Linear Algebra Quiz: Similarity And Change Of Basis
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Similarity And Change Of BasisQuestion 1 of 19

Let B={(1−1),(11)}\mathcal{B} = \{\begin{pmatrix} 1 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \end{pmatrix}\} and C={(12),(23)}\mathcal{C} = \{\begin{pmatrix} 1 \\ 2 \end{pmatrix}, \begin{pmatrix} 2 \\ 3 \end{pmatrix}\} be two bases for R2\mathbb{R}^2. If the coordinate vector of x⃗\vec{x} relative to basis B\mathcal{B} is [x⃗]B=(31)[\vec{x}]_{\mathcal{B}} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}, what is [x⃗]C[\vec{x}]_{\mathcal{C}}?

(−27)\begin{pmatrix} -2 \\ 7 \end{pmatrix}
(4−2)\begin{pmatrix} 4 \\ -2 \end{pmatrix}
(−1610)\begin{pmatrix} -16 \\ 10 \end{pmatrix}
(−212)\begin{pmatrix} -2 \\ 12 \end{pmatrix}
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Linear Algebra Quiz

Linear Algebra Quiz: Similarity And Change Of Basis

Practice Similarity And Change Of Basis in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Similarity And Change Of Basis, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Let B={(1−1),(11)}\mathcal{B} = \{\begin{pmatrix} 1 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \end{pmatrix}\} and C={(12),(23)}\mathcal{C} = \{\begin{pmatrix} 1 \\ 2 \end{pmatrix}, \begin{pmatrix} 2 \\ 3 \end{pmatrix}\} be two bases for R2\mathbb{R}^2. If the coordinate vector of x⃗\vec{x} relative to basis B\mathcal{B} is [x⃗]B=(31)[\vec{x}]_{\mathcal{B}} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}, what is [x⃗]C[\vec{x}]_{\mathcal{C}}?

  1. (−27)\begin{pmatrix} -2 \\ 7 \end{pmatrix}
  2. (4−2)\begin{pmatrix} 4 \\ -2 \end{pmatrix}
  3. (−1610)\begin{pmatrix} -16 \\ 10 \end{pmatrix} (correct answer)
  4. (−212)\begin{pmatrix} -2 \\ 12 \end{pmatrix}
Explanation: First, find the vector x⃗\vec{x} in the standard basis E\mathcal{E}. x⃗=PB[x⃗]B=(11−11)(31)=(3+1−3+1)=(4−2)\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 x⃗\vec{x} relative to the basis C\mathcal{C}. We need to solve PC[x⃗]C=x⃗P_{\mathcal{C}} [\vec{x}]_{\mathcal{C}} = \vec{x}, which means [x⃗]C=PC−1x⃗[\vec{x}]_{\mathcal{C}} = P_{\mathcal{C}}^{-1} \vec{x}. The matrix for basis C\mathcal{C} is PC=(1223)P_{\mathcal{C}} = \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}. Its inverse is PC−1=13−4(3−2−21)=(−322−1)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 [x⃗]C=(−322−1)(4−2)=(−12−48+2)=(−1610)[\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 x⃗\vec{x} in standard coordinates.

Question 2

Let AA and BB be n×nn \times n matrices such that B=P−1APB = P^{-1}AP for some invertible matrix PP. Which of the following properties is NOT necessarily shared by both AA and BB?

  1. They have the same determinant.
  2. They have the same trace.
  3. They have the same eigenvalues.
  4. They have the same eigenspaces. (correct answer)
Explanation: Similarity transformations preserve determinant, trace, and eigenvalues. However, eigenspaces are not generally preserved. If v⃗\vec{v} is an eigenvector of AA corresponding to eigenvalue λ\lambda, then P−1v⃗P^{-1}\vec{v} is the corresponding eigenvector of BB. The eigenspace of BB for λ\lambda is the set of all vectors P−1v⃗P^{-1}\vec{v} where v⃗\vec{v} is in the eigenspace of AA for λ\lambda. This new set of vectors, P−1(EigA(λ))P^{-1}(\mathrm{Eig}_A(\lambda)), is generally a different subspace from EigA(λ)\mathrm{Eig}_A(\lambda) unless PP is a multiple of the identity matrix.

