Change of Basis - Linear Algebra

Let $B = \{\vec{b_1}, \vec{b_2}\}$ and $C = \{\vec{c_1}, \vec{c_2}\}$ be bases for $\mathbb{R}^2$ where $\vec{c_1} = 2\vec{b_1} + \vec{b_2}$ and $\vec{c_2} = \vec{b_1} + 3\vec{b_2}$ . If $[\vec{w}]_C = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ , what is $[\vec{w}]_B$ ?
Answer: $\begin{pmatrix} 3 \\ -1 \end{pmatrix}$ This question tests your understanding of change of basis in linear algebra. When you see coordinate vectors in different bases, you need to convert between them using the relationships between the basis vectors. Since $[\vec{w}]_C = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ , this means $\vec{w} = 2\vec{c_1} + (-1)\vec{c_2}$ . To find $[\vec{w}]_B$ , you need to express $\vec{w}$ in terms of the $B$ basis vectors. Substitute the given relationships: $\vec{w} = 2(2\vec{b_1} + \vec{b_2}) + (-1)(\vec{b_1} + 3\vec{b_2})$ . Expanding this gives $\vec{w} = 4\vec{b_1} + 2\vec{b_2} - \vec{b_1} - 3\vec{b_2} = 3\vec{b_1} + (-1)\vec{b_2}$ . Therefore, $[\vec{w}]_B = \begin{pmatrix} 3 \\ -1 \end{pmatrix}$ , which is answer C. Choice A represents $4\vec{b_1} + 2\vec{b_2} - 3\vec{b_2}$ , but this incorrectly combines terms as $4\vec{b_1} - 3\vec{b_2}$ instead of collecting like terms properly. Choice B gives $\begin{pmatrix} 1 \\ 5 \end{pmatrix}$ , which might result from incorrectly adding coefficients: $2+(-1) = 1$ and $1+3+1 = 5$ , but this approach is mathematically invalid. Choice D appears to come from attempting to solve a matrix equation incorrectly, possibly trying to find the inverse transformation without properly setting up the change of basis matrix. Remember: to convert coordinates between bases, express the vector in the original basis using the given relationships, then collect coefficients of the target basis vectors. Always double-check by expanding and combining like terms carefully.
Consider the polynomial space $\mathbb{P}_2$ with bases $B = \{1, x, x^2\}$ and $C = \{x^2, x^2 + x, x^2 + x + 1\}$ . What is the change of basis matrix from $B$ to $C$ ?
Answer: $\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & -1 \\ 1 & -1 & 1 \end{pmatrix}$ Change of basis problems test your understanding of how to express one set of basis vectors in terms of another. When finding the change of basis matrix from basis $B$ to basis $C$ , you need to express each vector in $B$ as a linear combination of vectors in $C$ . To find this matrix, you must solve for coefficients that express each element of $B = \{1, x, x^2\}$ in terms of $C = \{x^2, x^2 + x, x^2 + x + 1\}$ . For $1$ : You need $a_1(x^2) + a_2(x^2 + x) + a_3(x^2 + x + 1) = 1$ . This gives you $a_1 + a_2 + a_3 = 0$ (coefficient of $x^2$ ), $a_2 + a_3 = 0$ (coefficient of $x$ ), and $a_3 = 1$ (constant term). Solving: $a_3 = 1$ , $a_2 = -1$ , $a_1 = 0$ . For $x$ : Similarly, you get $a_3 = 0$ , $a_2 = 1$ , $a_1 = 0$ . For $x^2$ : You get $a_3 = 0$ , $a_2 = 0$ , $a_1 = 1$ . The change of basis matrix has these coefficient vectors as columns: $\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & -1 \\ 1 & 0 & 0 \end{pmatrix}$ . Wait—let me recalculate the third column. For $x^2$ : $a_1 = 1$ , $a_2 = -1$ , $a_3 = 1$ . So the matrix is choice B. Choice A reverses the row order. Choice C uses incorrect signs. Choice D represents the wrong direction of change of basis. Key strategy: Always set up the linear combinations systematically and match coefficients of like terms. The columns of your matrix are the coordinate vectors of the original basis in terms of the new basis.
Let $S = \{\vec{v_1}, \vec{v_2}, \vec{v_3}\}$ and $T = \{\vec{u_1}, \vec{u_2}, \vec{u_3}\}$ be two bases for a vector space $V$ . If the change of basis matrix $P$ from $S$ to $T$ satisfies $P\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ 2 \end{pmatrix}$ , what can we conclude?
