Similarity & Change of Basis - Linear Algebra

1. Let $A$ and $B$ be $n \times n$ matrices such that $B = P^{-1}AP$ for some invertible matrix $P$. Which of the following properties is NOT necessarily shared by both $A$ and $B$?
   Answer: They have the same eigenspaces. Similarity transformations preserve determinant, trace, and eigenvalues. However, eigenspaces are not generally preserved. If $\vec{v}$ is an eigenvector of $A$ corresponding to eigenvalue $\lambda$, then $P^{-1}\vec{v}$ is the corresponding eigenvector of $B$. The eigenspace of $B$ for $\lambda$ is the set of all vectors $P^{-1}\vec{v}$ where $\vec{v}$ is in the eigenspace of $A$ for $\lambda$. This new set of vectors, $P^{-1}(\mathrm{Eig}_A(\lambda))$, is generally a different subspace from $\mathrm{Eig}_A(\lambda)$ unless $P$ is a multiple of the identity matrix.
2. Consider two $2 \times 2$ matrices $A$ and $B$ where $B = P^{-1}AP$ with $P = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$ for some scalar $k$. If $A = \begin{pmatrix} 3 & 2 \\ 0 & 3 \end{pmatrix}$, which statement about $B$ is always true regardless of the value of $k$?
   Answer: $B$ has the same eigenvalues as $A$ with the same algebraic multiplicities When you see matrices related by $B = P^{-1}AP$, you're dealing with similarity transformations — one of the most important concepts in linear algebra. Similar matrices represent the same linear transformation under different coordinate systems. Let's examine what similarity preserves. First, we'll compute $P^{-1} = \begin{pmatrix} 1 & -k \\ 0 & 1 \end{pmatrix}$ and find $B = P^{-1}AP = \begin{pmatrix} 3 & 2-3k \\ 0 & 3 \end{pmatrix}$. Notice that $B$ depends on $k$, but its eigenvalues come from $\det(B - \lambda I) = (3-\lambda)^2$, giving $\lambda = 3$ with algebraic multiplicity 2 — exactly the same as matrix $A$. This confirms (C) is correct: similar matrices always have identical eigenvalues with the same algebraic multiplicities. Here's why the other options fail: (A) is wrong because $B = \begin{pmatrix} 3 & 2-3k \\ 0 & 3 \end{pmatrix} \neq A$ when $k \neq 0$. (B) is incorrect because eigenvectors change under similarity transformations. If $v$ is an eigenvector of $A$, then $P^{-1}v$ is the corresponding eigenvector of $B$. (D) is backwards. Matrix $A$ has only one linearly independent eigenvector $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$, making it non-diagonalizable. Since similarity preserves diagonalizability, $B$ is also non-diagonalizable regardless of $k$. Key takeaway: Similar matrices are like the same person wearing different clothes — the fundamental properties (eigenvalues, determinant, trace, diagonalizability) remain unchanged, but the specific representation varies.
3. A $2 \times 2$ matrix $A$ is similar to $D = \begin{pmatrix} 3 & 0 \\ 0 & -1 \end{pmatrix}$. If $P$ is the matrix such that $A = PDP^{-1}$ and $P = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}$, what is the characteristic polynomial of $A$?
   Answer: $\lambda^2 - 2\lambda - 3$ When you encounter matrix similarity problems, remember that similar matrices share the same eigenvalues and therefore the same characteristic polynomial. This is a powerful property that often provides the quickest path to the solution. Since $A$ is similar to $D = \begin{pmatrix} 3 & 0 \\ 0 & -1 \end{pmatrix}$, matrix $A$ has the same eigenvalues as $D$. The eigenvalues of a diagonal matrix are simply the diagonal entries, so the eigenvalues are $\lambda_1 = 3$ and $\lambda_2 = -1$. For any $2 \times 2$ matrix with eigenvalues $\lambda_1$ and $\lambda_2$, the characteristic polynomial is $(\lambda - \lambda_1)(\lambda - \lambda_2)$. Substituting our eigenvalues: $(\lambda - 3)(\lambda - (-1)) = (\lambda - 3)(\lambda + 1) = \lambda^2 + \lambda - 3\lambda - 3 = \lambda^2 - 2\lambda - 3$. Looking at the wrong answers: Choice A gives $\lambda^2 - 3\lambda + 1$, which would correspond to eigenvalues that don't match our diagonal matrix. Choice B gives $\lambda^2 + 2\lambda - 3$, which has the wrong sign on the middle term. Choice D gives $\lambda^2 + 3\lambda - 1$, which would correspond to completely different eigenvalues. The answer is C: $\lambda^2 - 2\lambda - 3$. Study tip: When you see matrix similarity, immediately think "same eigenvalues, same characteristic polynomial." You don't need to compute $A = PDP^{-1}$ explicitly—the diagonal matrix $D$ gives you the eigenvalues directly, which is all you need for the characteristic polynomial.
