Linear Algebra Practice Test: Practice Test 1

Question 1

1. An additive identity element does not exist in $V$.
2. An additive inverse does not exist in $V$ for every element.
3. The set is not closed under 'vector addition'.
4. The set is not closed under 'scalar multiplication'. (correct answer)

Explanation: Closure under scalar multiplication requires that for any $c \in \mathbb{R}$ and $x \in V$, the result $c \odot x$ must be in $V$. Let $x=2 \in \mathbb{R}^+$ and $c=-1 \in \mathbb{R}$. Then $c \odot x = (-1)(2) = -2$. Since $-2$ is not a positive real number, $-2 \notin V$. Thus, the set is not closed under scalar multiplication. The other axioms do not fail: The additive identity is $1$ since $x \cdot 1 = x$. The additive inverse of $x$ is $1/x$ since $x \cdot (1/x) = 1$. The product of two positive numbers is positive, so it's closed under vector addition.

Question 2

1. $\begin{bmatrix} 6 \\ 5 \\ 11 \end{bmatrix}$
2. $\begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}$
3. $\begin{bmatrix} 6 \\ -11 \\ 7 \end{bmatrix}$ (correct answer)
4. $\begin{bmatrix} 6 \\ -8 \\ 2 \end{bmatrix}$

Explanation: By the properties of linear transformations, we can write the input vector as a linear combination of the basis vectors and apply the transformation. Specifically, $T(c\vec{u} + d\vec{v}) = cT(\vec{u}) + dT(\vec{v})$. We have $\begin{bmatrix} 3 \\ -2 \end{bmatrix} = 3\begin{bmatrix} 1 \\ 0 \end{bmatrix} - 2\begin{bmatrix} 0 \\ 1 \end{bmatrix}$. Therefore, $T\left(\begin{bmatrix} 3 \\ -2 \end{bmatrix}\right) = T\left(3\begin{bmatrix} 1 \\ 0 \end{bmatrix} - 2\begin{bmatrix} 0 \\ 1 \end{bmatrix}\right) = 3T\left(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\right) - 2T\left(\begin{bmatrix} 0 \\ 1 \end{bmatrix}\right)$. Substituting the given values: $3\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} - 2\begin{bmatrix} 0 \\ 4 \\ 1 \end{bmatrix} = \begin{bmatrix} 6 \\ -3 \\ 9 \end{bmatrix} - \begin{bmatrix} 0 \\ 8 \\ 2 \end{bmatrix} = \begin{bmatrix} 6 \\ -11 \\ 7 \end{bmatrix}$.

Question 3

1. $\vec{y}$ must be the zero vector.
2. $\vec{y}$ must be in $W$.
3. $\vec{y}$ must be in the orthogonal complement of $W$. (correct answer)
4. $\vec{y}$ must be a basis vector for $W$.

Explanation: The projection of $\vec{y}$ onto $W$ is zero if and only if $\vec{y}$ is orthogonal to every vector in $W$. The set of all vectors orthogonal to $W$ is, by definition, the orthogonal complement of $W$, denoted $W^\perp$. Therefore, if $\mathrm{proj}_W(\vec{y}) = \vec{0}$, it must be that $\vec{y} \in W^\perp$.\nA is a possible case, since if $\vec{y}=\vec{0}$, its projection is $\vec{0}$. However, any non-zero vector in $W^\perp$ also has a zero projection onto $W$, so this is not a necessary condition.\nB is incorrect. If $\vec{y}$ is in $W$ (and $\vec{y} \neq \vec{0}$), its projection onto $W$ is $\vec{y}$ itself, not $\vec{0}$.\nD is a specific case of B and is also incorrect.