Answer: The vector with $S$ -coordinates $(1,2,1)$ has $T$ -coordinates $(3,0,2)$ When working with change of basis problems, you need to understand what the change of basis matrix $P$ actually does: it converts coordinate vectors from one basis to another. Specifically, if $P$ is the change of basis matrix from basis $S$ to basis $T$ , then $P$ takes $S$ -coordinates and outputs the corresponding $T$ -coordinates of the same vector. Given that $P\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ 2 \end{pmatrix}$ , this means the vector that has coordinates $(1,2,1)$ in basis $S$ has coordinates $(3,0,2)$ in basis $T$ . This confirms answer choice D. Let's examine why the other options are incorrect: A) This equation reverses the direction of the transformation and incorrectly mixes the coordinate representations with the actual vector equation. B) This has the transformation backwards—it suggests that $T$ -coordinates $(1,2,1)$ become $S$ -coordinates $(3,0,2)$ , which would be true if we had $P^{-1}$ instead of $P$ . C) This equation also reverses the relationship. The vector $\vec{v_1} + 2\vec{v_2} + \vec{v_3}$ (with $S$ -coordinates $(1,2,1)$ ) should equal $3\vec{u_1} + 0\vec{u_2} + 2\vec{u_3}$ , not $3\vec{u_1} + 2\vec{u_3}$ . Study tip: Always remember the direction of your change of basis matrix. If $P$ changes from basis $S$ to basis $T$ , then $P$ maps $S$ -coordinates to $T$ -coordinates, not the reverse.
In $\mathbb{R}^2$ , consider the basis $\mathcal{B} = \left\{\begin{pmatrix} 3 \\ 1 \end{pmatrix}, \begin{pmatrix} -2 \\ 2 \end{pmatrix}\right\}$ . Let $\mathcal{E}$ be the standard basis. What is the change-of-coordinates matrix $P_{\mathcal{E} \leftarrow \mathcal{B}}$ ?
Answer: $\begin{pmatrix} 3 & -2 \\ 1 & 2 \end{pmatrix}$ The change-of-coordinates matrix from a basis $\mathcal{B}$ to the standard basis $\mathcal{E}$ is formed by using the vectors of $\mathcal{B}$ as its columns. This matrix, often denoted $P_\mathcal{B}$ , directly converts $\mathcal{B}$ -coordinates to standard coordinates. Therefore, $P_{\mathcal{E} \leftarrow \mathcal{B}} = [\vec{b}_1 \; \vec{b}_2] = \begin{pmatrix} 3 & -2 \\ 1 & 2 \end{pmatrix}$ . Choice (B) is the inverse matrix, $P_{\mathcal{B} \leftarrow \mathcal{E}}$ . Choice (C) incorrectly places the basis vectors as rows. Choice (D) is the identity matrix, which would imply $\mathcal{B}$ is the standard basis.
Let $\mathcal{B} = \{\vec{b}_1, \vec{b}_2\}$ and $\mathcal{C} = \{\vec{c}_1, \vec{c}_2\}$ be two bases for a vector space $V$ . Let $P_{\mathcal{C} \leftarrow \mathcal{B}}$ be the change-of-coordinates matrix from $\mathcal{B}$ to $\mathcal{C}$ . Which equation correctly relates the coordinate vector $[\vec{x}]_\mathcal{B}$ of a vector $\vec{x} \in V$ to its coordinate vector $[\vec{x}]_\mathcal{C}$ ?
Answer: $[\vec{x}]_\mathcal{C} = P_{\mathcal{C} \leftarrow \mathcal{B}} [\vec{x}]_\mathcal{B}$ When you encounter change-of-coordinates matrices, remember that the notation $P_{\mathcal{B} \leftarrow \mathcal{C}}$ tells you exactly what the matrix should contain: it changes coordinates FROM basis $\mathcal{C}$ TO basis $\mathcal{B}$ . The key insight is that $P_{\mathcal{B} \leftarrow \mathcal{C}}$ has the vectors of basis $\mathcal{C}$ written as columns, but expressed in terms of basis $\mathcal{B}$ . Since you're given that $\vec{c}_1 = 2\vec{b}_1 + \vec{b}_2$ and $\vec{c}_2 = \vec{b}_1 - 3\vec{b}_2$ , you can read the coordinates directly: - $\vec{c}_1$ in $\mathcal{B}$ -coordinates is $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ - $\vec{c}_2$ in $\mathcal{B}$ -coordinates is $\begin{pmatrix} 1 \\ -3 \end{pmatrix}$ Therefore, $P_{\mathcal{B} \leftarrow \mathcal{C}} = \begin{pmatrix} 2 & 1 \\ 1 & -3 \end{pmatrix}$ , which is answer C. Looking at the wrong answers: A) $\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ incorrectly uses $+1$ instead of $-3$ in the bottom-right entry. B) takes the inverse of the correct matrix, which would give you $P_{\mathcal{C} \leftarrow \mathcal{B}}$ instead. D) $\begin{pmatrix} 1 & -3 \\ 2 & 1 \end{pmatrix}$ transposes the correct answer, mixing up which coefficients go in which positions. Study tip: The change-of-coordinates matrix $P_{\mathcal{B} \leftarrow \mathcal{C}}$ always has the basis vectors of $\mathcal{C}$ as its columns, expressed in $\mathcal{B}$ -coordinates. The arrow direction in the notation tells you which basis goes where.
