W4. Basis, Coordinates, Change-of-Basis Matrix, Matrix Transpose

Author

Salman Ahmadi-Asl

Published

September 23, 2025

Quiz | Flashcards

1. Summary

1.1 Basis of a Vector Space

A basis of a vector space is a fundamental concept in linear algebra. Think of it as a set of building blocks or a coordinate system for that space. For a set of vectors to be a basis, it must satisfy two crucial conditions:

  1. Linearly Independent: None of the vectors in the set can be written as a linear combination of the others. This means each vector adds a unique direction, and there is no redundancy. Imagine three arrows in a 2D plane (like a sheet of paper); one of them is always unnecessary because you can form it from the other two. A basis for a 2D plane will have exactly two such independent arrows.
  2. Spans the Space: Every vector in the entire vector space can be created by taking a specific linear combination of the basis vectors. This ensures that the basis is “complete” and can reach any point in the space.

A common example is the standard basis for \(\mathbb{R}^2\), which consists of two vectors: \[ \mathcal{S} = \left\{ \vec{e}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \vec{e}_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right\} \] Here, \(\vec{e}_1\) is a unit vector along the x-axis, and \(\vec{e}_2\) is a unit vector along the y-axis. Any vector \(\begin{pmatrix} a \\ b \end{pmatrix}\) can be uniquely written as \(a\vec{e}_1 + b\vec{e}_2\).

1.2 Coordinates Relative to a Basis

While a vector is a geometric object (like an arrow), its representation depends on the chosen basis. The coordinate vector of a vector \(\vec{v}\) relative to a basis \(\mathcal{B} = \{\vec{b}_1, \vec{b}_2, \dots, \vec{b}_n\}\) is the unique set of scalars (weights) \(c_1, c_2, \dots, c_n\) such that: \[ \vec{v} = c_1\vec{b}_1 + c_2\vec{b}_2 + \dots + c_n\vec{b}_n \] This set of scalars is written as a column vector and denoted by \([\vec{v}]_{\mathcal{B}}\): \[ [\vec{v}]_{\mathcal{B}} = \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix} \] The key idea is that the same geometric vector will have different coordinate representations in different bases. Changing the basis is like changing the grid system on which you measure the vector’s position.

1.3 Change-of-Basis Matrix
1.3.1 Definition and Purpose

A change-of-basis matrix is a special matrix that systematically converts the coordinates of a vector from one basis to another. The matrix that changes coordinates from a basis \(\mathcal{B}\) (the “old” basis) to a basis \(\mathcal{C}\) (the “new” basis) is denoted \(P_{\mathcal{C} \leftarrow \mathcal{B}}\).

Its job is to perform the following transformation: \[ [\vec{v}]_{\mathcal{C}} = P_{\mathcal{C} \leftarrow \mathcal{B}} [\vec{v}]_{\mathcal{B}} \] To construct this matrix, its columns are the coordinate vectors of the old basis vectors (\(\mathcal{B}\)) written relative to the new basis (\(\mathcal{C}\)). \[ P_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{bmatrix} [\vec{b}_1]_{\mathcal{C}} & [\vec{b}_2]_{\mathcal{C}} & \dots & [\vec{b}_n]_{\mathcal{C}} \end{bmatrix} \]

1.3.2 Changing to the Standard Basis

The process becomes much simpler when converting from a basis \(\mathcal{B} = \{\vec{b}_1, \vec{b}_2, \dots, \vec{b}_n\}\) to the standard basis \(\mathcal{S}\). Since the basis vectors \(\vec{b}_i\) are already expressed in standard coordinates, the change-of-basis matrix \(P_{\mathcal{S} \leftarrow \mathcal{B}}\) is simply the matrix whose columns are the vectors of basis \(\mathcal{B}\). \[ P_{\mathcal{S} \leftarrow \mathcal{B}} = \begin{bmatrix} \vec{b}_1 & \vec{b}_2 & \dots & \vec{b}_n \end{bmatrix} \]

1.3.3 Inverting the Change-of-Basis Matrix

To reverse the process and convert coordinates from basis \(\mathcal{C}\) back to basis \(\mathcal{B}\), you need the inverse of the change-of-basis matrix. \[ P_{\mathcal{B} \leftarrow \mathcal{C}} = (P_{\mathcal{C} \leftarrow \mathcal{B}})^{-1} \] This gives the transformation: \[ [\vec{v}]_{\mathcal{B}} = P_{\mathcal{B} \leftarrow \mathcal{C}} [\vec{v}]_{\mathcal{C}} \]

1.4 The Transpose of a Matrix
1.4.1 Definition

The transpose of a matrix \(A\), denoted \(A^T\), is the matrix obtained by flipping the matrix over its main diagonal. This operation swaps the rows and columns. If \(A\) is an \(m \times n\) matrix, then \(A^T\) will be an \(n \times m\) matrix. The element at position \((i, j)\) in \(A\) moves to position \((j, i)\) in \(A^T\).

For example: If \(A = \begin{pmatrix} a & b \\ c & d \\ e & f \end{pmatrix}\), then \(A^T = \begin{pmatrix} a & c & e \\ b & d & f \end{pmatrix}\).

1.4.2 Properties of the Transpose

The transpose operation has several useful properties:

  • Double Transpose: Taking the transpose twice returns the original matrix: \((A^T)^T = A\).
  • Transpose of a Sum: The transpose of a sum of matrices is the sum of their transposes: \((A + B)^T = A^T + B^T\).
  • Transpose of a Scalar Multiple: A scalar can be factored out of a transpose operation: \((\lambda A)^T = \lambda A^T\).

2. Definitions

  • Basis: A set of linearly independent vectors that spans a vector space.
  • Linearly Independent: A set of vectors where no vector can be expressed as a linear combination of the others.
  • Spanning Set: A set of vectors from which every vector in the space can be created through a linear combination.
  • Coordinate Vector: The column vector of scalars (coefficients) that represents a vector as a linear combination of basis vectors. Its notation is \([\vec{v}]_{\mathcal{B}}\).
  • Change-of-Basis Matrix: A matrix, denoted \(P_{\mathcal{C} \leftarrow \mathcal{B}}\), that transforms coordinate vectors from basis \(\mathcal{B}\) to basis \(\mathcal{C}\).
  • Standard Basis: The most common basis for a vector space, consisting of vectors with one component equal to 1 and all other components equal to 0.
  • Matrix Transpose: The matrix, denoted \(A^T\), formed by interchanging the rows and columns of the original matrix \(A\).