Question 3

Let AA be a 3×33 \times 3 matrix representing a linear transformation TT with respect to the standard basis. Let B={b⃗1,b⃗2,b⃗3}\mathcal{B} = \{\vec{b}_1, \vec{b}_2, \vec{b}_3\} be another basis for R3\mathbb{R}^3. The matrix of the same transformation TT with respect to basis B\mathcal{B} is B=(2000−10002)B = \begin{pmatrix} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 2 \end{pmatrix}. Which conclusion is guaranteed to be true?

  1. AA must be a diagonal matrix.
  2. The basis vectors b⃗1,b⃗2,b⃗3\vec{b}_1, \vec{b}_2, \vec{b}_3 are orthogonal.
  3. The determinant of AA is −4-4. (correct answer)
  4. The matrix AA has three distinct eigenvalues.
Explanation: The matrices AA and BB represent the same linear transformation with respect to different bases, which means they are similar. Similar matrices have the same determinant. The determinant of BB is the product of its diagonal entries: det⁡(B)=(2)(−1)(2)=−4\det(B) = (2)(-1)(2) = -4. Therefore, det⁡(A)\det(A) must also be −4-4. (A) is incorrect; AA 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 TT, but eigenvectors corresponding to different eigenvalues are only guaranteed to be linearly independent, not necessarily orthogonal. (D) is incorrect; the eigenvalues of BB (and thus AA) are 2,−1,22, -1, 2. There are only two distinct eigenvalues, not three.

Question 4

The matrix A=(7−63−2)A = \begin{pmatrix} 7 & -6 \\ 3 & -2 \end{pmatrix} is similar to a diagonal matrix D=(4001)D = \begin{pmatrix} 4 & 0 \\ 0 & 1 \end{pmatrix}. Which of the following could be an invertible matrix PP such that A=PDP−1A = PDP^{-1}?

  1. (4001)\begin{pmatrix} 4 & 0 \\ 0 & 1 \end{pmatrix}
  2. (2111)\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} (correct answer)
  3. (1211)\begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}
  4. (73−6−2)\begin{pmatrix} 7 & 3 \\ -6 & -2 \end{pmatrix}
Explanation: If A=PDP−1A = PDP^{-1}, the columns of PP must be the eigenvectors of AA, and the diagonal entries of DD must be the corresponding eigenvalues. The eigenvalues are given as 4 and 1. We need to find the eigenvectors of AA. For λ=4\lambda = 4: (A−4I)v⃗=(3−63−6)(xy)=(00)(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=03x - 6y = 0, or x=2yx=2y. An eigenvector is (21)\begin{pmatrix} 2 \\ 1 \end{pmatrix}. For λ=1\lambda = 1: (A−1I)v⃗=(6−63−3)(xy)=(00)(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=0x - y = 0, or x=yx=y. An eigenvector is (11)\begin{pmatrix} 1 \\ 1 \end{pmatrix}. The matrix PP is formed by these eigenvectors. The order must correspond to the order of eigenvalues in DD. Since D11=4D_{11}=4 and D22=1D_{22}=1, the first column of PP must be an eigenvector for λ=4\lambda=4, and the second column must be an eigenvector for λ=1\lambda=1. Thus, P=(2111)P = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}. Choice C reverses the order of the eigenvectors, which would correspond to D=(1004)D = \begin{pmatrix} 1 & 0 \\ 0 & 4 \end{pmatrix}.

Question 5

Let T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 be the linear transformation that reflects vectors across the line y=2xy=2x. What is the matrix of TT with respect to the basis B={(12),(−21)}\mathcal{B} = \{\begin{pmatrix} 1 \\ 2 \end{pmatrix}, \begin{pmatrix} -2 \\ 1 \end{pmatrix}\}?

  1. (100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} (correct answer)
  2. (−1001)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}
  4. (1−221)\begin{pmatrix} 1 & -2 \\ 2 & 1 \end{pmatrix}
Explanation: The basis B\mathcal{B} is chosen to be geometrically significant. The first basis vector, b⃗1=(12)\vec{b}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}, lies on the line of reflection y=2xy=2x. Any vector on the line of reflection is unchanged by the reflection, so T(b⃗1)=b⃗1T(\vec{b}_1) = \vec{b}_1. In B\mathcal{B}-coordinates, this is [T(b⃗1)]B=(10)[T(\vec{b}_1)]_{\mathcal{B}} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}. The second basis vector, b⃗2=(−21)\vec{b}_2 = \begin{pmatrix} -2 \\ 1 \end{pmatrix}, is orthogonal to the line of reflection (its dot product with b⃗1\vec{b}_1 is 1(−2)+2(1)=01(-2) + 2(1) = 0). A vector orthogonal to the line of reflection is mapped to its negative, so T(b⃗2)=−b⃗2T(\vec{b}_2) = -\vec{b}_2. In B\mathcal{B}-coordinates, this is [T(b⃗2)]B=(0−1)[T(\vec{b}_2)]_{\mathcal{B}} = \begin{pmatrix} 0 \\ -1 \end{pmatrix}. The matrix of the transformation with respect to this basis, [T]B[T]_{\mathcal{B}}, has these coordinate vectors as its columns: [T]B=([T(b⃗1)]B[T(b⃗2)]B)=(100−1)[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.