Consider the bases $B = \{\vec{b_1}, \vec{b_2}, \vec{b_3}\}$ and $C = \{\vec{c_1}, \vec{c_2}, \vec{c_3}\}$ for $\mathbb{R}^3$ . If the change of basis matrix from $B$ to $C$ is $P = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 1 & 3 \\ 1 & 0 & 2 \end{pmatrix}$ , what is the change of basis matrix from $C$ to $B$ ?
Answer: $\begin{pmatrix} \frac{2}{3} & -\frac{2}{3} & 1 \\ \frac{1}{3} & \frac{4}{3} & -2 \\ -\frac{1}{3} & \frac{1}{3} & \frac{2}{3} \end{pmatrix}$ The change of basis matrix from $C$ to $B$ is $P^{-1}$ . Computing the inverse: $\det(P) = 2(2) - 1(0-3) + 0 = 4 + 3 = 7$ . Using the adjugate method and dividing by 7 gives choice A. Choice B results from incorrectly computing cofactors. Choice C is $3P - I$ from a conceptual error. Choice D results from dividing the correct adjugate by the wrong determinant value.
Let $\mathcal{B} = \{\vec{b}_1, \vec{b}_2\}$ and $\mathcal{C} = \{\vec{c}_1, \vec{c}_2\}$ be bases for $\mathbb{R}^2$ , where $\vec{c}_1 = 2\vec{b}_1 + \vec{b}_2$ and $\vec{c}_2 = \vec{b}_1 - 3\vec{b}_2$ . Which matrix is the change-of-coordinates matrix from $\mathcal{C}$ to $\mathcal{B}$ , denoted $P_{\mathcal{B} \leftarrow \mathcal{C}}$ ?
Answer: $\begin{pmatrix} 2 & 1 \\ 1 & -3 \end{pmatrix}$ When you encounter change-of-coordinates matrices, remember that the notation $P_{\mathcal{B} \leftarrow \mathcal{C}}$ tells you exactly what the matrix should contain: it changes coordinates FROM basis $\mathcal{C}$ TO basis $\mathcal{B}$ . The key insight is that $P_{\mathcal{B} \leftarrow \mathcal{C}}$ has the vectors of basis $\mathcal{C}$ written as columns, but expressed in terms of basis $\mathcal{B}$ . Since you're given that $\vec{c}_1 = 2\vec{b}_1 + \vec{b}_2$ and $\vec{c}_2 = \vec{b}_1 - 3\vec{b}_2$ , you can read the coordinates directly: - $\vec{c}_1$ in $\mathcal{B}$ -coordinates is $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ - $\vec{c}_2$ in $\mathcal{B}$ -coordinates is $\begin{pmatrix} 1 \\ -3 \end{pmatrix}$ Therefore, $P_{\mathcal{B} \leftarrow \mathcal{C}} = \begin{pmatrix} 2 & 1 \\ 1 & -3 \end{pmatrix}$ , which is answer C. Looking at the wrong answers: A) $\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ incorrectly uses $+1$ instead of $-3$ in the bottom-right entry. B) takes the inverse of the correct matrix, which would give you $P_{\mathcal{C} \leftarrow \mathcal{B}}$ instead. D) $\begin{pmatrix} 1 & -3 \\ 2 & 1 \end{pmatrix}$ transposes the correct answer, mixing up which coefficients go in which positions. Study tip: The change-of-coordinates matrix $P_{\mathcal{B} \leftarrow \mathcal{C}}$ always has the basis vectors of $\mathcal{C}$ as its columns, expressed in $\mathcal{B}$ -coordinates. The arrow direction in the notation tells you which basis goes where.