3. Formulas

  • Vector from Coordinates: \[ \vec{v} = c_1\vec{b}_1 + c_2\vec{b}_2 + \dots + c_n\vec{b}_n \]
  • Change-of-Basis Transformation: \[ [\vec{v}]_{\mathcal{C}} = P_{\mathcal{C} \leftarrow \mathcal{B}} [\vec{v}]_{\mathcal{B}} \]
  • Change-of-Basis Matrix Composition: \[ P_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{bmatrix} [\vec{b}_1]_{\mathcal{C}} & [\vec{b}_2]_{\mathcal{C}} & \dots & [\vec{b}_n]_{\mathcal{C}} \end{bmatrix} \]
  • Inverse Change-of-Basis: \[ P_{\mathcal{B} \leftarrow \mathcal{C}} = (P_{\mathcal{C} \leftarrow \mathcal{B}})^{-1} \]
  • General Change-of-Basis via Standard Basis: \[ P_{\mathcal{B} \leftarrow \mathcal{C}} = (P_{\mathcal{S} \leftarrow \mathcal{B}})^{-1} P_{\mathcal{S} \leftarrow \mathcal{C}} \]
  • Inverse of a 2x2 Matrix: For \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), \[ A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]
  • Condition for Basis using Determinant: Vectors \(\{\vec{v_1}, \dots, \vec{v_n}\}\) form a basis if \(det([\vec{v_1} \dots \vec{v_n}]) \neq 0\).
  • Matrix Multiplication Element: \((AB)_{ij} = \sum_{k} a_{ik}b_{kj}\)

4. Examples

4.1. Transition Matrices and Coordinate Conversion (Lab 4, Task 1)

Two bases are given in the plane: \(\mathcal{B} = \{\vec{e_1}, \vec{e_2}\}\) (old basis) and \(\mathcal{B}' = \{\vec{e'_1}, \vec{e'_2}\}\) (new basis). The vector \(\vec{e_1}\) has coordinates \((-1, 3)\) and vector \(\vec{e_2}\) has coordinates \((2, -7)\) in the second basis.

  1. Compose transition matrices from the old basis to the new and vice versa.
  2. Find the coordinates of a vector in the old basis given that it has coordinates \(\alpha'_1, \alpha'_2\) in the new basis.
  3. Find the coordinates of a vector in the new basis given that it has coordinates \(\alpha_1, \alpha_2\) in the old basis.
Click to see the solution

Part (a): Transition Matrices

  1. Old to New Matrix (\(P_{\mathcal{B}' \leftarrow \mathcal{B}}\)): The columns of this matrix are the coordinate vectors of the old basis vectors with respect to the new basis.
    • Given: \([\vec{e_1}]_{\mathcal{B}'} = \begin{bmatrix} -1 \\ 3 \end{bmatrix}\) and \([\vec{e_2}]_{\mathcal{B}'} = \begin{bmatrix} 2 \\ -7 \end{bmatrix}\).
    • \(P_{\mathcal{B}' \leftarrow \mathcal{B}} = \begin{bmatrix} -1 & 2 \\ 3 & -7 \end{bmatrix}\).
  2. New to Old Matrix (\(P_{\mathcal{B} \leftarrow \mathcal{B}'}\)): This matrix is the inverse of the matrix from the previous step.
    • \(P_{\mathcal{B} \leftarrow \mathcal{B}'} = (P_{\mathcal{B}' \leftarrow \mathcal{B}})^{-1} = \begin{bmatrix} -1 & 2 \\ 3 & -7 \end{bmatrix}^{-1}\).
    • Calculate the determinant: \(\det = (-1)(-7) - (2)(3) = 7 - 6 = 1\).
    • Calculate the inverse: \(P_{\mathcal{B} \leftarrow \mathcal{B}'} = \frac{1}{1}\begin{bmatrix} -7 & -2 \\ -3 & -1 \end{bmatrix} = \begin{bmatrix} -7 & -2 \\ -3 & -1 \end{bmatrix}\).

Part (b): New Coordinates to Old Coordinates

  1. Use the appropriate transition matrix: To convert from the new basis (\(\mathcal{B}'\)) to the old basis (\(\mathcal{B}\)), we use \(P_{\mathcal{B} \leftarrow \mathcal{B}'}\).
  2. Apply the formula: \([\vec{v}]_{\mathcal{B}} = P_{\mathcal{B} \leftarrow \mathcal{B}'} [\vec{v}]_{\mathcal{B}'}\).
    • Let \([\vec{v}]_{\mathcal{B}'} = \begin{bmatrix} \alpha'_1 \\ \alpha'_2 \end{bmatrix}\).
    • \([\vec{v}]_{\mathcal{B}} = \begin{bmatrix} -7 & -2 \\ -3 & -1 \end{bmatrix} \begin{bmatrix} \alpha'_1 \\ \alpha'_2 \end{bmatrix} = \begin{bmatrix} -7\alpha'_1 - 2\alpha'_2 \\ -3\alpha'_1 - \alpha'_2 \end{bmatrix}\).

Part (c): Old Coordinates to New Coordinates

  1. Use the appropriate transition matrix: To convert from the old basis (\(\mathcal{B}\)) to the new basis (\(\mathcal{B}'\)), we use \(P_{\mathcal{B}' \leftarrow \mathcal{B}}\).
  2. Apply the formula: \([\vec{v}]_{\mathcal{B}'} = P_{\mathcal{B}' \leftarrow \mathcal{B}} [\vec{v}]_{\mathcal{B}}\).
    • Let \([\vec{v}]_{\mathcal{B}} = \begin{bmatrix} \alpha_1 \\ \alpha_2 \end{bmatrix}\).
    • \([\vec{v}]_{\mathcal{B}'} = \begin{bmatrix} -1 & 2 \\ 3 & -7 \end{bmatrix} \begin{bmatrix} \alpha_1 \\ \alpha_2 \end{bmatrix} = \begin{bmatrix} -\alpha_1 + 2\alpha_2 \\ 3\alpha_1 - 7\alpha_2 \end{bmatrix}\).