4. A matrix $A$ has eigenvalues $\lambda_1 = 2$ and $\lambda_2 = -1$ with corresponding eigenvectors $v_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $v_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$. If $P$ is the matrix whose columns are these eigenvectors, what is the $(2,1)$ entry of $P^{-1}AP$?
   Answer: 0 When P has eigenvectors as columns, P⁻¹AP is the diagonal matrix with eigenvalues on the diagonal: [[2,0],[0,-1]]. The (2,1) entry is 0. This is a fundamental property of diagonalization. Choice B gives the second eigenvalue. Choice C gives the first eigenvalue. Choice D might result from confusing row and column indexing or misunderstanding the diagonalization process.
5. Matrices $A$ and $B$ are similar. If $A$ is invertible, what can be concluded about matrix $B$?
   Answer: $B$ is also invertible, and its inverse $B^{-1}$ is similar to $A^{-1}$. If $A$ and $B$ are similar, then $B = P^{-1}AP$ for some invertible matrix $P$. The determinant is an invariant of similarity, so $\det(B) = \det(P^{-1}AP) = \det(P^{-1})\det(A)\det(P) = \det(A)$. Since $A$ is invertible, $\det(A) \neq 0$. Therefore, $\det(B) \neq 0$, which means $B$ is also invertible. To find the relationship between the inverses, we take the inverse of the similarity equation: $B^{-1} = (P^{-1}AP)^{-1} = P^{-1}A^{-1}(P^{-1})^{-1} = P^{-1}A^{-1}P$. This equation shows, by definition, that $B^{-1}$ is similar to $A^{-1}$.
6. A linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ is defined by $T(\begin{pmatrix} x \\ y \end{pmatrix}) = \begin{pmatrix} x + y \\ -2x + y \end{pmatrix}$. Consider the basis $\mathcal{B} = \{\begin{pmatrix} 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 1 \end{pmatrix}\}$. What is the matrix representation of $T$ with respect to the basis $\mathcal{B}$, denoted $[T]_{\mathcal{B}}$?
   Answer: $\begin{pmatrix} -4 & -9 \\ 3 & 6 \end{pmatrix}$ The standard matrix for $T$ is $A = \begin{pmatrix} 1 & 1 \\ -2 & 1 \end{pmatrix}$. The change of basis matrix from $\mathcal{B}$ to the standard basis $\mathcal{E}$ is $P = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}$. Its inverse is $P^{-1} = \frac{1}{1-2}\begin{pmatrix} 1 & -2 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix}$. The matrix $[T]_{\mathcal{B}}$ is given by the similarity transformation $[T]_{\mathcal{B}} = P^{-1}AP$. First, compute $AP$: $AP = \begin{pmatrix} 1 & 1 \\ -2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ -1 & -3 \end{pmatrix}$. Then, compute $P^{-1}(AP)$: $[T]_{\mathcal{B}} = \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ -1 & -3 \end{pmatrix} = \begin{pmatrix} -2-2 & -3-6 \\ 2+1 & 3+3 \end{pmatrix} = \begin{pmatrix} -4 & -9 \\ 3 & 6 \end{pmatrix}$. Distractor C is the standard matrix $A$. Distractor B is the intermediate matrix $AP$.
7. Let $A$ be a $2 \times 2$ matrix with eigenvalues $\lambda_1 = 3$ and $\lambda_2 = -1$. Let $B$ be a matrix similar to $A$. What is the trace of the matrix $B+2I$, where $I$ is the $2 \times 2$ identity matrix?