In $\mathbb{R}^2$ , consider the basis $\mathcal{B} = \left\{\begin{pmatrix} 4 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \end{pmatrix}\right\}$ . Let $\mathcal{E}$ be the standard basis. Find the change-of-coordinates matrix $P_{\mathcal{B} \leftarrow \mathcal{E}}$ .
Answer: $\frac{1}{5}\begin{pmatrix} 1 & -1 \\ 1 & 4 \end{pmatrix}$ The matrix $P_{\mathcal{B} \leftarrow \mathcal{E}}$ converts standard coordinates to $\mathcal{B}$ -coordinates. It is the inverse of the matrix $P_{\mathcal{E} \leftarrow \mathcal{B}}$ , whose columns are the vectors of $\mathcal{B}$ . First, we find $P_{\mathcal{E} \leftarrow \mathcal{B}} = \begin{pmatrix} 4 & 1 \\ -1 & 1 \end{pmatrix}$ . The determinant is $(4)(1) - (1)(-1) = 5$ . The inverse is $P_{\mathcal{B} \leftarrow \mathcal{E}} = (P_{\mathcal{E} \leftarrow \mathcal{B}})^{-1} = \frac{1}{5}\begin{pmatrix} 1 & -1 \\ 1 & 4 \end{pmatrix}$ . Choice (A) is the matrix for the reverse transformation, $P_{\mathcal{E} \leftarrow \mathcal{B}}$ . Choice (B) is the correct inverse but omits the scalar factor $1/5$ . Choice (C) is the transpose of the correct answer.
Let $\mathcal{B}$ , $\mathcal{C}$ , and $\mathcal{D}$ be three different bases for a finite-dimensional vector space $V$ . Let $P_{\mathcal{C} \leftarrow \mathcal{B}}$ , $P_{\mathcal{D} \leftarrow \mathcal{C}}$ , and $P_{\mathcal{D} \leftarrow \mathcal{B}}$ be the corresponding change-of-coordinates matrices. Which of the following statements is always true?
Answer: $P_{\mathcal{D} \leftarrow \mathcal{B}} = P_{\mathcal{D} \leftarrow \mathcal{C}} P_{\mathcal{C} \leftarrow \mathcal{B}}$ Changing coordinates from $\mathcal{B}$ to $\mathcal{D}$ can be done in two steps: first from $\mathcal{B}$ to $\mathcal{C}$ , and then from $\mathcal{C}$ to $\mathcal{D}$ . The combined transformation is the product of the individual transformation matrices. For a vector $\vec{x}$ , we have $[\vec{x}]_\mathcal{C} = P_{\mathcal{C} \leftarrow \mathcal{B}}[\vec{x}]_\mathcal{B}$ and $[\vec{x}]_\mathcal{D} = P_{\mathcal{D} \leftarrow \mathcal{C}}[\vec{x}]_\mathcal{C}$ . Substituting the first equation into the second gives $[\vec{x}]_\mathcal{D} = P_{\mathcal{D} \leftarrow \mathcal{C}}(P_{\mathcal{C} \leftarrow \mathcal{B}}[\vec{x}]_\mathcal{B}) = (P_{\mathcal{D} \leftarrow \mathcal{C}} P_{\mathcal{C} \leftarrow \mathcal{B}})[\vec{x}]_\mathcal{B}$ . Thus, $P_{\mathcal{D} \leftarrow \mathcal{B}} = P_{\mathcal{D} \leftarrow \mathcal{C}} P_{\mathcal{C} \leftarrow \mathcal{B}}$ . Choice (A) has the matrices in the wrong order. Choice (C) is incorrect; the relationship is $P_{\mathcal{C} \leftarrow \mathcal{B}} = (P_{\mathcal{B} \leftarrow \mathcal{C}})^{-1}$ . Choice (D) is incorrect as transformations are combined via multiplication, not addition.
In the space $P_1$ , consider the bases $\mathcal{B} = \{1+t, 1-t\}$ and $\mathcal{C} = \{2, 3t\}$ . Find the change-of-coordinates matrix $P_{\mathcal{C} \leftarrow \mathcal{B}}$ .