Answer:

  1. \(P_{\mathcal{B}' \leftarrow \mathcal{B}} = \begin{bmatrix} -1 & 2 \\ 3 & -7 \end{bmatrix}\), \(P_{\mathcal{B} \leftarrow \mathcal{B}'} = \begin{bmatrix} -7 & -2 \\ -3 & -1 \end{bmatrix}\).
  2. The coordinates in the old basis are \((-7\alpha'_1 - 2\alpha'_2, -3\alpha'_1 - \alpha'_2)\).
  3. The coordinates in the new basis are \((-\alpha_1 + 2\alpha_2, 3\alpha_1 - 7\alpha_2)\).
4.2. Find Coordinates in a Custom Basis (Lab 4, Task 2)

Given vectors: \(\vec{a} = \begin{bmatrix} -5 \\ -1 \end{bmatrix}\), \(\vec{b} = \begin{bmatrix} -1 \\ 3 \end{bmatrix}\), \(\vec{c} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}\), \(\vec{d} = \begin{bmatrix} 2 \\ -6 \end{bmatrix}\). If vectors \(\vec{a}\) and \(\vec{b}\) form a basis (you should check it), find the coordinates of \(\vec{c}\) and \(\vec{d}\) in this basis.

Click to see the solution
  1. Check if \(\{\vec{a}, \vec{b}\}\) is a basis: Two vectors in \(\mathbb{R}^2\) form a basis if they are linearly independent. This is true if the determinant of the matrix formed by them is non-zero.
    • Let \(M = [\vec{a} \ \vec{b}] = \begin{bmatrix} -5 & -1 \\ -1 & 3 \end{bmatrix}\).
    • \(\det(M) = (-5)(3) - (-1)(-1) = -15 - 1 = -16\).
    • Since \(\det(M) \neq 0\), the vectors are linearly independent and form a basis.
  2. Find the coordinates of \(\vec{c}\): We need to find scalars \(k_1, k_2\) such that \(\vec{c} = k_1\vec{a} + k_2\vec{b}\).
    • \(\begin{bmatrix} -1 \\ 2 \end{bmatrix} = k_1\begin{bmatrix} -5 \\ -1 \end{bmatrix} + k_2\begin{bmatrix} -1 \\ 3 \end{bmatrix}\).
    • This gives the system of equations:
      • \(-5k_1 - k_2 = -1\)
      • \(-k_1 + 3k_2 = 2\)
    • Solving this system yields \(k_1 = 1/16\) and \(k_2 = 11/16\).
  3. Find the coordinates of \(\vec{d}\): We need to find scalars \(m_1, m_2\) such that \(\vec{d} = m_1\vec{a} + m_2\vec{b}\).
    • \(\begin{bmatrix} 2 \\ -6 \end{bmatrix} = m_1\begin{bmatrix} -5 \\ -1 \end{bmatrix} + m_2\begin{bmatrix} -1 \\ 3 \end{bmatrix}\).
    • This gives the system of equations:
      • \(-5m_1 - m_2 = 2\)
      • \(-m_1 + 3m_2 = -6\)
    • Solving this system yields \(m_1 = 0\) and \(m_2 = -2\).
Answer: The coordinates of \(\vec{c}\) in the basis \(\{\vec{a}, \vec{b}\}\) are \(\begin{bmatrix} 1/16 \\ 11/16 \end{bmatrix}\). The coordinates of \(\vec{d}\) are \(\begin{bmatrix} 0 \\ -2 \end{bmatrix}\).
4.3. Find Coordinates in a 3D Basis (Lab 4, Task 3)

Given 4 vectors \(\vec{f_1}, \vec{f_2}, \vec{f_3}, \vec{x}\) and the standard basis \(\vec{e_1}, \vec{e_2}, \vec{e_3}\). Find the coordinates of \(\vec{x}\) in the basis \(\{\vec{f_1}, \vec{f_2}, \vec{f_3}\}\), where \(\vec{f_1} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\), \(\vec{f_2} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}\), \(\vec{f_3} = \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}\), \(\vec{x} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}\).

Click to see the solution
  1. Set up the vector equation: We need to find scalars \(c_1, c_2, c_3\) such that \(\vec{x} = c_1\vec{f_1} + c_2\vec{f_2} + c_3\vec{f_3}\).
    • \(\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = c_1\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + c_2\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} + c_3\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}\).
  2. Write the system of linear equations:
      1. \(c_1 + c_2 + c_3 = 1\)
      1. \(c_1 + 2c_2 + 2c_3 = 1\)
      1. \(c_1 + c_2 + 2c_3 = 0\)
  3. Solve the system:
    • Subtract equation (1) from equation (3): \((c_1 + c_2 + 2c_3) - (c_1 + c_2 + c_3) = 0 - 1 \implies c_3 = -1\).
    • Subtract equation (1) from equation (2): \((c_1 + 2c_2 + 2c_3) - (c_1 + c_2 + c_3) = 1 - 1 \implies c_2 + c_3 = 0\).
    • Substitute \(c_3 = -1\) into the new equation: \(c_2 + (-1) = 0 \implies c_2 = 1\).
    • Substitute \(c_2 = 1\) and \(c_3 = -1\) into equation (1): \(c_1 + 1 + (-1) = 1 \implies c_1 = 1\).
  4. Write the coordinate vector: The coordinates of \(\vec{x}\) are the scalars \(c_1, c_2, c_3\).
    • \([\vec{x}]_{\mathcal{F}} = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}\).
Answer: The coordinates of \(\vec{x}\) in the basis \(\{\vec{f_1}, \vec{f_2}, \vec{f_3}\}\) are \(\begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}\).
4.4. Multiplying 2x2 Matrices (Lecture 4, Example 1)

Let’s compute \(C = AB\) where: \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\), \(B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}\).