   Answer: 6 When you encounter eigenvalues and similar matrices, remember that similarity preserves eigenvalues—this is one of the most important properties in linear algebra. If matrix B is similar to matrix A, then B has exactly the same eigenvalues as A. Since A has eigenvalues $\lambda_1 = 3$ and $\lambda_2 = -1$, matrix B also has eigenvalues 3 and -1. The trace of any matrix equals the sum of its eigenvalues, so $\text{tr}(B) = 3 + (-1) = 2$. Now, when you add $2I$ to matrix B, you're adding 2 to each diagonal element. This shifts each eigenvalue by 2, giving B + 2I eigenvalues of $3 + 2 = 5$ and $-1 + 2 = 1$. Therefore, $\text{tr}(B + 2I) = 5 + 1 = 6$. Looking at the wrong answers: Choice (A) gives 2, which is just the trace of B itself—this ignores the $+2I$ part entirely. Choice (B) gives 4, which you might get if you incorrectly thought adding $2I$ means adding 2 to the trace rather than adding 2 to each eigenvalue. Choice (C) suggests the answer can't be determined, but similarity relationships give us enough information since they preserve eigenvalues. Key strategy: Remember that similar matrices share eigenvalues, and adding a scalar multiple of the identity matrix shifts all eigenvalues by that scalar. The trace always equals the sum of eigenvalues, making these calculations straightforward once you recognize the pattern.
8. Let $\mathcal{B} = \{\begin{pmatrix} 1 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \end{pmatrix}\}$ and $\mathcal{C} = \{\begin{pmatrix} 1 \\ 2 \end{pmatrix}, \begin{pmatrix} 2 \\ 3 \end{pmatrix}\}$ be two bases for $\mathbb{R}^2$. If the coordinate vector of $\vec{x}$ relative to basis $\mathcal{B}$ is $[\vec{x}]_{\mathcal{B}} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$, what is $[\vec{x}]_{\mathcal{C}}$?
   Answer: $\begin{pmatrix} -16 \\ 10 \end{pmatrix}$ First, find the vector $\vec{x}$ in the standard basis $\mathcal{E}$. $\vec{x} = P_{\mathcal{B}}[\vec{x}]_{\mathcal{B}} = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 3+1 \\ -3+1 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$. Next, find the coordinates of $\vec{x}$ relative to the basis $\mathcal{C}$. We need to solve $P_{\mathcal{C}} [\vec{x}]_{\mathcal{C}} = \vec{x}$, which means $[\vec{x}]_{\mathcal{C}} = P_{\mathcal{C}}^{-1} \vec{x}$. The matrix for basis $\mathcal{C}$ is $P_{\mathcal{C}} = \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}$. Its inverse is $P_{\mathcal{C}}^{-1} = \frac{1}{3-4} \begin{pmatrix} 3 & -2 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} -3 & 2 \\ 2 & -1 \end{pmatrix}$. Finally, calculate $[\vec{x}]_{\mathcal{C}} = \begin{pmatrix} -3 & 2 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} -12-4 \\ 8+2 \end{pmatrix} = \begin{pmatrix} -16 \\ 10 \end{pmatrix}$. Distractor B is the vector $\vec{x}$ in standard coordinates.
9. Let matrix $A = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix}$. An eigenvector of $A$ is $\vec{v} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$. Let $P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $B = P^{-1}AP$. Which of the following is an eigenvector of matrix $B$?
   Answer: $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ First, find the eigenvalue corresponding to $\vec{v}$ for matrix $A$: $A\vec{v} = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \end{pmatrix} = 3\begin{pmatrix} 2 \\ 1 \end{pmatrix}$. The eigenvalue is $\lambda = 3$. If $\vec{v}$ is an eigenvector of $A$ for eigenvalue $\lambda$, then the corresponding eigenvector of $B = P^{-1}AP$ is $\vec{w} = P^{-1}\vec{v}$. First, we find the inverse of $P$: $P^{-1} = \frac{1}{1(1)-1(0)} \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$. Now, we compute $\vec{w}$: $\vec{w} = P^{-1}\vec{v} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2-1 \\ 0+1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$. Distractor C is the result of mistakenly calculating $P\vec{v}$.
10. 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 $P$ be the change-of-coordinates matrix from $\mathcal{B}$ to $\mathcal{C}$, denoted $P_{\mathcal{C} \leftarrow \mathcal{B}}$. Which of the following correctly describes the columns of $P$?