Answer: $\begin{pmatrix} 1/2 & 1/2 \\ 1/3 & -1/3 \end{pmatrix}$ Using the standard basis $\mathcal{E} = \{1, t\}$ , we write the coordinate matrices for $\mathcal{B}$ and $\mathcal{C}$ : $P_\mathcal{B} = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ and $P_\mathcal{C} = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$ . The required matrix is $P_{\mathcal{C} \leftarrow \mathcal{B}} = (P_\mathcal{C})^{-1}P_\mathcal{B}$ . The inverse of the diagonal matrix $P_\mathcal{C}$ is $(P_\mathcal{C})^{-1} = \begin{pmatrix} 1/2 & 0 \\ 0 & 1/3 \end{pmatrix}$ . Multiplying gives $P_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{pmatrix} 1/2 & 0 \\ 0 & 1/3 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 1/2 & 1/2 \\ 1/3 & -1/3 \end{pmatrix}$ . Choice (A) is $P_\mathcal{C} P_\mathcal{B}$ . Choice (B) is $P_{\mathcal{B} \leftarrow \mathcal{C}}$ . Choice (C) results from computing $P_\mathcal{B} (P_\mathcal{C})^{-1}$ .
In $\mathbb{R}^2$ , let $\mathcal{B} = \left\{\begin{pmatrix} 1 \\ 2 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \end{pmatrix}\right\}$ and let $\mathcal{C} = \{\vec{c}_1, \vec{c}_2\}$ be another basis. The change-of-coordinates matrix from $\mathcal{B}$ to $\mathcal{C}$ is $P_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{pmatrix} 2 & 1 \\ -1 & 1 \end{pmatrix}$ . What is the basis vector $\vec{c}_1$ ?
Answer: $\begin{pmatrix} 1/3 \\ 1 \end{pmatrix}$ The matrix $P_{\mathcal{C} \leftarrow \mathcal{B}}$ is given by $[[\vec{b}_1]_\mathcal{C} \; [\vec{b}_2]_\mathcal{C}]$ . We also know the relationship $P_\mathcal{C} = P_\mathcal{B} (P_{\mathcal{C} \leftarrow \mathcal{B}})^{-1}$ , where $P_\mathcal{B}$ and $P_\mathcal{C}$ have the basis vectors as columns. First, find $(P_{\mathcal{C} \leftarrow \mathcal{B}})^{-1}$ . The determinant is $(2)(1) - (1)(-1) = 3$ . So, $(P_{\mathcal{C} \leftarrow \mathcal{B}})^{-1} = \frac{1}{3}\begin{pmatrix} 1 & -1 \\ 1 & 2 \end{pmatrix}$ . Now, compute $P_\mathcal{C} = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \frac{1}{3}\begin{pmatrix} 1 & -1 \\ 1 & 2 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 1 & -1 \\ 3 & 0 \end{pmatrix} = \begin{pmatrix} 1/3 & -1/3 \\ 1 & 0 \end{pmatrix}$ . The columns of $P_\mathcal{C}$ are the vectors $\vec{c}_1$ and $\vec{c}_2$ . Thus, $\vec{c}_1 = \begin{pmatrix} 1/3 \\ 1 \end{pmatrix}$ . Choice (B) results from multiplying $P_\mathcal{B} P_{\mathcal{C} \leftarrow \mathcal{B}}$ . Choice (C) is the first column of $P_{\mathcal{C} \leftarrow \mathcal{B}}$ . Choice (D) is the result of using a sign error when finding the inverse.
In $\mathbb{R}^2$ , consider the bases $\mathcal{B} = \left\{\begin{pmatrix} 1 \\ 2 \end{pmatrix}, \begin{pmatrix} 1 \\ 3 \end{pmatrix}\right\}$ and $\mathcal{C} = \left\{\begin{pmatrix} 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 1 \end{pmatrix}\right\}$ . A vector $\vec{x}$ has coordinate vector $[\vec{x}]_\mathcal{B} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ . What is the coordinate vector $[\vec{x}]_\mathcal{C}$ ?
Answer: $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ First, find the change-of-coordinates matrix $P_{\mathcal{C} \leftarrow \mathcal{B}} = (P_\mathcal{C})^{-1}P_\mathcal{B}$ . We have $P_\mathcal{B} = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}$ and $P_\mathcal{C} = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}$ . The inverse of $P_\mathcal{C}$ is $(P_\mathcal{C})^{-1} = \frac{1}{1-2}\begin{pmatrix} 1 & -2 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix}$ . Then, $P_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 3 & 5 \\ -1 & -2 \end{pmatrix}$ . Finally, compute $[\vec{x}]_\mathcal{C} = P_{\mathcal{C} \leftarrow \mathcal{B}}[\vec{x}]_\mathcal{B} = \begin{pmatrix} 3 & 5 \\ -1 & -2 \end{pmatrix}\begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 6-5 \\ -2+2 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ . Choice (B) results from using $P_{\mathcal{B} \leftarrow \mathcal{C}}$ instead. Choice (C) comes from multiplying $P_\mathcal{C}[\vec{x}]_\mathcal{B}$ . Choice (D) is the vector $\vec{x}$ in standard coordinates, not $\mathcal{C}$ -coordinates.