Question 4

1. $D(c\vec{r}_1, c\vec{r}_2, c\vec{r}_3) = c D(\vec{r}_1, \vec{r}_2, \vec{r}_3)$
2. $D(\vec{r}_1 + \vec{s}, \vec{r}_2 + \vec{s}, \vec{r}_3 + \vec{s}) = D(\vec{r}_1, \vec{r}_2, \vec{r}_3) + D(\vec{s}, \vec{s}, \vec{s})$
3. $D(\vec{r}_1, \vec{r}_2, \vec{r}_3) = D(\vec{r}_1, \vec{r}_2, \vec{r}_3) - D(\vec{r}_1, \vec{r}_1, \vec{r}_3)$
4. $D(\vec{r}_1, c\vec{r}_2 + \vec{s}_2, \vec{r}_3) = c D(\vec{r}_1, \vec{r}_2, \vec{r}_3) + D(\vec{r}_1, \vec{s}_2, \vec{r}_3)$ (correct answer)

Explanation: When you encounter questions about determinant properties, focus on understanding that determinants are multilinear functions. This means they are linear in each row (or column) individually, but not necessarily linear across all rows simultaneously. The linearity property states that if you have a linear combination in any single row, the determinant distributes over that combination while keeping other rows fixed. Option D perfectly demonstrates this: $D(\vec{r}_1, c\vec{r}_2 + \vec{s}_2, \vec{r}_3) = c D(\vec{r}_1, \vec{r}_2, \vec{r}_3) + D(\vec{r}_1, \vec{s}_2, \vec{r}_3)$. The determinant is linear in the second row while the first and third rows remain unchanged. Option A is incorrect because scaling all rows by factor $c$ scales the determinant by $c^3$, not $c$. The correct relationship would be $D(c\vec{r}_1, c\vec{r}_2, c\vec{r}_3) = c^3 D(\vec{r}_1, \vec{r}_2, \vec{r}_3)$. Option B fails because determinants aren't linear across multiple rows simultaneously. Adding the same vector to all rows doesn't follow any standard determinant property, and $D(\vec{s}, \vec{s}, \vec{s}) = 0$ since identical rows make the determinant zero. Option C makes no mathematical sense—it's not even a proper equation since it claims something equals itself minus something else (unless that "something else" is zero, which isn't generally true). Study tip: Remember that determinants are linear in each individual row or column, but not across multiple rows/columns at once. Practice identifying which single row or column is being modified in linearity problems.

Question 5

1. When the vector $\mathbf{b}$ is orthogonal to the column space of $A$.
2. When the columns of the matrix $A$ are linearly dependent. (correct answer)
3. When the number of rows in $A$ is greater than the number of its columns.
4. When the system $A\mathbf{x} = \mathbf{b}$ is consistent (i.e., has an exact solution).

Explanation: The set of least-squares solutions is unique if and only if the matrix $A^T A$ is invertible. This occurs if and only if the columns of $A$ are linearly independent. If the columns of $A$ are linearly dependent, then $A^T A$ is singular, and the system $A^T A \hat{\mathbf{x}} = A^T \mathbf{b}$ will have infinitely many solutions. Geometrically, this means that a vector in the column space can be formed as a linear combination of the columns in more than one way.

Question 6

1. $T_1 \circ T_3$
2. $T_2 \circ T_1$
3. $T_3 \circ T_4$ (correct answer)
4. $T_4 \circ T_1$

Explanation: For a composition $S \circ T$ to be well-defined, the codomain of the inner transformation ($T$) must be the same as the domain of the outer transformation ($S$). We check each option: A) $T_1 \circ T_3$: The inner transformation is $T_3: \mathbb{R}^3 \to \mathbb{R}^3$. Its codomain is $\mathbb{R}^3$. The outer transformation is $T_1: \mathbb{R}^3 \to \mathbb{R}^2$. Its domain is $\mathbb{R}^3$. Since the codomain of $T_3$ matches the domain of $T_1$, this composition is well-defined. The resulting transformation maps $\mathbb{R}^3 \to \mathbb{R}^2$. B) $T_2 \circ T_1$: The inner transformation is $T_1: \mathbb{R}^3 \to \mathbb{R}^2$. Its codomain is $\mathbb{R}^2$. The outer transformation is $T_2: \mathbb{R}^2 \to \mathbb{R}^3$. Its domain is $\mathbb{R}^2$. Since the codomain of $T_1$ matches the domain of $T_2$, this composition is well-defined. The resulting transformation maps $\mathbb{R}^3 \to \mathbb{R}^3$. C) $T_3 \circ T_4$: The inner transformation is $T_4: \mathbb{R}^2 \to \mathbb{R}^2$. Its codomain is $\mathbb{R}^2$. The outer transformation is $T_3: \mathbb{R}^3 \to \mathbb{R}^3$. Its domain is $\mathbb{R}^3$. Since the codomain of $T_4$ ($\mathbb{R}^2$) does not match the domain of $T_3$ ($\mathbb{R}^3$), this composition is NOT well-defined. D) $T_4 \circ T_1$: The inner transformation is $T_1: \mathbb{R}^3 \to \mathbb{R}^2$. Its codomain is $\mathbb{R}^2$. The outer transformation is $T_4: \mathbb{R}^2 \to \mathbb{R}^2$. Its domain is $\mathbb{R}^2$. Since the codomain of $T_1$ matches the domain of $T_4$, this composition is well-defined. The resulting transformation maps $\mathbb{R}^3 \to \mathbb{R}^2$.