Click to see the solution
  1. Recall the rule for matrix multiplication: The element in the \(i\)-th row and \(j\)-th column of the product matrix C is the dot product of the \(i\)-th row of A and the \(j\)-th column of B.
  2. Calculate the element \(c_{11}\): First row of A \(\cdot\) First column of B.
    • \(c_{11} = (1)(5) + (2)(7) = 5 + 14 = 19\).
  3. Calculate the element \(c_{12}\): First row of A \(\cdot\) Second column of B.
    • \(c_{12} = (1)(6) + (2)(8) = 6 + 16 = 22\).
  4. Calculate the element \(c_{21}\): Second row of A \(\cdot\) First column of B.
    • \(c_{21} = (3)(5) + (4)(7) = 15 + 28 = 43\).
  5. Calculate the element \(c_{22}\): Second row of A \(\cdot\) Second column of B.
    • \(c_{22} = (3)(6) + (4)(8) = 18 + 32 = 50\).
  6. Assemble the final matrix C:
    • \(C = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix}\).
Answer: \(C = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix}\).
4.5. Multiplying 3x3 Matrices (Lecture 4, Example 2)

Let’s compute just one element, \(c_{23}\), of \(C = AB\) where: \(A = \begin{bmatrix} 1 & 0 & 4 \\ -2 & 3 & 1 \\ 5 & -1 & 2 \end{bmatrix}\), \(B = \begin{bmatrix} 2 & 1 & 3 \\ 0 & 4 & 1 \\ -1 & 2 & 5 \end{bmatrix}\).

Click to see the solution
  1. Identify the required row and column: The element \(c_{23}\) is the result of the dot product of the second row of matrix A and the third column of matrix B.
  2. Extract the row and column:
    • Row 2 of A: \([-2, 3, 1]\)
    • Column 3 of B: \([3, 1, 5]\)
  3. Calculate the dot product:
    • \(c_{23} = (-2)(3) + (3)(1) + (1)(5)\)
    • \(= -6 + 3 + 5 = 2\).
Answer: The element \(c_{23}\) of the product matrix C is \(2\).
4.6. Coordinates in a Non-Standard Basis (Lecture 4, Example 3)

Let’s consider \(\mathbb{R}^2\) with:

  • Standard Basis: \(S = \left\{\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}\right\}\)
  • New Basis: \(\mathcal{B} = \left\{\mathbf{b}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \mathbf{b}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\right\}\)

What are the coordinates of \(\mathbf{v} = \begin{bmatrix} 4 \\ 2 \end{bmatrix}_S\) in the new basis \(\mathcal{B}\)?

Click to see the solution
  1. Set up the equation: We are looking for scalars \(c_1\) and \(c_2\) such that \(\mathbf{v} = c_1\mathbf{b}_1 + c_2\mathbf{b}_2\). This gives the vector equation:
    • \(\begin{bmatrix} 4 \\ 2 \end{bmatrix} = c_1\begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2\begin{bmatrix} 1 \\ -1 \end{bmatrix}\)
  2. Write the system of linear equations:
    • \(4 = 1c_1 + 1c_2\)
    • \(2 = 1c_1 - 1c_2\)
  3. Solve the system:
    • Add the two equations together: \((4+2) = (c_1+c_1) + (c_2-c_2) \implies 6 = 2c_1 \implies c_1 = 3\).
    • Substitute \(c_1=3\) into the first equation: \(4 = 3 + c_2 \implies c_2 = 1\).
  4. Write the coordinate vector: The coordinates of \(\mathbf{v}\) in the basis \(\mathcal{B}\) are the scalars we found. This is written as \([\mathbf{v}]_{\mathcal{B}}\).
    • \([\mathbf{v}]_{\mathcal{B}} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}\).
Answer: The coordinates of \(\mathbf{v}\) in the new basis \(\mathcal{B}\) are \(\begin{bmatrix} 3 \\ 1 \end{bmatrix}\).
4.7. Find a Change-of-Basis Matrix (Lecture 4, Example 4)

Given the bases from the previous example:

  • \(\mathcal{B} = \left\{\begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ -1 \end{bmatrix}\right\}\)
  • \(S = \left\{\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}\right\}\) (the standard basis)

Find \(P_{S \leftarrow \mathcal{B}}\), the matrix that changes from \(\mathcal{B}\)-coordinates to standard \(S\)-coordinates.

Click to see the solution
  1. Recall the definition: The change-of-basis matrix from a basis \(\mathcal{B}\) to the standard basis \(S\) is formed by simply using the vectors of the basis \(\mathcal{B}\) as its columns.
  2. Identify the basis vectors of \(\mathcal{B}\):
    • \(\mathbf{b}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\)
    • \(\mathbf{b}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\)
  3. Construct the matrix: Place the basis vectors as columns in the matrix.
    • \(P_{S \leftarrow \mathcal{B}} = [\mathbf{b}_1 \ | \ \mathbf{b}_2] = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\).
  4. Verification (Using previous example): The relationship is \([\mathbf{v}]_S = P_{S \leftarrow \mathcal{B}} [\mathbf{v}]_{\mathcal{B}}\). From the previous example, we found \([\mathbf{v}]_{\mathcal{B}} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}\).
    • \(\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 3 \\ 1 \end{bmatrix} = \begin{bmatrix} 1(3)+1(1) \\ 1(3)-1(1) \end{bmatrix} = \begin{bmatrix} 4 \\ 2 \end{bmatrix}\), which is the original vector \([\mathbf{v}]_S\). The matrix is correct.
Answer: The change-of-basis matrix is \(P_{S \leftarrow \mathcal{B}} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\).
4.8. Convert from a Custom Basis to Standard Coordinates (Tutorial 4, Task 1)

Consider the basis \(\mathcal{B} = \left\{\mathbf{b}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix}, \mathbf{b}_2 = \begin{bmatrix} 3 \\ 1 \end{bmatrix}\right\}\) for \(\mathbb{R}^2\). The vector \(\mathbf{v}\) has coordinates \([\mathbf{v}]_{\mathcal{B}} = \begin{bmatrix} 2 \\ -1 \end{bmatrix}\). What is \(\mathbf{v}\) in the standard coordinates \(([\mathbf{v}]_S)\)?