The basis $\mathcal{B} = \left\{ \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix}, \begin{pmatrix} -1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix} \right\}$ is obtained by rotating the standard basis vectors in $\mathbb{R}^2$ counter-clockwise by $45^\circ$ . Which matrix represents the change-of-coordinates matrix $P_{\mathcal{B} \leftarrow \mathcal{E}}$ that converts standard coordinates to $\mathcal{B}$ -coordinates?
Answer: $\begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{2} \\ -1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix}$ The matrix $P_{\mathcal{B} \leftarrow \mathcal{E}}$ converts standard coordinates to $\mathcal{B}$ -coordinates. This is the inverse of $P_{\mathcal{E} \leftarrow \mathcal{B}}$ , which has the vectors of $\mathcal{B}$ as columns. So, $P_{\mathcal{E} \leftarrow \mathcal{B}} = \begin{pmatrix} 1/\sqrt{2} & -1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix}$ . Since the basis $\mathcal{B}$ is orthonormal, this matrix is orthogonal, and its inverse is simply its transpose. Thus, $P_{\mathcal{B} \leftarrow \mathcal{E}} = (P_{\mathcal{E} \leftarrow \mathcal{B}})^{-1} = (P_{\mathcal{E} \leftarrow \mathcal{B}})^T = \begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{2} \\ -1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix}$ . Geometrically, converting standard coordinates to the rotated basis is equivalent to rotating the vector by $-45^\circ$ , and this is the matrix for that rotation. Choice (A) is the matrix $P_{\mathcal{E} \leftarrow \mathcal{B}}$ .
Let $V$ be a vector space with bases $A = \{\vec{a_1}, \vec{a_2}, \vec{a_3}\}$ and $B = \{\vec{b_1}, \vec{b_2}, \vec{b_3}\}$ . If the change of basis matrix from $A$ to $B$ has determinant $-2$ , and $[\vec{v}]_A = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$ , what can be concluded about the relationship between the bases?
Answer: Basis $B$ has opposite orientation to basis $A$ , and both span the same subspace When you encounter questions about change of basis matrices, focus on what the determinant tells you about the geometric relationship between the bases. The determinant of a change of basis matrix reveals two crucial pieces of information: orientation and volume scaling. Since the determinant is $-2$ , the negative sign indicates that basis $B$ has opposite orientation to basis $A$ - imagine flipping one basis to create a mirror image of the other. The absolute value $|-2| = 2$ tells you about volume scaling, which we'll address shortly. Crucially, both bases span the entire vector space $V$ because they are both bases of the same space. Change of basis matrices are always invertible transformations between equivalent representations of the same space. Choice B correctly identifies that basis $B$ has opposite orientation to basis $A$ and both span the same subspace (the entire vector space $V$ ). Choice A is wrong because bases of the same vector space always span the same subspace - the entire space. Also, the negative determinant indicates opposite orientation, not the same orientation. Choice C misinterprets the determinant. While $|-2| = 2$ does scale volumes by a factor of 2, this choice ignores the crucial orientation information from the negative sign. Choice D confuses "change of basis" with geometric orthogonality. The bases themselves aren't orthogonal; the determinant measures orientation and scaling, not perpendicularity between basis vectors. Study tip: Remember that determinant sign = orientation relationship, and determinant magnitude = volume scaling factor. Both pieces matter for complete geometric understanding.
In the vector space $P_2$ of polynomials of degree at most 2, consider the standard basis $\mathcal{E} = \{1, t, t^2\}$ and another basis $\mathcal{B} = \{1, t-1, (t-1)^2\}$ . Let $p(t) = 2t^2 + 3t + 5$ . Find the coordinate vector $[p(t)]_\mathcal{B}$ .