    Answer: The columns of $P$ are the coordinate vectors of $\vec{b}_1$ and $\vec{b}_2$ with respect to the basis $\mathcal{C}$. 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 ($\mathcal{C}$).
11. Consider the change of basis from the standard basis $\{e_1, e_2\}$ to the basis $\{u_1, u_2\}$ where $u_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ and $u_2 = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$. If a linear transformation $T$ has matrix $\begin{pmatrix} 4 & -2 \\ 1 & 3 \end{pmatrix}$ with respect to the standard basis, what is the sum of the diagonal entries of the matrix representing $T$ with respect to the basis $\{u_1, u_2\}$?
    Answer: 7 The trace is invariant under similarity transformations. If A represents T in the standard basis and B represents T in the new basis, then B = P⁻¹AP where P = [[2,1],[1,3]]. Since tr(B) = tr(A), and tr(A) = 4 + 3 = 7, the sum of diagonal entries in the new basis is 7. Choice B subtracts instead of adding the diagonal entries. Choice C adds an arithmetic error. Choice D might come from incorrectly computing with the change of basis matrix entries.
12. Which of the following statements is always true for any two $n \times n$ similar matrices $A$ and $B$?
    Answer: $A^k$ and $B^k$ are similar for any positive integer $k$. When you encounter questions about similar matrices, focus on properties that are preserved under similarity transformations. Two matrices $A$ and $B$ are similar if there exists an invertible matrix $P$ such that $B = P^{-1}AP$. The correct answer is D because similarity is preserved under matrix powers. If $B = P^{-1}AP$, then $B^k = (P^{-1}AP)^k = P^{-1}A^kP$, making $A^k$ and $B^k$ similar. This follows from the fact that $(P^{-1}AP)^k = P^{-1}A^kP$, which you can verify by repeatedly applying the associative property of matrix multiplication. Let's examine why the other options fail. Option A is incorrect because similar matrices don't necessarily share eigenvectors. While they have the same eigenvalues, their eigenvectors are generally different. For example, if $v$ is an eigenvector of $A$, then $P^{-1}v$ (not $v$) is the corresponding eigenvector of $B$. Option B is wrong because similar matrices don't preserve null spaces. The null space of $B$ is $P^{-1}(\text{null space of } A)$, not the null space of $A$ itself. The dimensions are equal, but the actual subspaces differ. Option C fails because adding different matrices ($I$ to one, $-I$ to another) breaks the similarity relationship. If $B = P^{-1}AP$, then $B - I = P^{-1}AP - I \neq P^{-1}(A + I)P$. Remember: similar matrices preserve eigenvalues, determinants, traces, and ranks, but not eigenvectors or specific subspaces. Operations that "commute" with similarity transformations (like taking powers) preserve similarity.
13. Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be the linear transformation that reflects vectors across the line $y=2x$. What is the matrix of $T$ with respect to the basis $\mathcal{B} = \{\begin{pmatrix} 1 \\ 2 \end{pmatrix}, \begin{pmatrix} -2 \\ 1 \end{pmatrix}\}$?
    Answer: $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ The basis $\mathcal{B}$ is chosen to be geometrically significant. The first basis vector, $\vec{b}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$, lies on the line of reflection $y=2x$. Any vector on the line of reflection is unchanged by the reflection, so $T(\vec{b}_1) = \vec{b}_1$. In $\mathcal{B}$-coordinates, this is $[T(\vec{b}_1)]_{\mathcal{B}} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$. The second basis vector, $\vec{b}_2 = \begin{pmatrix} -2 \\ 1 \end{pmatrix}$, is orthogonal to the line of reflection (its dot product with $\vec{b}_1$ is $1(-2) + 2(1) = 0$). A vector orthogonal to the line of reflection is mapped to its negative, so $T(\vec{b}_2) = -\vec{b}_2$. In $\mathcal{B}$-coordinates, this is $[T(\vec{b}_2)]_{\mathcal{B}} = \begin{pmatrix} 0 \\ -1 \end{pmatrix}$. The matrix of the transformation with respect to this basis, $[T]_{\mathcal{B}}$, has these coordinate vectors as its columns: $[T]_{\mathcal{B}} = \begin{pmatrix} [T(\vec{b}_1)]_{\mathcal{B}} & [T(\vec{b}_2)]_{\mathcal{B}} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$. Choice C is the standard matrix for this transformation.