Question 7

1. $\{u_1, u_2, u_3\}$
2. $\{u_1, u_2, u_5\}$ (correct answer)
3. $\{u_1, u_2, u_4\}$
4. $\{u_2, u_3, u_4\}$

Explanation: To find a basis, we need to identify which vectors are linearly independent. Notice that $u_3 = u_1 + u_2$ since $(1, 0, 1, 1, 0, 1) + (0, 1, 1, 0, 1, 1) = (1, 1, 2, 1, 1, 2) = u_3$. Also, $u_4 = 2u_1 + u_2$ since $2(1, 0, 1, 1, 0, 1) + (0, 1, 1, 0, 1, 1) = (2, 1, 3, 2, 1, 3) = u_4$. We can verify that $u_1$, $u_2$, and $u_5$ are linearly independent by checking that no nontrivial linear combination equals zero. Since $u_3$ and $u_4$ are linear combinations of $u_1$ and $u_2$, and $u_5$ is not in the span of $u_1$ and $u_2$, the vectors $\{u_1, u_2, u_5\}$ form a basis for the 3-dimensional subspace V. The other options either include dependent vectors or cannot span the full space.

Question 8

1. $\mathbf{q}_1 = [0, 1, 0]^T$ and $r_{11} = 1$ (correct answer)
2. $\mathbf{q}_1 = [0, 1, 0]^T$ and $r_{11} = 5$
3. $\mathbf{q}_1 = [0, 1/\sqrt{50}, 0]^T$ and $r_{11} = \sqrt{50}$
4. $\mathbf{q}_1 = [0, 1, 0]^T$ and $r_{11} = 0$

Explanation: The first step in the Gram-Schmidt process is to set $\mathbf{v}_1 = \mathbf{a}_1 = [0, 1, 0]^T$. Next, we find the norm of $\mathbf{v}_1$, which is $\|\mathbf{v}_1\| = \sqrt{0^2+1^2+0^2} = 1$. The first diagonal entry of $R$ is this norm, so $r_{11} = 1$. The first column of $Q$ is found by normalizing $\mathbf{v}_1$: $\mathbf{q}_1 = \frac{\mathbf{v}_1}{\|\mathbf{v}_1\|} = \frac{[0, 1, 0]^T}{1} = [0, 1, 0]^T$. Since the first column of $A$ was already a unit vector, the process is simplified for the first step.

Question 9

1. $\mathrm{nullity}(A) \ge 1$ (correct answer)
2. $\mathrm{nullity}(A) = 1$
3. $\mathrm{rank}(A) = 3$
4. The columns of $A$ are linearly independent.

Explanation: This question tests your understanding of linear dependence and the rank-nullity theorem. When you see a linear relationship between columns of a matrix, immediately think about what this tells you about the matrix's rank and nullity. The given condition $3\mathbf{v}_1 - \mathbf{v}_3 = \mathbf{0}$ can be rewritten as $3\mathbf{v}_1 + 0\mathbf{v}_2 + (-1)\mathbf{v}_3 + 0\mathbf{v}_4 = \mathbf{0}$. This is a nontrivial linear combination of the columns that equals zero (the coefficients aren't all zero), which means the columns are linearly dependent. This linear dependence relation gives us a vector in the null space of $A$: the vector $\mathbf{x} = \begin{bmatrix} 3 \\ 0 \\ -1 \\ 0 \end{bmatrix}$ satisfies $A\mathbf{x} = \mathbf{0}$. Since the null space contains at least this nonzero vector, we know $\mathrm{nullity}(A) \geq 1$. This makes choice A correct. Choice B is wrong because we only know there's at least one linearly dependent relationship—there could be more, making the nullity greater than 1. Choice C is incorrect for the same reason: while the rank is at most 3 (since the columns are dependent), it could be less if there are additional dependencies. Choice D directly contradicts our finding that the columns are linearly dependent. Remember: whenever you find a nontrivial linear combination of matrix columns equaling zero, you've discovered that the nullity is at least 1. Look for the weakest statement that must be true—it's often the correct answer in "must be true" questions.