Question 6

Let AA be a 2×22 \times 2 matrix with eigenvalues λ1=3\lambda_1 = 3 and λ2=−1\lambda_2 = -1. Let BB be a matrix similar to AA. What is the trace of the matrix B+2IB+2I, where II is the 2×22 \times 2 identity matrix?

  1. 2
  2. 4
  3. Cannot be determined from the given information.
  4. 6 (correct answer)
Explanation: 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 λ1=3\lambda_1 = 3 and λ2=−1\lambda_2 = -1, matrix B also has eigenvalues 3 and -1. The trace of any matrix equals the sum of its eigenvalues, so tr(B)=3+(−1)=2\text{tr}(B) = 3 + (-1) = 2. Now, when you add 2I2I to matrix B, you're adding 2 to each diagonal element. This shifts each eigenvalue by 2, giving B + 2I eigenvalues of 3+2=53 + 2 = 5 and −1+2=1-1 + 2 = 1. Therefore, tr(B+2I)=5+1=6\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+2I part entirely. Choice (B) gives 4, which you might get if you incorrectly thought adding 2I2I 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.

Question 7

Let B={b⃗1,b⃗2}\mathcal{B} = \{\vec{b}_1, \vec{b}_2\} and C={c⃗1,c⃗2}\mathcal{C} = \{\vec{c}_1, \vec{c}_2\} be two bases for a vector space VV. Let PP be the change-of-coordinates matrix from B\mathcal{B} to C\mathcal{C}, denoted PC←BP_{\mathcal{C} \leftarrow \mathcal{B}}. Which of the following correctly describes the columns of PP?

  1. The columns of PP are the vectors b⃗1\vec{b}_1 and b⃗2\vec{b}_2.
  2. The columns of PP are the vectors c⃗1\vec{c}_1 and c⃗2\vec{c}_2.
  3. The columns of PP are the coordinate vectors of b⃗1\vec{b}_1 and b⃗2\vec{b}_2 with respect to the basis C\mathcal{C}. (correct answer)
  4. The columns of PP are the coordinate vectors of c⃗1\vec{c}_1 and c⃗2\vec{c}_2 with respect to the basis B\mathcal{B}.
Explanation: By definition, the change-of-coordinates matrix PC←BP_{\mathcal{C} \leftarrow \mathcal{B}} transforms B\mathcal{B}-coordinates into C\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 B\mathcal{B}. Specifically, the jj-th column of PC←BP_{\mathcal{C} \leftarrow \mathcal{B}} is the coordinate vector [b⃗j]C[\vec{b}_j]_{\mathcal{C}}. Therefore, the columns are the coordinate vectors of the vectors from the 'from' basis (B\mathcal{B}) expressed in the 'to' basis (C\mathcal{C}).

Question 8

Let matrix A=(4−211)A = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix}. An eigenvector of AA is v⃗=(21)\vec{v} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}. Let P=(1101)P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} and B=P−1APB = P^{-1}AP. Which of the following is an eigenvector of matrix BB?

  1. (21)\begin{pmatrix} 2 \\ 1 \end{pmatrix}
  2. (11)\begin{pmatrix} 1 \\ 1 \end{pmatrix} (correct answer)
  3. (31)\begin{pmatrix} 3 \\ 1 \end{pmatrix}
  4. (1−1)\begin{pmatrix} 1 \\ -1 \end{pmatrix}
Explanation: First, find the eigenvalue corresponding to v⃗\vec{v} for matrix AA: Av⃗=(4−211)(21)=(63)=3(21)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 λ=3\lambda = 3. If v⃗\vec{v} is an eigenvector of AA for eigenvalue λ\lambda, then the corresponding eigenvector of B=P−1APB = P^{-1}AP is w⃗=P−1v⃗\vec{w} = P^{-1}\vec{v}. First, we find the inverse of PP: P−1=11(1)−1(0)(1−101)=(1−101)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 w⃗\vec{w}: w⃗=P−1v⃗=(1−101)(21)=(2−10+1)=(11)\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 Pv⃗P\vec{v}.