14. The fundamental reason that similar matrices share the same eigenvalues is that:
    Answer: they represent the same linear transformation, and eigenvalues are geometric properties of the transformation itself, independent of the coordinate system. Similarity of matrices is fundamentally about representing the same linear transformation with respect to different bases (coordinate systems). An eigenvalue of a linear transformation is a scalar $\lambda$ such that for some nonzero vector $\vec{v}$, $T(\vec{v}) = \lambda \vec{v}$. This property—that the transformation $T$ acts on $\vec{v}$ by simple scaling—is an intrinsic, geometric feature of the transformation. It does not depend on how we choose to write down coordinates for vectors or the matrix for the transformation. Since similar matrices represent the same transformation, they must capture the same intrinsic scaling factors, which are the eigenvalues. The other options are either incorrect statements (C is false, $P$ is not always orthogonal) or describe properties that are consequences, not the fundamental reason (A, D).
15. The matrix $A = \begin{pmatrix} 7 & -6 \\ 3 & -2 \end{pmatrix}$ is similar to a diagonal matrix $D = \begin{pmatrix} 4 & 0 \\ 0 & 1 \end{pmatrix}$. Which of the following could be an invertible matrix $P$ such that $A = PDP^{-1}$?
    Answer: $\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ If $A = PDP^{-1}$, the columns of $P$ must be the eigenvectors of $A$, and the diagonal entries of $D$ must be the corresponding eigenvalues. The eigenvalues are given as 4 and 1. We need to find the eigenvectors of $A$. For $\lambda = 4$: $(A-4I)\vec{v} = \begin{pmatrix} 3 & -6 \\ 3 & -6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$. This gives $3x - 6y = 0$, or $x=2y$. An eigenvector is $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$. For $\lambda = 1$: $(A-1I)\vec{v} = \begin{pmatrix} 6 & -6 \\ 3 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$. This gives $x - y = 0$, or $x=y$. An eigenvector is $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$. The matrix $P$ is formed by these eigenvectors. The order must correspond to the order of eigenvalues in $D$. Since $D_{11}=4$ and $D_{22}=1$, the first column of $P$ must be an eigenvector for $\lambda=4$, and the second column must be an eigenvector for $\lambda=1$. Thus, $P = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$. Choice C reverses the order of the eigenvectors, which would correspond to $D = \begin{pmatrix} 1 & 0 \\ 0 & 4 \end{pmatrix}$.
16. Let $A$ and $B$ be $3 \times 3$ matrices where $B = Q^{-1}AQ$ for some invertible matrix $Q$. If $A$ has eigenvalues $2, 3, 5$ and $\det(A - 4I) = -6$, what is $\det(B - 4I)$?
    Answer: -6 When you encounter matrices related by similarity ($B = Q^{-1}AQ$), remember that similar matrices share many fundamental properties, including determinants of shifted matrices. Since $B = Q^{-1}AQ$, we can find $\det(B - 4I)$ by substituting: $B - 4I = Q^{-1}AQ - 4I = Q^{-1}AQ - 4Q^{-1}IQ = Q^{-1}(A - 4I)Q$. Using the determinant property $\det(XYZ) = \det(X)\det(Y)\det(Z)$, we get: $\det(B - 4I) = \det(Q^{-1})\det(A - 4I)\det(Q)$. Since $\det(Q^{-1}) = \frac{1}{\det(Q)}$, this simplifies to $\det(B - 4I) = \det(A - 4I) = -6$. The wrong answers represent common misconceptions. Choice (A) -24 likely comes from incorrectly multiplying $\det(A - 4I) = -6$ by 4, perhaps thinking the shift affects the determinant multiplicatively. Choice (B) 6 makes the sign error of taking the absolute value of the given determinant. Choice (C) 24 combines both errors: taking the absolute value and then multiplying by 4. The key insight is that similarity transformations preserve the determinant of any shifted matrix. You don't need to calculate $\det(A - 4I)$ from the eigenvalues since it's already given. When studying similar matrices, focus on which properties are invariant (eigenvalues, determinant, trace) versus which can change (eigenvectors, individual entries). This will help you recognize when complex calculations aren't necessary.