Question 10

1. $14$ (correct answer)
2. $\frac{14}{9}$
3. $126$
4. $42$

Explanation: For any orthonormal basis, Parseval's identity states that $\|\mathbf{v}\|^2 = \alpha^2 + \beta^2 + \gamma^2$ where $\alpha, \beta, \gamma$ are the coordinates of $\mathbf{v}$ in that basis. Since $\{\mathbf{q}_1, \mathbf{q}_2, \mathbf{q}_3\}$ is orthonormal and $\|\mathbf{v}\|^2 = 14$, we immediately get $\alpha^2 + \beta^2 + \gamma^2 = 14$. Choice B incorrectly applies $\frac{\|\mathbf{v}\|^2}{9}$ thinking the basis vectors have length $\frac{1}{3}$ rather than 1. Choice C computes $14 \times 9 = 126$, incorrectly scaling up. Choice D uses $14 \times 3 = 42$, another scaling error.

Question 11

1. $\begin{pmatrix} 0 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \end{pmatrix}$
2. $\begin{pmatrix} 0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ (correct answer)
3. $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$
4. $\begin{pmatrix} 0 & 2 & 3 \\ 2 & 0 & 3 \\ 3 & 2 & 0 \end{pmatrix}$

Explanation: Since the $3 \times 3$ matrix $A$ has three distinct eigenvalues (0, 2, and 3), it is guaranteed to be diagonalizable. This means $A$ is similar to a diagonal matrix $D$ whose diagonal entries are the eigenvalues of $A$. Two matrices are similar if one can be obtained from the other by a similarity transformation ($B = P^{-1}AP$). Therefore, $A$ must be similar to the diagonal matrix formed by its eigenvalues. The matrix in choice B is the correct diagonal matrix. The order of the eigenvalues on the diagonal can vary, but any such diagonal matrix is similar to any other permutation.

Question 12

1. The geometric multiplicity is $2$ since the algebraic multiplicity is $2$
2. The geometric multiplicity is $1$ regardless of the values of $a, b, c$ (correct answer)
3. The geometric multiplicity is $2$ if and only if $c = 0$
4. The geometric multiplicity depends on whether $ac = b$

Explanation: The matrix $A - 3I = \begin{pmatrix} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 2 \end{pmatrix}$. The geometric multiplicity of eigenvalue $3$ equals $3 - \text{rank}(A - 3I)$. Since the second and third columns are linearly independent (the third column has a nonzero entry in position $(3,3)$ while the second column has zero there, and $c \neq 0$), and the first column is zero, we have $\text{rank}(A - 3I) = 2$. Therefore, the geometric multiplicity is $3 - 2 = 1$, regardless of the specific nonzero values of $a, b, c$.

Question 13

1. The RREF of $A$ has a pivot in every column.
2. The RREF of $A$ contains a row of zeros.
3. The nullity of $A$ is 0.
4. The RREF of $A$ has a pivot in every row. (correct answer)

Explanation: For $T$ to be surjective, its range must be all of the codomain, $\mathbb{R}^3$. This means the dimension of the range, which is the rank of matrix $A$, must be 3. The rank of a matrix is equal to the number of pivots in its RREF. Since $A$ is a $3 \times 4$ matrix, having a rank of 3 means there must be 3 pivots. As there are only 3 rows, this implies there must be a pivot position in every row.

Question 14

1. $k = -2$
2. $k = 0$
3. $k = 2$ (correct answer)
4. $k = 4$

Explanation: A set of three vectors in $\mathbb{R}^3$ fails to be a generating set (i.e., does not span $\mathbb{R}^3$) if and only if the vectors are linearly dependent. This occurs when the matrix formed by these vectors as columns has a determinant of zero. The determinant of the matrix $A = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & -2 \\ 2 & 1 & k \end{pmatrix}$ is calculated as $1(k - (-2)) - 0(...) + 2(0 - 2) = k + 2 - 4 = k - 2$. Setting the determinant to zero, we get $k - 2 = 0$, which gives $k = 2$. When $k=2$, the third vector is a linear combination of the first two ($2\vec{v}_1 - 2\vec{v}_2$), so the set is linearly dependent and spans only a plane.