Question 9

A linear transformation T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 is defined by T((xy))=(x+y−2x+y)T(\begin{pmatrix} x \\ y \end{pmatrix}) = \begin{pmatrix} x + y \\ -2x + y \end{pmatrix}. Consider the basis B={(11),(21)}\mathcal{B} = \{\begin{pmatrix} 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 1 \end{pmatrix}\}. What is the matrix representation of TT with respect to the basis B\mathcal{B}, denoted [T]B[T]_{\mathcal{B}}?

  1. (−4−936)\begin{pmatrix} -4 & -9 \\ 3 & 6 \end{pmatrix} (correct answer)
  2. (23−1−3)\begin{pmatrix} 2 & 3 \\ -1 & -3 \end{pmatrix}
  3. (11−21)\begin{pmatrix} 1 & 1 \\ -2 & 1 \end{pmatrix}
  4. (−1221−1017)\begin{pmatrix} -12 & 21 \\ -10 & 17 \end{pmatrix}
Explanation: The standard matrix for TT is A=(11−21)A = \begin{pmatrix} 1 & 1 \\ -2 & 1 \end{pmatrix}. The change of basis matrix from B\mathcal{B} to the standard basis E\mathcal{E} is P=(1211)P = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}. Its inverse is P−1=11−2(1−2−11)=(−121−1)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]B[T]_{\mathcal{B}} is given by the similarity transformation [T]B=P−1AP[T]_{\mathcal{B}} = P^{-1}AP. First, compute APAP: AP=(11−21)(1211)=(23−1−3)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)P^{-1}(AP): [T]B=(−121−1)(23−1−3)=(−2−2−3−62+13+3)=(−4−936)[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 AA. Distractor B is the intermediate matrix APAP.

Question 10

A matrix AA has eigenvalues λ1=2\lambda_1 = 2 and λ2=−1\lambda_2 = -1 with corresponding eigenvectors v1=(11)v_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} and v2=(1−2)v_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix} . If PP is the matrix whose columns are these eigenvectors, what is the (2,1)(2,1) entry of P−1APP^{-1}AP?

  1. 0 (correct answer)
  2. -1
  3. 2
  4. 1
Explanation: 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.

Question 11

Consider the change of basis from the standard basis {e1,e2}\{e_1, e_2\} to the basis {u1,u2}\{u_1, u_2\} where u1=(21)u_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix} and u2=(13)u_2 = \begin{pmatrix} 1 \\ 3 \end{pmatrix} . If a linear transformation TT has matrix (4−213)\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 TT with respect to the basis {u1,u2}\{u_1, u_2\}?

  1. 7 (correct answer)
  2. 5
  3. 9
  4. 12
Explanation: 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.

Question 12

Consider two 2×22 \times 2 matrices AA and BB where B=P−1APB = P^{-1}AP with P=(1k01)P = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} for some scalar kk. If A=(3203)A = \begin{pmatrix} 3 & 2 \\ 0 & 3 \end{pmatrix} , which statement about BB is always true regardless of the value of kk?

  1. B=AB = A for all values of kk
  2. BB has the same eigenvectors as AA up to scalar multiples
  3. BB has the same eigenvalues as AA with the same algebraic multiplicities (correct answer)
  4. BB is diagonalizable if and only if k≠0k \neq 0
Explanation: When you see matrices related by B=P−1APB = 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=(1−k01)P^{-1} = \begin{pmatrix} 1 & -k \\ 0 & 1 \end{pmatrix} and find B=P−1AP=(32−3k03)B = P^{-1}AP = \begin{pmatrix} 3 & 2-3k \\ 0 & 3 \end{pmatrix} . Notice that BB depends on kk, but its eigenvalues come from det⁡(B−λI)=(3−λ)2\det(B - \lambda I) = (3-\lambda)^2, giving λ=3\lambda = 3 with algebraic multiplicity 2 — exactly the same as matrix AA. 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=(32−3k03)≠AB = \begin{pmatrix} 3 & 2-3k \\ 0 & 3 \end{pmatrix} \neq A when k≠0k \neq 0. (B) is incorrect because eigenvectors change under similarity transformations. If vv is an eigenvector of AA, then P−1vP^{-1}v is the corresponding eigenvector of BB. (D) is backwards. Matrix AA has only one linearly independent eigenvector (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix}, making it non-diagonalizable. Since similarity preserves diagonalizability, BB is also non-diagonalizable regardless of kk. 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.