Click to see the solution
  1. Recall the definition of coordinates: The coordinate vector \([\mathbf{v}]_{\mathcal{B}} = \begin{bmatrix} 2 \\ -1 \end{bmatrix}\) means that \(\mathbf{v}\) is a linear combination of the basis vectors of \(\mathcal{B}\) with these scalars.
    • \(\mathbf{v} = 2\mathbf{b}_1 + (-1)\mathbf{b}_2\).
  2. Substitute the basis vectors:
    • \(\mathbf{v} = 2\begin{bmatrix} 1 \\ 2 \end{bmatrix} - 1\begin{bmatrix} 3 \\ 1 \end{bmatrix}\).
  3. Perform the vector arithmetic:
    • \(\mathbf{v} = \begin{bmatrix} 2 \cdot 1 \\ 2 \cdot 2 \end{bmatrix} - \begin{bmatrix} 1 \cdot 3 \\ 1 \cdot 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \end{bmatrix} - \begin{bmatrix} 3 \\ 1 \end{bmatrix} = \begin{bmatrix} 2-3 \\ 4-1 \end{bmatrix} = \begin{bmatrix} -1 \\ 3 \end{bmatrix}\).
Answer: The vector in standard coordinates is \(\mathbf{v} = \begin{bmatrix} -1 \\ 3 \end{bmatrix}\).
4.9. Find Coordinates Relative to a Basis (Tutorial 4, Task 2)

Consider the basis \(\mathcal{C} = \left\{\mathbf{c}_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \mathbf{c}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\right\}\) for \(\mathbb{R}^2\). Find the coordinates of the vector \(\mathbf{v} = \begin{bmatrix} 5 \\ 0 \end{bmatrix}\) relative to the basis \(\mathcal{C}\), i.e., find \([\mathbf{v}]_{\mathcal{C}}\).

Click to see the solution
  1. Set up the vector equation: We are looking for scalars \(k_1\) and \(k_2\) such that \(\mathbf{v} = k_1\mathbf{c}_1 + k_2\mathbf{c}_2\). The coordinate vector will then be \([\mathbf{v}]_{\mathcal{C}} = \begin{bmatrix} k_1 \\ k_2 \end{bmatrix}\).
    • \(\begin{bmatrix} 5 \\ 0 \end{bmatrix} = k_1\begin{bmatrix} 2 \\ 1 \end{bmatrix} + k_2\begin{bmatrix} 1 \\ -1 \end{bmatrix}\).
  2. Write the corresponding system of linear equations:
    • \(5 = 2k_1 + 1k_2\)
    • \(0 = 1k_1 - 1k_2\)
  3. Solve the system:
    • From the second equation, we get \(k_1 = k_2\).
    • Substitute this into the first equation: \(5 = 2k_1 + k_1 \implies 5 = 3k_1 \implies k_1 = 5/3\).
    • Since \(k_1 = k_2\), we have \(k_2 = 5/3\).
  4. Write the coordinate vector:
    • \([\mathbf{v}]_{\mathcal{C}} = \begin{bmatrix} 5/3 \\ 5/3 \end{bmatrix}\).
Answer: The coordinates of \(\mathbf{v}\) relative to the basis \(\mathcal{C}\) are \(\begin{bmatrix} 5/3 \\ 5/3 \end{bmatrix}\).
4.10. Find a Change-of-Basis Matrix (Tutorial 4, Task 3a)

Let two bases for \(\mathbb{R}^2\) be:

  • \(\mathcal{B} = \left\{\mathbf{b}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \mathbf{b}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\right\}\)
  • \(\mathcal{C} = \left\{\mathbf{c}_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \mathbf{c}_2 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\right\}\)

Find the change-of-basis matrix \(P_{\mathcal{B} \leftarrow \mathcal{C}}\).

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  1. Recall the formula for the change-of-basis matrix: The matrix \(P_{\mathcal{B} \leftarrow \mathcal{C}}\) can be found using the formula \(P_{\mathcal{B} \leftarrow \mathcal{C}} = (P_{S \leftarrow \mathcal{B}})^{-1} P_{S \leftarrow \mathcal{C}}\), where \(S\) is the standard basis.
  2. Construct the matrices for changing to the standard basis: These matrices have the basis vectors as their columns.
    • \(P_{S \leftarrow \mathcal{B}} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\)
    • \(P_{S \leftarrow \mathcal{C}} = \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix}\)
  3. Find the inverse of \(P_{S \leftarrow \mathcal{B}}\): For a \(2 \times 2\) matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\), the inverse is \(\frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\).
    • \(\det(P_{S \leftarrow \mathcal{B}}) = (1)(-1) - (1)(1) = -2\).
    • \((P_{S \leftarrow \mathcal{B}})^{-1} = \frac{1}{-2}\begin{bmatrix} -1 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{bmatrix}\).
  4. Multiply the matrices to find \(P_{\mathcal{B} \leftarrow \mathcal{C}}\):
    • \(P_{\mathcal{B} \leftarrow \mathcal{C}} = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix}\)
    • \(= \begin{bmatrix} (1/2)(2)+(1/2)(1) & (1/2)(1)+(1/2)(0) \\ (1/2)(2)+(-1/2)(1) & (1/2)(1)+(-1/2)(0) \end{bmatrix}\)
    • \(= \begin{bmatrix} 1+1/2 & 1/2 \\ 1-1/2 & 1/2 \end{bmatrix} = \begin{bmatrix} 3/2 & 1/2 \\ 1/2 & 1/2 \end{bmatrix}\).
Answer: The change-of-basis matrix is \(P_{\mathcal{B} \leftarrow \mathcal{C}} = \begin{bmatrix} 3/2 & 1/2 \\ 1/2 & 1/2 \end{bmatrix}\).
4.11. Use a Change-of-Basis Matrix (Tutorial 4, Task 3b)

Let \([\mathbf{v}]_{\mathcal{C}} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}\). Find \([\mathbf{v}]_{\mathcal{B}}\) using the matrix from the previous part.

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  1. Recall the conversion formula: The coordinates in basis \(\mathcal{B}\) can be found by multiplying the change-of-basis matrix by the coordinates in basis \(\mathcal{C}\).
    • \([\mathbf{v}]_{\mathcal{B}} = P_{\mathcal{B} \leftarrow \mathcal{C}} [\mathbf{v}]_{\mathcal{C}}\).
  2. Use the matrix from part (a):
    • \(P_{\mathcal{B} \leftarrow \mathcal{C}} = \begin{bmatrix} 3/2 & 1/2 \\ 1/2 & 1/2 \end{bmatrix}\).
  3. Perform the matrix-vector multiplication:
    • \([\mathbf{v}]_{\mathcal{B}} = \begin{bmatrix} 3/2 & 1/2 \\ 1/2 & 1/2 \end{bmatrix} \begin{bmatrix} 3 \\ 1 \end{bmatrix}\)
    • \(= \begin{bmatrix} (3/2)(3) + (1/2)(1) \\ (1/2)(3) + (1/2)(1) \end{bmatrix} = \begin{bmatrix} 9/2 + 1/2 \\ 3/2 + 1/2 \end{bmatrix} = \begin{bmatrix} 10/2 \\ 4/2 \end{bmatrix} = \begin{bmatrix} 5 \\ 2 \end{bmatrix}\).
Answer: \([\mathbf{v}]_{\mathcal{B}} = \begin{bmatrix} 5 \\ 2 \end{bmatrix}\).
4.12. Find Coordinates for a Polynomial Basis (Tutorial 4, Task 4)