Question 15

1. $\begin{pmatrix} 0.5 \\ 0.5 \end{pmatrix}$ (correct answer)
2. $\begin{pmatrix} 0.8 \\ 0.2 \end{pmatrix}$
3. $\begin{pmatrix} 0.4 \\ 0.1 \end{pmatrix}$
4. $\begin{pmatrix} 0.65 \\ 0.35 \end{pmatrix}$

Explanation: This is a discrete dynamical system problem where you apply matrix multiplication iteratively to track how a state vector evolves over time. When you see $\vec{x}_{k+1} = P\vec{x}_k$, you're looking at a linear transformation that gets applied repeatedly. To find $\vec{x}_2$, you need to apply the transformation matrix $P$ twice. First, calculate $\vec{x}_1 = P\vec{x}_0$: $\vec{x}_1 = \begin{pmatrix} 0.5 & 0.5 \\ 0.5 & 0.5 \end{pmatrix} \begin{pmatrix} 0.8 \\ 0.2 \end{pmatrix} = \begin{pmatrix} 0.5(0.8) + 0.5(0.2) \\ 0.5(0.8) + 0.5(0.2) \end{pmatrix} = \begin{pmatrix} 0.5 \\ 0.5 \end{pmatrix}$ Then calculate $\vec{x}_2 = P\vec{x}_1$: $\vec{x}_2 = \begin{pmatrix} 0.5 & 0.5 \\ 0.5 & 0.5 \end{pmatrix} \begin{pmatrix} 0.5 \\ 0.5 \end{pmatrix} = \begin{pmatrix} 0.5 \\ 0.5 \end{pmatrix}$ Choice A is correct. Notice that this matrix has identical rows, which means any vector gets transformed into $\begin{pmatrix} 0.5 \\ 0.5 \end{pmatrix}$ after one step and stays there. Choice B represents the initial state $\vec{x}_0$, suggesting the system doesn't change—a common misconception. Choice C looks like someone multiplied the initial vector by 0.5 element-wise rather than using proper matrix multiplication. Choice D appears to be an intermediate calculation error, possibly averaging incorrectly or stopping partway through the matrix multiplication. Study tip: With discrete systems, always compute step-by-step rather than trying shortcuts. Also, look for patterns—matrices with identical rows quickly converge to a fixed point, which can help you check your work.

Question 16

1. the rows of the matrix $P-I$ are linearly dependent.
2. the Perron-Frobenius theorem for positive matrices guarantees it.
3. the columns of the matrix $P-I$ sum to the zero vector, implying they are linearly dependent. (correct answer)
4. the trace of a stochastic matrix is always equal to 1.

Explanation: An eigenvalue $\lambda$ exists if and only if the matrix $(P-\lambda I)$ is singular, meaning its determinant is zero. For $\lambda=1$, we consider $P-I$. The columns of $P$ each sum to 1. When we subtract the identity matrix $I$, we are subtracting 1 from each diagonal element. This means that for each column of $P-I$, the sum of its elements becomes $1-1=0$. If the elements of each column vector sum to zero, the sum of the row vectors of $(P-I)$ is the zero vector, meaning the rows (and columns) are linearly dependent. A matrix with linearly dependent columns is singular, so $\det(P-I)=0$, proving that $\lambda=1$ is an eigenvalue.

Question 17

1. $\begin{pmatrix} 1 & 2 & 0 & 0 & | & 4 \\ 0 & 0 & 1 & -3 & | & -1 \\ 0 & 0 & 0 & 0 & | & 0 \end{pmatrix}$ (correct answer)
2. $\begin{pmatrix} 1 & -2 & 0 & 0 & | & 4 \\ 0 & 0 & 1 & 3 & | & -1 \\ 0 & 0 & 0 & 0 & | & 0 \end{pmatrix}$
3. $\begin{pmatrix} 1 & 0 & 0 & 0 & | & 4 \\ 0 & 1 & 0 & 0 & | & 0 \\ 0 & 0 & 1 & 0 & | & -1 \end{pmatrix}$
4. $\begin{pmatrix} 1 & 2 & 0 & 0 & | & 4 \\ 0 & 0 & 1 & -3 & | & -1 \\ 0 & 0 & 0 & 0 & | & 1 \end{pmatrix}$

