Matrix Representation - Linear Algebra

1. The linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ represented by the matrix $A = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}$ can be described as a composition of which two geometric operations?
   Answer: A rotation by $90^\circ$ clockwise and a uniform scaling by a factor of 2. The matrix can be factored as $A = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$. The first matrix represents a uniform scaling by a factor of 2. The second matrix represents a rotation by $270^\circ$ counterclockwise, which is equivalent to a $90^\circ$ clockwise rotation. The determinant of A is 4, which rules out reflections (det=-1) and projections (det=0).
2. Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be a linear transformation defined by $T\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3x - y \\ x + 2y \end{pmatrix}$. Let $\mathcal{B} = \left\{ \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ -1 \end{pmatrix} \right\}$ be a basis for $\mathbb{R}^2$. Find the matrix representation of $T$ with respect to the basis $\mathcal{B}$, denoted $[T]_{\mathcal{B}}$.
   Answer: $\begin{pmatrix} 5/2 & 3/2 \\ -1/2 & 5/2 \end{pmatrix}$ Let $\mathbf{b}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $\mathbf{b}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$. First, find the images of the basis vectors: $T(\mathbf{b}_1) = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and $T(\mathbf{b}_2) = \begin{pmatrix} 4 \\ -1 \end{pmatrix}$. Next, express these images in $\mathcal{B}$-coordinates. For $T(\mathbf{b}_1)$, solve $c_1\mathbf{b}_1 + c_2\mathbf{b}_2 = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$, which yields $c_1=5/2, c_2=-1/2$. This is the first column. For $T(\mathbf{b}_2)$, solve $d_1\mathbf{b}_1 + d_2\mathbf{b}_2 = \begin{pmatrix} 4 \\ -1 \end{pmatrix}$, which yields $d_1=3/2, d_2=5/2$. This is the second column. So, $[T]_{\mathcal{B}} = \begin{pmatrix} 5/2 & 3/2 \\ -1/2 & 5/2 \end{pmatrix}$.
3. Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be a linear transformation whose standard matrix is $A$. If $T(\mathbf{e}_1 + \mathbf{e}_2) = \begin{pmatrix} 3 \\ 5 \end{pmatrix}$ and $T(\mathbf{e}_1 - \mathbf{e}_2) = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$, what is the matrix $A$?
   Answer: $\begin{pmatrix} 2 & 1 \\ 2 & 3 \end{pmatrix}$ When you encounter a linear transformation problem where you're given the outputs for specific input vectors, your goal is to find what the transformation does to the standard basis vectors $\mathbf{e}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\mathbf{e}_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$, since these form the columns of the standard matrix. You have two equations: $T(\mathbf{e}_1 + \mathbf{e}_2) = \begin{pmatrix} 3 \\ 5 \end{pmatrix}$ and $T(\mathbf{e}_1 - \mathbf{e}_2) = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$. Since $T$ is linear, you can write these as $T(\mathbf{e}_1) + T(\mathbf{e}_2) = \begin{pmatrix} 3 \\ 5 \end{pmatrix}$ and $T(\mathbf{e}_1) - T(\mathbf{e}_2) = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$. Adding these equations: $2T(\mathbf{e}_1) = \begin{pmatrix} 4 \\ 4 \end{pmatrix}$, so $T(\mathbf{e}_1) = \begin{pmatrix} 2 \\ 2 \end{pmatrix}$. Subtracting the second from the first: $2T(\mathbf{e}_2) = \begin{pmatrix} 2 \\ 6 \end{pmatrix}$, so $T(\mathbf{e}_2) = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$. Therefore, $A = \begin{pmatrix} 2 & 1 \\ 2 & 3 \end{pmatrix}$, which is choice D. Choice A simply places the given output vectors as columns, ignoring that these aren't the images of the standard basis vectors. Choice B contains a computational error in solving the system. Choice C reverses some entries, likely from incorrectly setting up the linear system. Study tip: When given transformations of non-standard vectors, always use linearity to set up a system of equations to find $T(\mathbf{e}_1)$ and $T(\mathbf{e}_2)$ first—these become your matrix columns.
4. A linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ first reflects a vector across the line $y=x$ and then projects the resulting vector orthogonally onto the x-axis. What is the standard matrix for this transformation $T$?
   Answer: $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ Let $T_1$ be the reflection across $y=x$, and $T_2$ be the projection onto the x-axis. The matrix for $T_1$ is $A_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. The matrix for $T_2$ is $A_2 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$. The composite transformation is $T = T_2 \circ T_1$, so its matrix is $A = A_2 A_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$.