The set \(\mathcal{B} = \{1 - t^2, t - t^2, 2 - t + t^2\}\) is a basis for \(\mathbb{P}_2\) (the space of polynomials of degree \(\le 2\)). Find the coordinate vector of the polynomial \(\mathbf{p}(t) = 3 + t - 6t^2\) relative to \(\mathcal{B}\).

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  1. Set up the equation: We are looking for scalars \(c_1, c_2, c_3\) such that:
    • \(\mathbf{p}(t) = c_1(1 - t^2) + c_2(t - t^2) + c_3(2 - t + t^2)\).
    • \(3 + t - 6t^2 = c_1 - c_1t^2 + c_2t - c_2t^2 + 2c_3 - c_3t + c_3t^2\).
  2. Group terms by powers of \(t\):
    • \(3 + t - 6t^2 = (c_1 + 2c_3) + (c_2 - c_3)t + (-c_1 - c_2 + c_3)t^2\).
  3. Equate coefficients to form a system of linear equations:
    • Constant term: \(c_1 + 2c_3 = 3\)
    • Coefficient of \(t\): \(c_2 - c_3 = 1\)
    • Coefficient of \(t^2\): \(-c_1 - c_2 + c_3 = -6\)
  4. Solve the system:
    • From the second equation, \(c_2 = 1 + c_3\).
    • Substitute this into the third equation: \(-c_1 - (1 + c_3) + c_3 = -6 \implies -c_1 - 1 = -6 \implies c_1 = 5\).
    • Substitute \(c_1=5\) into the first equation: \(5 + 2c_3 = 3 \implies 2c_3 = -2 \implies c_3 = -1\).
    • Finally, find \(c_2\): \(c_2 = 1 + c_3 = 1 + (-1) = 0\).
  5. Write the coordinate vector:
    • \([\mathbf{p}]_{\mathcal{B}} = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 5 \\ 0 \\ -1 \end{bmatrix}\).
Answer: The coordinate vector is \(\begin{bmatrix} 5 \\ 0 \\ -1 \end{bmatrix}\).
4.13. Angle Between Vectors (Test 1, Task 1)

What is the angle between \(\mathbf{a} = (3, 2, 4)\) and \(\mathbf{b} = (2, 3, -3)\)?

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Key Concept: Use the dot product to find the angle between vectors.

  1. Calculate the dot product:

    \[\mathbf{a} \cdot \mathbf{b} = (3)(2) + (2)(3) + (4)(-3) = 6 + 6 - 12 = 0\]

  2. Interpret the result:

    Since \(\mathbf{a} \cdot \mathbf{b} = 0\), the vectors are perpendicular.

  3. Find the angle:

    When two vectors are perpendicular, the angle between them is \(90°\).

Answer: \(90°\)

4.14. Vector Space Identification (Test 1, Task 2)

Which of the following cases is not a vector space?

A. \(\mathbb{R}^2\) B. \(\mathbb{R}^3\) C. The set of all polynomials of at most degree \(n\) D. The set \(S = \left\{ \begin{bmatrix} x \\ y \end{bmatrix} \in \mathbb{R}^2 \mid x \ge 0, y \ge 0 \right\}\)

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Key Concept: A vector space must be closed under addition and scalar multiplication.

  1. Check option A:

    \(\mathbb{R}^2\) is a standard vector space. ✓

  2. Check option B:

    \(\mathbb{R}^3\) is a standard vector space. ✓

  3. Check option C:

    The set of all polynomials of at most degree \(n\) is a vector space. ✓

  4. Check option D:

    Consider the set \(S = \left\{ \begin{bmatrix} x \\ y \end{bmatrix} \in \mathbb{R}^2 \mid x \ge 0, y \ge 0 \right\}\).

    Take a vector \(\begin{bmatrix} 1 \\ 1 \end{bmatrix} \in S\) and multiply by scalar \(-1\): \[-1 \cdot \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \end{bmatrix}\]

    This result has negative components, so it’s NOT in \(S\). Therefore, \(S\) is not closed under scalar multiplication.

Answer: D. The set \(S = \left\{ \begin{bmatrix} x \\ y \end{bmatrix} \in \mathbb{R}^2 \mid x \ge 0, y \ge 0 \right\}\)

4.15. Equivalent Vectors (Test 1, Task 3)

Determine the unknown coordinate \((x, y)\) if \(\vec{AB} = \vec{CD}\), where \(A(3, 5)\), \(B(4, 6)\), \(C(-2, 5)\), \(D(x, y)\).

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Key Concept: Two vectors are equivalent if they have the same components.

  1. Find vector \(\vec{AB}\):

    \[\vec{AB} = B - A = (4 - 3, 6 - 5) = (1, 1)\]

  2. Find vector \(\vec{CD}\) in terms of \(x\) and \(y\):

    \[\vec{CD} = D - C = (x - (-2), y - 5) = (x + 2, y - 5)\]

  3. Set the vectors equal:

    Since \(\vec{AB} = \vec{CD}\): \[(1, 1) = (x + 2, y - 5)\]

  4. Solve for \(x\) and \(y\):

    \[x + 2 = 1 \quad \Rightarrow \quad x = -1\] \[y - 5 = 1 \quad \Rightarrow \quad y = 6\]

Answer: \((-1, 6)\)

4.16. Vector Projection (Test 1, Task 4)

What is the vector projection of \(\mathbf{v} = (5, 3)\) onto \(\mathbf{w} = (4, 0)\)?

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Key Concept: The vector projection formula is \(\text{proj}_{\mathbf{w}}\mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}}\mathbf{w}\).