Answer: $\begin{pmatrix} 10 \\ 7 \\ 2 \end{pmatrix}$ We need to find coefficients $c_1, c_2, c_3$ such that $2t^2 + 3t + 5 = c_1(1) + c_2(t-1) + c_3(t-1)^2$ . Expanding the right side gives $c_1 + c_2t - c_2 + c_3(t^2 - 2t + 1) = c_3t^2 + (c_2 - 2c_3)t + (c_1 - c_2 + c_3)$ . By equating coefficients of powers of $t$ : (coefficient of $t^2$ ) $c_3 = 2$ ; (coefficient of $t$ ) $c_2 - 2c_3 = 3 \Rightarrow c_2 - 4 = 3 \Rightarrow c_2 = 7$ ; (constant term) $c_1 - c_2 + c_3 = 5 \Rightarrow c_1 - 7 + 2 = 5 \Rightarrow c_1 = 10$ . Thus, the coordinate vector is $\begin{pmatrix} 10 \\ 7 \\ 2 \end{pmatrix}$ . Choice (A) is the coordinate vector in the standard basis $\mathcal{E}$ . Choices (B) and (C) result from common algebraic errors during the expansion and solving process.
In $\mathbb{R}^2$ , consider the bases $B_1 = \{(1,1), (1,-1)\}$ and $B_2 = \{(2,0), (0,3)\}$ . If $[\vec{v}]_{B_1} = \begin{pmatrix} 4 \\ -1 \end{pmatrix}$ , what is the first component of $[\vec{v}]_{B_2}$ ?
Answer: $\frac{3}{2}$ When you encounter coordinate representation problems with different bases, you need to convert between coordinate systems by first finding the actual vector, then expressing it in the new basis. Start by finding the actual vector $\vec{v}$ in standard coordinates. Since $[\vec{v}]_{B_1} = \begin{pmatrix} 4 \\ -1 \end{pmatrix}$ , this means $\vec{v} = 4(1,1) + (-1)(1,-1) = (4,4) + (-1,-1) = (3,3)$ . Now express $(3,3)$ in terms of basis $B_2 = \{(2,0), (0,3)\}$ . You need to find coefficients $a$ and $b$ such that $(3,3) = a(2,0) + b(0,3) = (2a, 3b)$ . This gives you the system: $2a = 3$ and $3b = 3$ . Solving: $a = \frac{3}{2}$ and $b = 1$ . Therefore, the first component of $[\vec{v}]_{B_2}$ is $\frac{3}{2}$ . Looking at the wrong answers: Choice A ( $\frac{5}{3}$ ) likely comes from incorrectly solving $2a = 3$ as $a = \frac{2+3}{3}$ . Choice B ( $3$ ) results from mistakenly thinking the first component should match the x-coordinate of $\vec{v}$ . Choice D ( $\frac{5}{2}$ ) could arise from arithmetic errors when combining the basis vectors or solving the linear system. Remember this two-step process: first convert coordinate representation to the actual vector using the original basis, then find the new coordinate representation by expressing that vector as a linear combination of the new basis vectors. Always verify your answer by checking that your coefficients actually reconstruct the original vector.
In $\mathbb{R}^3$ , let $B = \{(1,0,1), (0,1,1), (1,1,0)\}$ be a basis. If the coordinates of vector $\vec{w}$ with respect to the standard basis are $(5,3,2)$ , what is the sum of the coordinates in $[\vec{w}]_B$ ?
Answer: $5$ When you encounter basis conversion problems, you're working with two different coordinate systems for the same vector. The key is understanding that $[\vec{w}]_B$ represents the coefficients needed to express $\vec{w}$ as a linear combination of the basis vectors in $B$ . To find $[\vec{w}]_B$ , you need to solve: $\vec{w} = c_1(1,0,1) + c_2(0,1,1) + c_3(1,1,0) = (5,3,2)$ This gives you the system: - $c_1 + c_3 = 5$ - $c_2 + c_3 = 3$ - $c_1 + c_2 = 2$ From the third equation: $c_2 = 2 - c_1$ . Substituting into the second equation: $(2 - c_1) + c_3 = 3$ , so $c_3 = 1 + c_1$ . Substituting into the first equation: $c_1 + (1 + c_1) = 5$ , giving $c_1 = 2$ . Therefore: $c_1 = 2$ , $c_2 = 0$ , $c_3 = 3$ . So $[\vec{w}]_B = (2, 0, 3)$ , and the sum is $2 + 0 + 3 = 5$ . Choice A ( $\frac{20}{3}$ ) likely comes from incorrectly averaging or combining the original coordinates. Choice B ( $10$ ) is simply the sum of the standard coordinates $5 + 3 + 2$ , which ignores the basis conversion entirely. Choice C ( $\frac{10}{3}$ ) might result from arithmetic errors in solving the system or incorrectly manipulating fractions. Remember: basis conversion requires solving a linear system, not just manipulating the original coordinates. Always verify your solution by checking that your linear combination actually produces the target vector.