17. Let $A$ be a $3 \times 3$ matrix representing a linear transformation $T$ with respect to the standard basis. Let $\mathcal{B} = \{\vec{b}_1, \vec{b}_2, \vec{b}_3\}$ be another basis for $\mathbb{R}^3$. The matrix of the same transformation $T$ with respect to basis $\mathcal{B}$ is $B = \begin{pmatrix} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 2 \end{pmatrix}$. Which conclusion is guaranteed to be true?
    Answer: The determinant of $A$ is $-4$. The matrices $A$ and $B$ represent the same linear transformation with respect to different bases, which means they are similar. Similar matrices have the same determinant. The determinant of $B$ is the product of its diagonal entries: $\det(B) = (2)(-1)(2) = -4$. Therefore, $\det(A)$ must also be $-4$. (A) is incorrect; $A$ is similar to a diagonal matrix, meaning it is diagonalizable, but it is not necessarily diagonal itself. (B) is incorrect; the basis vectors are eigenvectors of $T$, but eigenvectors corresponding to different eigenvalues are only guaranteed to be linearly independent, not necessarily orthogonal. (D) is incorrect; the eigenvalues of $B$ (and thus $A$) are $2, -1, 2$. There are only two distinct eigenvalues, not three.
18. A linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ has matrix $A$ in the standard basis with $\text{tr}(A) = 9$ and $\det(A) = 24$. After a change of basis, $T$ has matrix $B$ where $B$ is upper triangular with diagonal entries $2, 3, 4$. What is $\text{tr}(A^2)$?
    Answer: 29 This question tests your understanding of how matrix properties behave under change of basis, particularly the relationship between eigenvalues and matrix traces. When a matrix undergoes a change of basis, it becomes similar to its original form, meaning $B = P^{-1}AP$ for some invertible matrix $P$. Crucially, similar matrices share the same eigenvalues. Since $B$ is upper triangular with diagonal entries 2, 3, 4, these are the eigenvalues of both $B$ and $A$. You can verify this makes sense with the given information: the trace equals the sum of eigenvalues ($2 + 3 + 4 = 9$ ✓), and the determinant equals their product ($2 \times 3 \times 4 = 24$ ✓). For $A^2$, the eigenvalues are the squares of $A$'s eigenvalues: $2^2 = 4$, $3^2 = 9$, and $4^2 = 16$. Therefore, $\text{tr}(A^2) = 4 + 9 + 16 = 29$, which is answer (B). The wrong answers represent common calculation errors: (A) 45 comes from incorrectly computing $9 + 24 + 12$ (perhaps mixing trace, determinant, and products). (C) 81 is simply $9^2$, the square of the trace—but $\text{tr}(A^2) \neq [\text{tr}(A)]^2$ in general. (D) 57 might result from adding the eigenvalues incorrectly or miscomputing their squares. Remember: when you see eigenvalue problems involving traces, use the fact that eigenvalues determine both the trace (sum) and determinant (product), and that $\text{tr}(A^k)$ equals the sum of the $k$th powers of the eigenvalues.
19. Let $A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$ and suppose $B = P^{-1}AP$ for some invertible matrix $P$. If the first column of $P$ is $\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$, what is the first column of $B$?
    Answer: $P^{-1}A\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$ When you encounter similarity transformations like $B = P^{-1}AP$, focus on what each column of $B$ represents. The key insight is that the $k$-th column of any matrix product can be found by multiplying the left matrix by the $k$-th column of the right matrix. To find the first column of $B = P^{-1}AP$, you need the first column of $AP$. Since $AP$ means "$A$ times $P$," the first column of $AP$ is $A$ times the first column of $P$. Given that the first column of $P$ is $\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$, the first column of $AP$ is $A\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$. Therefore, the first column of $B = P^{-1}AP$ is $P^{-1}$ times this result, which gives us $P^{-1}A\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$. This matches answer choice B. Answer choice A assumes $B$ is in some standard form, which isn't necessarily true. Answer choice C gives you the first column of $AP$, not $B = P^{-1}AP$ — you're missing the crucial $P^{-1}$ multiplication. Answer choice D appears to be the result of computing $A\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$, but again, this ignores the $P^{-1}$ factor. Remember: for matrix products, think column by column. The $k$-th column of $XY$ equals $X$ times the $k$-th column of $Y$. This systematic approach prevents you from getting lost in complex similarity transformations.

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