5. Let $T_1: \mathbb{R}^2 \to \mathbb{R}^2$ be a rotation counterclockwise by $90^\circ$, and let $T_2: \mathbb{R}^2 \to \mathbb{R}^2$ be a shear transformation that maps $(x,y)$ to $(x+2y, y)$. What is the standard matrix for the composite transformation $T_1 \circ T_2$?
   Answer: $\begin{pmatrix} 0 & -1 \\ 1 & 2 \end{pmatrix}$ The matrix for the rotation $T_1$ is $A_1 = \begin{pmatrix} \cos(90^\circ) & -\sin(90^\circ) \\ \sin(90^\circ) & \cos(90^\circ) \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. The matrix for the shear $T_2$ is found by transforming the basis vectors: $T_2(1,0)=(1,0)$ and $T_2(0,1)=(2,1)$, so $A_2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$. The matrix for the composition $T_1 \circ T_2$ is $A_1 A_2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 2 \end{pmatrix}$.
6. Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be the linear transformation that reflects vectors across the plane $x + y - z = 0$. If $A$ is the standard matrix representation of $T$, which of the following statements about $A$ is correct?
   Answer: $A$ has eigenvalues $1, 1, -1$ and $\det(A) = -1$ For a reflection across a plane, vectors in the plane are unchanged (eigenvalue $1$) and vectors perpendicular to the plane are reversed (eigenvalue $-1$). Since the plane has dimension $2$, there are two linearly independent eigenvectors with eigenvalue $1$ and one with eigenvalue $-1$. The normal vector to the plane $x + y - z = 0$ is $(1,1,-1)$, which corresponds to eigenvalue $-1$. Two vectors in the plane (eigenvalue $1$) can be found, such as $(1,-1,0)$ and $(1,0,1)$. Therefore the eigenvalues are $1, 1, -1$ and $\det(A) = 1 \cdot 1 \cdot (-1) = -1$. Choice B has the wrong eigenvalue multiplicities. Choice C is incorrect because $A$ is orthogonal (reflections preserve lengths and angles). Choice D is wrong because not all eigenvalues are distinct ($1$ has multiplicity $2$).
7. A linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ has eigenvectors $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ with eigenvalue $\lambda_1 = 3$, and $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$ with eigenvalue $\lambda_2 = -1$. What is the standard matrix $A$ for this transformation?
   Answer: $\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$ When you're given eigenvectors and eigenvalues of a linear transformation, you need to reconstruct the standard matrix using the relationship $A\mathbf{v} = \lambda\mathbf{v}$ for each eigenvector-eigenvalue pair. To find matrix $A$, use the fact that it must satisfy both eigenvalue equations simultaneously. From the given information: $A\begin{pmatrix} 1 \\ 1 \end{pmatrix} = 3\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \end{pmatrix}$ and $A\begin{pmatrix} 1 \\ -1 \end{pmatrix} = (-1)\begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}$ Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. From the first equation: $a + b = 3$ and $c + d = 3$. From the second equation: $a - b = -1$ and $c - d = 1$. Solving these systems gives $a = 1$, $b = 2$, $c = 2$, $d = 1$, so $A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$. Option A is a diagonal matrix with the eigenvalues on the diagonal, but this would only be correct if the eigenvectors were the standard basis vectors, which they're not. Option B incorrectly places eigenvalues as entries without proper construction. Option C fails both eigenvalue tests—multiplying this matrix by either eigenvector doesn't yield the correct scalar multiple. Option D correctly satisfies both eigenvalue equations when you multiply it out. Study tip: Always verify your answer by checking that $A\mathbf{v}_i = \lambda_i\mathbf{v}_i$ for each given eigenvector-eigenvalue pair. The diagonal form only works when eigenvectors are orthonormal basis vectors.
8. A linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ maps the vector $\mathbf{u} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ to $T(\mathbf{u}) = \begin{pmatrix} 5 \\ 2 \end{pmatrix}$ and the vector $\mathbf{v} = \begin{pmatrix} 0 \\ -1 \end{pmatrix}$ to $T(\mathbf{v}) = \begin{pmatrix} -1 \\ -1 \end{pmatrix}$. What is the standard matrix $A$ for this transformation?
   Answer: $\begin{pmatrix} 4 & 1 \\ 1 & 1 \end{pmatrix}$ The columns of the standard matrix $A$ are $T(\mathbf{e}_1)$ and $T(\mathbf{e}_2)$. We must express $\mathbf{e}_1$ and $\mathbf{e}_2$ as linear combinations of $\mathbf{u}$ and $\mathbf{v}$. We see that $\mathbf{e}_1 = \mathbf{u} + \mathbf{v}$ and $\mathbf{e}_2 = -\mathbf{v}$. By linearity: $T(\mathbf{e}_1) = T(\mathbf{u}) + T(\mathbf{v}) = \begin{pmatrix} 5 \\ 2 \end{pmatrix} + \begin{pmatrix} -1 \\ -1 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$. And $T(\mathbf{e}_2) = -T(\mathbf{v}) = -\begin{pmatrix} -1 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$. Therefore, the matrix $A$ is $\begin{pmatrix} 4 & 1 \\ 1 & 1 \end{pmatrix}$.
9. Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be a linear transformation with matrix representation $A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}$. If $B = \{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a basis for $\mathbb{R}^3$ such that $[T]_B$ is diagonal, which condition must the basis vectors satisfy?
   Answer: No such basis $B$ exists because $A$ is not diagonalizable over $\mathbb{R}$ The matrix $A$ has characteristic polynomial $\det(A - \lambda I) = (2-\lambda)^3$, so the only eigenvalue is $\lambda = 2$ with algebraic multiplicity $3$. To determine if $A$ is diagonalizable, we check the geometric multiplicity by finding $\dim(\ker(A - 2I))$. We have $A - 2I = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$. The nullspace is spanned by $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, so $\dim(\ker(A - 2I)) = 1$. Since the geometric multiplicity ($1$) is less than the algebraic multiplicity ($3$), the matrix $A$ is not diagonalizable. Therefore, no basis $B$ exists such that $[T]_B$ is diagonal. Choice A is wrong because $A$ doesn't have distinct eigenvalues. Choice B correctly identifies that we need Jordan chains, but the question asks for diagonalization, not Jordan form. Choice D is incorrect because even if we had three orthogonal eigenvectors with eigenvalue $2$, we've shown only one linearly independent eigenvector exists.
10. A linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ is represented by the invertible standard matrix $A = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}$. Which matrix represents the inverse transformation $T^{-1}$?
    Answer: $\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$ The matrix of the inverse transformation $T^{-1}$ is the inverse of the matrix $A$. For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the inverse is $\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. The determinant of $A$ is $(3)(2) - (-5)(-1) = 6 - 5 = 1$. Therefore, $A^{-1} = \frac{1}{1}\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$.
11. Consider the linear transformation $T: P_2 \to P_2$ on the space of polynomials of degree at most 2, defined by $T(p(x)) = x p'(x) - p(x)$. What is the matrix representation of $T$ with respect to the standard basis $\mathcal{B} = \{1, x, x^2\}$?
    Answer: $\begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ We apply the transformation $T$ to each basis vector and write the result in coordinates with respect to the basis $\mathcal{B}$. $T(1) = x \cdot (0) - 1 = -1 = -1 \cdot 1 + 0 \cdot x + 0 \cdot x^2 \implies [-1, 0, 0]^T$. $T(x) = x \cdot (1) - x = 0 = 0 \cdot 1 + 0 \cdot x + 0 \cdot x^2 \implies [0, 0, 0]^T$. $T(x^2) = x \cdot (2x) - x^2 = 2x^2 - x^2 = x^2 = 0 \cdot 1 + 0 \cdot x + 1 \cdot x^2 \implies [0, 0, 1]^T$. These coordinate vectors form the columns of the matrix, which is $\begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$.
12. Let $T: \mathbb{R}^3 \to \mathbb{R}^2$ be a linear transformation. If the kernel of $T$ is the line spanned by the vector $(1, 1, 1)$, and $T(1, 0, 0) = (1, 2)$, which of the following could be the standard matrix for $T$?
    Answer: $\begin{pmatrix} 1 & 1 & -2 \\ 2 & -1 & -1 \end{pmatrix}$ When you encounter questions about linear transformations with given kernel and image conditions, you need to use both pieces of information as constraints to identify the correct matrix. Since the kernel of $T$ is spanned by $(1,1,1)$, this vector must map to the zero vector. Additionally, $T(1,0,0) = (1,2)$ gives us the first column of the standard matrix. Let's call the matrix $A = \begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix}$. From $T(1,0,0) = (1,2)$, we know the first column is $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$, so $a = 1$ and $d = 2$. From the kernel condition, $A(1,1,1) = (0,0)$, which means: - $a + b + c = 0$, so $1 + b + c = 0$, giving us $b + c = -1$ - $d + e + f = 0$, so $2 + e + f = 0$, giving us $e + f = -2$ Now let's check each option: Option A: $\begin{pmatrix} 1 & 1 & -2 \\ 2 & 0 & -1 \end{pmatrix}$ - First column is correct, but $1 + (-2) = -1$ ✓ and $0 + (-1) = -1 \neq -2$ ✗ Option B: $\begin{pmatrix} 2 & -1 & -1 \\ 1 & 1 & -2 \end{pmatrix}$ - First column is $(2,1) \neq (1,2)$ ✗ Option C: $\begin{pmatrix} 1 & -1 & 0 \\ 2 & -2 & 0 \end{pmatrix}$ - First column is correct, but $(-1) + 0 = -1$ ✓ and $(-2) + 0 = -2$ ✓, but this gives kernel dimension 2, not 1 ✗ Option D: $\begin{pmatrix} 1 & 1 & -2 \\ 2 & -1 & -1 \end{pmatrix}$ - First column is correct, $1 + (-2) = -1$ ✓, and $(-1) + (-1) = -2$ ✓ Strategy tip: Always verify both the given transformation values and kernel conditions systematically. The kernel constraint often provides the most restrictive conditions for elimination.