Let $B_1 = \{(1, 0), (0, 1)\}$ and $B_2 = \{(3, 1), (2, 1)\}$ be two bases for $\mathbb{R}^2$ . If the coordinate vector of $\vec{v}$ with respect to $B_2$ is $[\vec{v}]_{B_2} = \begin{pmatrix} -2 \\ 5 \end{pmatrix}$ , what is $[\vec{v}]_{B_1}$ ?
Answer: $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$ To find $[\vec{v}]_{B_1}$ , we first express $\vec{v}$ in standard coordinates: $\vec{v} = -2(3,1) + 5(2,1) = (-6,2) + (10,5) = (4,3)$ . Since $B_1$ is the standard basis, $[\vec{v}]_{B_1} = (4,3)$ . Choice B results from incorrectly computing $\vec{v} = -2(3,1) - 5(2,1)$ . Choice C comes from sign error in the second component. Choice D results from computing $2(3,1) + 5(2,1) - (4,1)$ through misreading the coefficients.
Consider the basis $\mathcal{B} = \left\{ \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \right\}$ for $\mathbb{R}^3$ . A vector $\vec{v}$ has coordinate vector $[\vec{v}]_\mathcal{B} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}$ . What is $\vec{v}$ in the standard basis?
Answer: $\begin{pmatrix} 1 \\ -2 \\ -3 \end{pmatrix}$ When you see coordinate vectors with respect to a basis, you're dealing with the fundamental relationship between different coordinate systems. The coordinate vector $[\vec{v}]_\mathcal{B}$ tells you how to combine the basis vectors to reconstruct the original vector $\vec{v}$ . To find $\vec{v}$ in the standard basis, you multiply each basis vector by its corresponding coordinate and sum them: $\vec{v} = 3\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} + 1\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} + (-2)\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$ $= \begin{pmatrix} 3 \\ -3 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} + \begin{pmatrix} -2 \\ 0 \\ -2 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ -3 \end{pmatrix}$ This confirms answer C is correct. Answer A represents a common misconception: thinking the coordinate vector $[\vec{v}]_\mathcal{B}$ is the same as $\vec{v}$ itself. The coordinate vector only has meaning relative to the specific basis. Answer B appears to result from computational errors in the linear combination, possibly from sign mistakes or incorrect coefficient placement. Answer D is close to the correct answer but has the wrong third component (-1 instead of -3), suggesting an error in handling the final basis vector's contribution. Remember: coordinate vectors are instructions for building the actual vector. Always perform the linear combination $c_1\vec{b_1} + c_2\vec{b_2} + \ldots$ where the $c_i$ values come from the coordinate vector and the $\vec{b_i}$ are your basis vectors.
Let $\mathcal{B} = \left\{\begin{pmatrix} 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ -1 \end{pmatrix}\right\}$ and $\mathcal{C} = \left\{\begin{pmatrix} 2 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \end{pmatrix}\right\}$ be two bases for $\mathbb{R}^2$ . Find the change-of-coordinates matrix $P_{\mathcal{C} \leftarrow \mathcal{B}}$ .
Answer: $\frac{1}{3}\begin{pmatrix} 1 & 3 \\ 1 & -3 \end{pmatrix}$ The change-of-coordinates matrix is found using the formula $P_{\mathcal{C} \leftarrow \mathcal{B}} = (P_{\mathcal{E} \leftarrow \mathcal{C}})^{-1} (P_{\mathcal{E} \leftarrow \mathcal{B}})$ . Let $P_\mathcal{C} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ and $P_\mathcal{B} = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ . First, find $(P_\mathcal{C})^{-1}$ . The determinant is $(2)(2)-(1)(1)=3$ . So, $(P_\mathcal{C})^{-1} = \frac{1}{3}\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}$ . Now, multiply: $P_{\mathcal{C} \leftarrow \mathcal{B}} = \frac{1}{3}\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 2-1 & 2+1 \\ -1+2 & -1-2 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 1 & 3 \\ 1 & -3 \end{pmatrix}$ . Choice (A) results from multiplying $P_\mathcal{C}P_\mathcal{B}$ without inverting. Choice (B) is the matrix for the opposite direction, $P_{\mathcal{B} \leftarrow \mathcal{C}}$ . Choice (C) results from multiplying in the wrong order, $P_\mathcal{B}(P_\mathcal{C})^{-1}$ .

Start this quiz