  1. Calculate dot products:

    \[\mathbf{v} \cdot \mathbf{w} = (5)(4) + (3)(0) = 20\] \[\mathbf{w} \cdot \mathbf{w} = (4)(4) + (0)(0) = 16\]

  2. Apply the projection formula:

    \[\text{proj}_{\mathbf{w}}\mathbf{v} = \frac{20}{16}(4, 0) = \frac{5}{4}(4, 0) = (5, 0)\]

Answer: \((5, 0)\)

4.17. Change-of-Basis Matrix (Test 1, Task 5)

Consider the following bases:

  • \(B = \left\{ \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ -1 \end{bmatrix} \right\}\)
  • \(S = \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right\}\) (the standard basis)

What is \(P_{S \leftarrow B}\), the matrix that changes from \(B\)-coordinates to standard \(S\)-coordinates?

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Key Concept: The change-of-basis matrix from a basis to the standard basis has the basis vectors as its columns.

  1. Identify the basis vectors:

    \[\mathbf{b}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \quad \mathbf{b}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\]

  2. Construct the matrix:

    The columns of \(P_{S \leftarrow B}\) are \([\mathbf{b}_1]_S\) and \([\mathbf{b}_2]_S\). Since these are already in standard coordinates:

    \[P_{S \leftarrow B} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\]

Answer: \(\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\)

4.18. Triangle Area (Test 1, Task 6)

What is the area of a triangle with vertices at \(A(1, 2, 0)\), \(B(3, 0, -3)\), and \(C(5, 2, 6)\)?

Click to see the solution

Key Concept: The area of a triangle is \(\frac{1}{2}||\vec{AB} \times \vec{AC}||\).

  1. Find the vectors:

    \[\vec{AB} = B - A = (3 - 1, 0 - 2, -3 - 0) = (2, -2, -3)\] \[\vec{AC} = C - A = (5 - 1, 2 - 2, 6 - 0) = (4, 0, 6)\]

  2. Calculate the cross product:

    \[\vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -2 & -3 \\ 4 & 0 & 6 \end{vmatrix}\]

    \[= \mathbf{i}((-2)(6) - (-3)(0)) - \mathbf{j}((2)(6) - (-3)(4)) + \mathbf{k}((2)(0) - (-2)(4))\]

    \[= \mathbf{i}(-12 - 0) - \mathbf{j}(12 + 12) + \mathbf{k}(0 + 8)\]

    \[= (-12, -24, 8)\]

  3. Find the magnitude:

    \[||\vec{AB} \times \vec{AC}|| = \sqrt{(-12)^2 + (-24)^2 + 8^2} = \sqrt{144 + 576 + 64} = \sqrt{784} = 28\]

  4. Calculate the area:

    \[\text{Area} = \frac{1}{2} \times 28 = 14\]

Answer: \(14\)

4.19. Linearly Independent Vectors (Test 1, Task 7)

Which of the following pairs of vectors are linearly independent?

A. \(\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, \begin{bmatrix} 3 \\ 6 \\ 9 \end{bmatrix}\)

B. \(\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 5 \end{bmatrix}\)

C. \(\begin{bmatrix} 2 \\ 1 \\ 5 \end{bmatrix}, \begin{bmatrix} 1 \\ 1/2 \\ 5/2 \end{bmatrix}\)

D. \(\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\)

Click to see the solution

Key Concept: Two vectors are linearly independent if neither is a scalar multiple of the other.

  1. Check option A:

    \[\begin{bmatrix} 3 \\ 6 \\ 9 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\]

    These vectors are linearly dependent. ✗

  2. Check option B:

    \[\begin{bmatrix} 0 \\ 0 \\ 5 \end{bmatrix} = 5 \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\]

    These vectors are linearly dependent. ✗

  3. Check option C:

    \[\begin{bmatrix} 1 \\ 1/2 \\ 5/2 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 2 \\ 1 \\ 5 \end{bmatrix}\]

    These vectors are linearly dependent. ✗

  4. Check option D:

    These vectors are not scalar multiples of each other, so they are linearly independent. ✓

Answer: D. \(\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\)

4.20. Vector Projection and Reflection (Test 1, Task 8)

Suppose \(\mathbf{a} = (2, 2, -1)\) and \(\mathbf{b} = (0, 4, 3)\). Compute the projection of \(\mathbf{a}\) on \(\mathbf{b}\) which is the vector \(\mathbf{a}'\). Also find \(\mathbf{a}''\) (the reflection of \(\mathbf{a}\) over \(\mathbf{b}\)).

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Key Concept: The projection is \(\mathbf{a}' = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}}\mathbf{b}\) and the reflection is \(\mathbf{a}'' = 2\mathbf{a}' - \mathbf{a}\).

  1. Calculate the projection \(\mathbf{a}'\):

    \[\mathbf{a} \cdot \mathbf{b} = (2)(0) + (2)(4) + (-1)(3) = 0 + 8 - 3 = 5\] \[\mathbf{b} \cdot \mathbf{b} = (0)^2 + (4)^2 + (3)^2 = 0 + 16 + 9 = 25\]

    \[\mathbf{a}' = \frac{5}{25}\mathbf{b} = \frac{1}{5}(0, 4, 3) = \left(0, \frac{4}{5}, \frac{3}{5}\right)\]

  2. Calculate the reflection \(\mathbf{a}''\):

    \[\mathbf{a}'' = 2\mathbf{a}' - \mathbf{a} = 2\left(0, \frac{4}{5}, \frac{3}{5}\right) - (2, 2, -1)\]

    \[= \left(0, \frac{8}{5}, \frac{6}{5}\right) - (2, 2, -1)\]

    \[= \left(0 - 2, \frac{8}{5} - 2, \frac{6}{5} - (-1)\right)\]

    \[= \left(-2, \frac{8}{5} - \frac{10}{5}, \frac{6}{5} + \frac{5}{5}\right)\]

    \[= \left(-2, -\frac{2}{5}, \frac{11}{5}\right)\]

Answer: \(\mathbf{a}' = \left(0, \frac{4}{5}, \frac{3}{5}\right)\), \(\mathbf{a}'' = \left(-2, -\frac{2}{5}, \frac{11}{5}\right)\)

4.21. Matrix Polynomial Verification (Test 1, Task 9)

If \(A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\), show that \(A^2 - 4A - 5I = 0\).

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Key Concept: Compute \(A^2\), then verify the equation.