13. Consider the linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^3$ defined by $T(x,y) = (2x-y, x+3y, -x+2y)$. If $B_1 = \{(1,1), (1,-1)\}$ is a basis for $\mathbb{R}^2$ and $B_2 = \{(1,0,0), (0,1,0), (0,0,1)\}$ is the standard basis for $\mathbb{R}^3$, what is the matrix representation $[T]_{B_1}^{B_2}$?
    Answer: $\begin{pmatrix} 1 & 3 \\ 4 & -2 \\ 1 & -3 \end{pmatrix}$ To find $[T]_{B_1}^{B_2}$, we need to compute $T$ applied to each vector in $B_1$ and express the results in coordinates relative to $B_2$. First, $T(1,1) = (2(1)-1, 1+3(1), -1+2(1)) = (1,4,1)$. Since $B_2$ is the standard basis, this is already in $B_2$-coordinates: $(1,4,1)^T$. Second, $T(1,-1) = (2(1)-(-1), 1+3(-1), -1+2(-1)) = (3,-2,-3)$. In $B_2$-coordinates: $(3,-2,-3)^T$. Therefore, $[T]_{B_1}^{B_2} = \begin{pmatrix} 1 & 3 \\ 4 & -2 \\ 1 & -3 \end{pmatrix}$. Choice B gives the standard matrix representation $[T]_{E_2}^{E_3}$ where $E_2, E_3$ are standard bases. Choice C has the wrong sign in the bottom-right entry. Choice D transposes some entries incorrectly.
14. What is the standard matrix for the linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ that orthogonally projects vectors onto the plane defined by the equation $x + 2y + 2z = 0$?
    Answer: $\frac{1}{9}\begin{pmatrix} 8 & -2 & -2 \\ -2 & 5 & -4 \\ -2 & -4 & 5 \end{pmatrix}$ The transformation is given by $T(\mathbf{x}) = \mathbf{x} - \text{proj}_{\mathbf{n}}\mathbf{x}$, where $\mathbf{n} = (1, 2, 2)$ is the normal vector to the plane. The matrix for the projection onto the normal vector is $P = \frac{1}{\mathbf{n}^T\mathbf{n}}\mathbf{n}\mathbf{n}^T = \frac{1}{9}\begin{pmatrix} 1 & 2 & 2 \\ 2 & 4 & 4 \\ 2 & 4 & 4 \end{pmatrix}$. The matrix for $T$ is $A = I - P = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} - \frac{1}{9}\begin{pmatrix} 1 & 2 & 2 \\ 2 & 4 & 4 \\ 2 & 4 & 4 \end{pmatrix} = \frac{1}{9}\begin{pmatrix} 8 & -2 & -2 \\ -2 & 5 & -4 \\ -2 & -4 & 5 \end{pmatrix}$. Choice A is the matrix for projection onto the normal line, not the plane.
15. Let $P_2$ denote the vector space of polynomials of degree at most $2$. Define the linear transformation $T: P_2 \to \mathbb{R}^3$ by $T(p(x)) = (p(0), p(1), p(-1))$. Using the basis $\{1, x, x^2\}$ for $P_2$ and the standard basis for $\mathbb{R}^3$, what is the rank of the matrix representation of $T$?
    Answer: $3$ because $T$ is injective on the finite-dimensional space $P_2$ To find the rank, we construct the matrix representation. For the basis polynomial $1$: $T(1) = (1, 1, 1)$. For $x$: $T(x) = (0, 1, -1)$. For $x^2$: $T(x^2) = (0, 1, 1)$. So the matrix is $A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{pmatrix}$. To find the rank, we row reduce: $\begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -1 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{pmatrix}$. This has $3$ pivots, so rank is $3$. Alternatively, $T$ is injective because if $T(p) = 0$, then $p(0) = p(1) = p(-1) = 0$, which means the polynomial $p$ of degree at most $2$ has three distinct roots, so $p \equiv 0$. Since $T$ is injective and $\dim(P_2) = 3$, we have $\text{rank}(T) = 3$. Choice A misunderstands that evaluation can still be injective. Choice B incorrectly assumes dependencies exist. Choice D incorrectly suggests codomain constraints reduce rank when the transformation is actually injective.

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