Explanation: From the parametric solution, we can identify the free and basic variables. The parameters $s$ and $t$ correspond to free variables. The vector for $s$ has a 1 in the second position, so $x_2 = s$. The vector for $t$ has a 1 in the fourth position, so $x_4 = t$. Thus, $x_2$ and $x_4$ are free variables, and $x_1$ and $x_3$ are basic variables. From the solution, we can write equations for the basic variables: $x_1 = 4 - 2s + 0t = 4 - 2x_2 \implies x_1 + 2x_2 = 4$ $x_3 = -1 + 0s + 3t = -1 + 3x_4 \implies x_3 - 3x_4 = -1$ These two equations correspond to the rows of the RREF matrix. The first equation gives the row $[1 \ 2 \ 0 \ 0 \ | \ 4]$. The second equation gives the row $[0 \ 0 \ 1 \ -3 \ | \ -1]$. The system is consistent, so any additional rows must be zero rows. This matches matrix A. Distractor B has incorrect signs in the non-pivot entries. Distractor C implies a unique solution with no free variables, which contradicts the given parametric form. Distractor D contains the row $[0 \ 0 \ 0 \ 0 \ | \ 1]$, which means the system is inconsistent and has no solution.

Question 18

1. $m - n$
2. $n - m$ (correct answer)
3. $0$
4. The dimension cannot be determined from the given information.

Explanation: By the Rank-Nullity Theorem, $\dim(\text{domain}) = \text{rank}(T) + \text{nullity}(T)$. The domain is $\mathbb{R}^n$, so its dimension is $n$. The transformation $T$ is surjective (onto), which means its range is the entire codomain $\mathbb{R}^m$. Therefore, the dimension of the range, $\text{rank}(T)$, is equal to $m$. Substituting these into the theorem gives $n = m + \text{nullity}(T)$. Solving for the nullity (the dimension of the kernel) yields $\text{nullity}(T) = n - m$.

Question 19

1. All solutions approach the origin along spiraling trajectories due to the dominant complex eigenvalues with negative real parts
2. Solutions grow exponentially and oscillate due to the positive real eigenvalue dominating the complex eigenvalues (correct answer)
3. Solutions approach a limit cycle because the complex eigenvalues have zero real parts and create periodic motion
4. The system exhibits saddle behavior with some solutions growing and others decaying depending on initial conditions

Explanation: The real eigenvalue $\lambda_1 = 2 > 0$ dominates the complex eigenvalues $\lambda_{2,3} = -1 \pm 3i$ which have negative real part $-1$. Since $2 > |-1|$, the positive real eigenvalue determines the long-term behavior. Solutions will grow exponentially in the direction of the eigenvector corresponding to $\lambda_1 = 2$, while also exhibiting oscillatory behavior from the complex eigenvalues. Choice A incorrectly focuses on complex eigenvalues. Choice C is wrong because real parts are not zero. Choice D describes saddle points which occur when eigenvalues have mixed signs in purely real cases.

Question 20

1. The columns of $P$ are the vectors $\vec{b}_1$ and $\vec{b}_2$.
2. The columns of $P$ are the vectors $\vec{c}_1$ and $\vec{c}_2$.
3. The columns of $P$ are the coordinate vectors of $\vec{b}_1$ and $\vec{b}_2$ with respect to the basis $\mathcal{C}$. (correct answer)
4. The columns of $P$ are the coordinate vectors of $\vec{c}_1$ and $\vec{c}_2$ with respect to the basis $\mathcal{B}$.

Explanation: By definition, the change-of-coordinates matrix $P_{\mathcal{C} \leftarrow \mathcal{B}}$ transforms $\mathcal{B}$-coordinates into $\mathcal{C}$-coordinates. The matrix is constructed by applying this transformation to the standard basis vectors in the coordinate space, which correspond to the basis vectors of $\mathcal{B}$. Specifically, the $j$-th column of $P_{\mathcal{C} \leftarrow \mathcal{B}}$ is the coordinate vector $[\vec{b}_j]_{\mathcal{C}}$. Therefore, the columns are the coordinate vectors of the vectors from the 'from' basis ($\mathcal{B}$) expressed in the 'to' basis ($\mathcal{C}$).

