(A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\) and compute \(S\Lambda^k S^{-1}\) to prove: \[A^k = \frac{1}{2}\begin{bmatrix} 3^k + 1 & 3^k - 1 \\ 3^k - 1 & 3^k + 1 \end{bmatrix}\]

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Key Concept: \(A^k = S\Lambda^k S^{-1}\) where \(\Lambda\) is diagonal with eigenvalues, and \(S\) has eigenvectors as columns.

1. Eigenvalues: \(\det(A - \lambda I) = (2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = (\lambda-1)(\lambda-3) = 0\), so \(\lambda_1 = 1\), \(\lambda_2 = 3\).

2. Eigenvectors:

   • \(\lambda_1 = 1\): \((A-I)\mathbf{v} = 0 \Rightarrow \mathbf{v}_1 = \begin{bmatrix}1\\-1\end{bmatrix}\)
   • \(\lambda_2 = 3\): \((A-3I)\mathbf{v} = 0 \Rightarrow \mathbf{v}_2 = \begin{bmatrix}1\\1\end{bmatrix}\)

3. Matrices: \[S = \begin{bmatrix}1&1\\-1&1\end{bmatrix}, \quad \Lambda = \begin{bmatrix}1&0\\0&3\end{bmatrix}, \quad S^{-1} = \frac{1}{2}\begin{bmatrix}1&-1\\1&1\end{bmatrix}\]

4. Compute \(A^k = S\Lambda^k S^{-1}\): \[A^k = \begin{bmatrix}1&1\\-1&1\end{bmatrix}\begin{bmatrix}1&0\\0&3^k\end{bmatrix}\frac{1}{2}\begin{bmatrix}1&-1\\1&1\end{bmatrix} = \frac{1}{2}\begin{bmatrix}1+3^k&-1+3^k\\-1+3^k&1+3^k\end{bmatrix}\]

Answer: \(A^k = \frac{1}{2}\begin{bmatrix} 3^k+1 & 3^k-1 \\ 3^k-1 & 3^k+1 \end{bmatrix}\) ✓

4.27. All Matrices That Diagonalize a Given Matrix (Tutorial 6, Task 7)

Describe all matrices \(S\) that diagonalize \(A = \begin{bmatrix} 4 & 0 \\ 1 & 2 \end{bmatrix}\). Then describe all matrices that diagonalize \(A^{-1}\).

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Key Concept: Any invertible matrix whose columns are eigenvectors of \(A\) (or scalar multiples thereof) diagonalizes \(A\).

1. Eigenvalues of \(A\): Diagonal entries \(\lambda_1 = 4\), \(\lambda_2 = 2\).
2. Eigenvectors:
   • \(\lambda_1 = 4\): \((A - 4I)\mathbf{v} = 0 \Rightarrow \begin{bmatrix}0&0\\1&-2\end{bmatrix}\mathbf{v}=0 \Rightarrow \mathbf{v}_1 = \begin{bmatrix}2\\1\end{bmatrix}\)
   • \(\lambda_2 = 2\): \((A - 2I)\mathbf{v} = 0 \Rightarrow \begin{bmatrix}2&0\\1&0\end{bmatrix}\mathbf{v}=0 \Rightarrow \mathbf{v}_2 = \begin{bmatrix}0\\1\end{bmatrix}\)
3. All diagonalizing matrices for \(A\): Any \(S = \begin{bmatrix}2\alpha & 0 \\ \alpha & \beta\end{bmatrix}\) with \(\alpha, \beta \neq 0\) (scalar multiples of eigenvectors in each column, keeping \(S\) invertible).
4. For \(A^{-1}\): \(A^{-1}\) has the same eigenvectors as \(A\) (with eigenvalues \(1/4\) and \(1/2\)). So the same set of matrices \(S\) diagonalizes \(A^{-1}\).

Answer: All invertible matrices with columns proportional to \(\begin{bmatrix}2\\1\end{bmatrix}\) and \(\begin{bmatrix}0\\1\end{bmatrix}\) diagonalize both \(A\) and \(A^{-1}\).

4.28. Which Matrix Cannot Be Diagonalized? (Tutorial 6, Task 8)

Which of the following matrices cannot be diagonalized? \[A_1 = \begin{bmatrix} 2 & -2 \\ 2 & -2 \end{bmatrix}, \quad A_2 = \begin{bmatrix} 2 & 0 \\ 2 & -2 \end{bmatrix}, \quad A_3 = \begin{bmatrix} 2 & 0 \\ 2 & 2 \end{bmatrix}\]

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Key Concept: A matrix is diagonalizable if and only if for each eigenvalue, the geometric multiplicity equals the algebraic multiplicity.

\(A_1\): \(\text{tr} = 0\), \(\det = 0\). Eigenvalues: \(\lambda = 0\) (double). \(\text{rank}(A_1) = 1\), so \(\text{nullity}(A_1) = 1\). Geometric multiplicity = 1 < algebraic multiplicity = 2. Not diagonalizable.

\(A_2\): \(\text{tr} = 0\), \(\det = -4\). Eigenvalues: \(\lambda^2 - 4 = 0 \Rightarrow \lambda = \pm 2\). Two distinct eigenvalues → always diagonalizable.

\(A_3\): \(\text{tr} = 4\), \(\det = 4\). Eigenvalues: \(\lambda = 2\) (double). \(\text{rank}(A_3 - 2I) = \text{rank}\begin{bmatrix}0&0\\2&0\end{bmatrix} = 1\), so \(\text{nullity} = 1\). Geometric multiplicity = 1 < 2. Not diagonalizable.

Answer: \(A_1\) and \(A_3\) cannot be diagonalized. \(A_2\) (with two distinct eigenvalues \(\pm 2\)) can be diagonalized.