  1. Calculate \(A^2\):

    \[A^2 = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\]

    \[= \begin{bmatrix} 1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1 \end{bmatrix}\]

    \[= \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}\]

  2. Calculate \(4A\):

    \[4A = \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix}\]

  3. Calculate \(5I\):

    \[5I = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}\]

  4. Verify \(A^2 - 4A - 5I = 0\):

    \[A^2 - 4A - 5I = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}\]

    \[= \begin{bmatrix} 9-4-5 & 8-8-0 & 8-8-0 \\ 8-8-0 & 9-4-5 & 8-8-0 \\ 8-8-0 & 8-8-0 & 9-4-5 \end{bmatrix}\]

    \[= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = 0\]

Answer: Verified: \(A^2 - 4A - 5I = 0\)

4.22. Vector Projection onto Vector (Test 1, Task 10)

Given \(\mathbf{a} = (1, -2, -1)\), \(\mathbf{b} = (1, 2, -1)\) and \(\mathbf{c} = (2, -1, -1)\), compute the projection of \(\mathbf{a} + \mathbf{b}\) onto \(\mathbf{c}\).

Click to see the solution

Key Concept: First find \(\mathbf{a} + \mathbf{b}\), then project onto \(\mathbf{c}\).

  1. Calculate \(\mathbf{a} + \mathbf{b}\):

    \[\mathbf{a} + \mathbf{b} = (1, -2, -1) + (1, 2, -1) = (2, 0, -2)\]

  2. Calculate the projection:

    \[(\mathbf{a} + \mathbf{b}) \cdot \mathbf{c} = (2)(2) + (0)(-1) + (-2)(-1) = 4 + 0 + 2 = 6\] \[\mathbf{c} \cdot \mathbf{c} = (2)^2 + (-1)^2 + (-1)^2 = 4 + 1 + 1 = 6\]

    \[\text{proj}_{\mathbf{c}}(\mathbf{a} + \mathbf{b}) = \frac{6}{6}\mathbf{c} = 1 \cdot (2, -1, -1) = (2, -1, -1)\]

Answer: \((2, -1, -1)\)

4.23. Dot Product Identity (Test 1, Task 11)

Assume \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{c}\) are three nonzero vectors. If \(\mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c}\), is it true to say \(\mathbf{b} = \mathbf{c}\)? Why?

Click to see the solution

Key Concept: The dot product provides only one piece of information about the relationship between vectors.

  1. Analyze the condition:

    The equation \(\mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c}\) can be rewritten as: \[\mathbf{a} \cdot \mathbf{b} - \mathbf{a} \cdot \mathbf{c} = 0\] \[\mathbf{a} \cdot (\mathbf{b} - \mathbf{c}) = 0\]

    This means \(\mathbf{a}\) is perpendicular to \((\mathbf{b} - \mathbf{c})\).

  2. Provide a counterexample:

    Let \(\mathbf{a} = (1, 0, 0)\), \(\mathbf{b} = (0, 1, 0)\), \(\mathbf{c} = (0, 0, 1)\).

    Then: \[\mathbf{a} \cdot \mathbf{b} = (1)(0) + (0)(1) + (0)(0) = 0\] \[\mathbf{a} \cdot \mathbf{c} = (1)(0) + (0)(0) + (0)(1) = 0\]

    So \(\mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c}\), but clearly \(\mathbf{b} \neq \mathbf{c}\).

Answer: No, it is not necessarily true. Counterexample: \(\mathbf{a} = (1, 0, 0)\), \(\mathbf{b} = (0, 1, 0)\), \(\mathbf{c} = (0, 0, 1)\).

4.24. Parallelepiped Volume (Test 1, Task 12)

Compute the volume of the parallelepiped formed by the vectors: \[\mathbf{a} = (1, 1, 0), \quad \mathbf{b} = (0, 1, 1), \quad \mathbf{c} = (1, 0, 1)\]

Click to see the solution

Key Concept: The volume is the absolute value of the scalar triple product.

  1. Calculate the scalar triple product:

    \[V = \left| \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \right| = \left| \det \begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix} \right|\]

  2. Expand the determinant:

    \[= \left| 1 \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} - 1 \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} \right|\]

    \[= |1(1 \cdot 1 - 1 \cdot 0) - 1(0 \cdot 1 - 1 \cdot 1) + 0|\]

    \[= |1(1) - 1(-1)| = |1 + 1| = |2| = 2\]

Answer: \(2\)

4.25. Matrix Determinant and Inverse (Test 1, Task 13)

Compute the determinant of the following matrix and also its inverse: \[A = \begin{bmatrix} 0 & 1 & 0 \\ 2 & 0 & 0 \\ 0 & 0 & 3 \end{bmatrix}\]

Click to see the solution

Key Concept: Use cofactor expansion for the determinant, then use the formula \(A^{-1} = \frac{1}{\det(A)}\text{adj}(A)\).

  1. Calculate the determinant:

    Expand along the first row: \[\det(A) = 0 \begin{vmatrix} 0 & 0 \\ 0 & 3 \end{vmatrix} - 1 \begin{vmatrix} 2 & 0 \\ 0 & 3 \end{vmatrix} + 0 \begin{vmatrix} 2 & 0 \\ 0 & 0 \end{vmatrix}\]

    \[= 0 - 1(2 \cdot 3 - 0 \cdot 0) + 0 = -6\]

  2. Find the inverse:

    Since this is a simple permutation-like matrix, we can verify by inspection or calculation:

    \[A^{-1} = \begin{bmatrix} 0 & 1/2 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1/3 \end{bmatrix}\]

    We can verify: \(AA^{-1} = I\).

Answer: \(\det(A) = -6\), \(A^{-1} = \begin{bmatrix} 0 & 1/2 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1/3 \end{bmatrix}\)

4.26. Non-Commutative Matrix Multiplication (Test 1, Task 14)

Give two matrices for which \(AB \neq BA\).

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Key Concept: Matrix multiplication is generally not commutative.

  1. Choose simple matrices:

    Let: \[A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\]

  2. Calculate \(AB\):

    \[AB = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\]

  3. Calculate \(BA\):

    \[BA = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\]

  4. Verify \(AB \neq BA\):

    Since \(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \neq \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\), we have \(AB \neq BA\). ✓

Answer: \(A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\), \(B = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\) (or many other examples)