Question 21

1. $0$ (correct answer)
2. $\frac{1}{2}$
3. $\frac{1}{\sqrt{3}}$
4. $1$

Explanation: First, we compute the vectors $\vec{u}$ and $\vec{v}$. $\vec{u} = \vec{a} + \vec{b} = \langle 1, 1, 0 \rangle + \langle 0, 1, 1 \rangle = \langle 1, 2, 1 \rangle$. $\vec{v} = \vec{a} - \vec{b} = \langle 1, 1, 0 \rangle - \langle 0, 1, 1 \rangle = \langle 1, 0, -1 \rangle$. The cosine of the angle $\theta$ between two vectors is given by the formula $\cos\theta = \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \|\vec{v}\|}$. Next, compute the dot product $\vec{u} \cdot \vec{v}$. $\vec{u} \cdot \vec{v} = (1)(1) + (2)(0) + (1)(-1) = 1 + 0 - 1 = 0$. Since the dot product is 0, the cosine of the angle is 0, which means the vectors are orthogonal. We do not need to calculate the magnitudes. $\|\vec{u}\| = \sqrt{1^2+2^2+1^2} = \sqrt{6}$. $\|\vec{v}\| = \sqrt{1^2+0^2+(-1)^2} = \sqrt{2}$. $\cos\theta = \frac{0}{\sqrt{6}\sqrt{2}} = 0$. Distractor B is the cosine of the angle between the original vectors $\vec{a}$ and $\vec{b}$. Distractor C would result from a sign error in calculating $\vec{v}$, leading to a non-zero dot product.

Question 22

1. The error vector $\mathbf{e}$ is orthogonal to the column space of $A$. (correct answer)
2. The error vector $\mathbf{e}$ is parallel to the projection of $\mathbf{b}$ onto the column space of $A$.
3. The error vector $\mathbf{e}$ is orthogonal to the solution vector $\hat{\mathbf{x}}$.
4. The error vector $\mathbf{e}$ is an element of the column space of $A$.

Explanation: The normal equations $A^T A \hat{\mathbf{x}} = A^T \mathbf{b}$ are derived from the orthogonality condition $A^T(\mathbf{b} - A\hat{\mathbf{x}}) = \mathbf{0}$. This condition means that the error vector $\mathbf{e} = \mathbf{b} - A\hat{\mathbf{x}}$ is orthogonal to every column of $A$. Therefore, the error vector is orthogonal to the entire column space of $A$.

Question 23

1. A horizontal shear that transforms the point $(x, y)$ to $(x+y, y)$. (correct answer)
2. A vertical shear that transforms the point $(x, y)$ to $(x, y-x)$.
3. A reflection across the line $y=x$.
4. The transformation is not invertible.

Explanation: The given matrix $A = \begin{pmatrix} 1 & -1 \ 0 & 1 \end{pmatrix}$ represents a horizontal shear that maps a point $(x, y)$ to $(x-y, y)$. To find the geometric interpretation of the inverse transformation, we first need to find the inverse matrix, $A^{-1}$. For a 2x2 matrix $\begin{pmatrix} a & b \ c & d \end{pmatrix}$, the inverse is $\frac{1}{ad-bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}$. Here, $ad-bc = (1)(1) - (-1)(0) = 1$. So, $A^{-1} = \frac{1}{1} \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}$. This inverse matrix transforms a point $(x, y)$ to $(x+y, y)$. This is a horizontal shear in the opposite direction of the original transformation. Thus, choice A is correct. Choice B describes a vertical shear. Choice C is an incorrect transformation type. Choice D is incorrect because the determinant is 1, so the matrix is invertible.

Question 24

1. Each $\mathbf{v}_i$ must be an eigenvector of $A$ with distinct eigenvalues $2, 4, 8$
2. The basis vectors must form a Jordan chain since $A$ has only one eigenvalue $\lambda = 2$
3. No such basis $B$ exists because $A$ is not diagonalizable over $\mathbb{R}$ (correct answer)
4. The basis vectors must satisfy $A\mathbf{v}_i = 2\mathbf{v}_i$ and be orthogonal to each other