Question 13

Let A=(110011001)A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} and suppose B=P−1APB = P^{-1}AP for some invertible matrix PP. If the first column of PP is (121)\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}, what is the first column of BB?

  1. (100)\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}
  2. (correct answer)
  4. (331)\begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix}
Explanation: When you encounter similarity transformations like B=P−1APB = P^{-1}AP, focus on what each column of BB represents. The key insight is that the kk-th column of any matrix product can be found by multiplying the left matrix by the kk-th column of the right matrix. To find the first column of B=P−1APB = P^{-1}AP, you need the first column of APAP. Since APAP means "AA times PP," the first column of APAP is AA times the first column of PP. Given that the first column of PP is (121)\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}, the first column of APAP 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.

Question 14

A linear transformation T:R3→R3T: \mathbb{R}^3 \to \mathbb{R}^3 has matrix AA in the standard basis with tr(A)=9\text{tr}(A) = 9 and det⁡(A)=24\det(A) = 24. After a change of basis, TT has matrix BB where BB is upper triangular with diagonal entries 2,3,42, 3, 4. What is tr(A2)\text{tr}(A^2)?

  1. 45
  2. 29 (correct answer)
  3. 81
  4. 57
Explanation: 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−1APB = P^{-1}AP for some invertible matrix PP. Crucially, similar matrices share the same eigenvalues. Since BB is upper triangular with diagonal entries 2, 3, 4, these are the eigenvalues of both BB and AA. You can verify this makes sense with the given information: the trace equals the sum of eigenvalues (2+3+4=92 + 3 + 4 = 9 ✓), and the determinant equals their product (2×3×4=242 \times 3 \times 4 = 24 ✓). For A2A^2, the eigenvalues are the squares of AA's eigenvalues: 22=42^2 = 4, 32=93^2 = 9, and 42=164^2 = 16. Therefore, tr(A2)=4+9+16=29\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+129 + 24 + 12 (perhaps mixing trace, determinant, and products). (C) 81 is simply 929^2, the square of the trace—but tr(A2)≠[tr(A)]2\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 tr(Ak)\text{tr}(A^k) equals the sum of the kkth powers of the eigenvalues.

Question 15

A 2×22 \times 2 matrix AA is similar to D=(300−1)D = \begin{pmatrix} 3 & 0 \\ 0 & -1 \end{pmatrix} . If PP is the matrix such that A=PDP−1A = PDP^{-1} and P=(1211)P = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix} , what is the characteristic polynomial of AA?

  1. λ2−3λ+1\lambda^2 - 3\lambda + 1
  2. λ2+2λ−3\lambda^2 + 2\lambda - 3
  3. λ2−2λ−3\lambda^2 - 2\lambda - 3 (correct answer)
  4. λ2+3λ−1\lambda^2 + 3\lambda - 1
Explanation: 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 AA is similar to D=(300−1)D = \begin{pmatrix} 3 & 0 \\ 0 & -1 \end{pmatrix} , matrix AA has the same eigenvalues as DD. The eigenvalues of a diagonal matrix are simply the diagonal entries, so the eigenvalues are λ1=3\lambda_1 = 3 and λ2=−1\lambda_2 = -1. For any 2×22 \times 2 matrix with eigenvalues λ1\lambda_1 and λ2\lambda_2, the characteristic polynomial is (λ−λ1)(λ−λ2)(\lambda - \lambda_1)(\lambda - \lambda_2). Substituting our eigenvalues: (λ−3)(λ−(−1))=(λ−3)(λ+1)=λ2+λ−3λ−3=λ2−2λ−3(\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 λ2−3λ+1\lambda^2 - 3\lambda + 1, which would correspond to eigenvalues that don't match our diagonal matrix. Choice B gives λ2+2λ−3\lambda^2 + 2\lambda - 3, which has the wrong sign on the middle term. Choice D gives λ2+3λ−1\lambda^2 + 3\lambda - 1, which would correspond to completely different eigenvalues. The answer is C: λ2−2λ−3\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−1A = PDP^{-1} explicitly—the diagonal matrix DD gives you the eigenvalues directly, which is all you need for the characteristic polynomial.

Question 16

Let AA and BB be 3×33 \times 3 matrices where B=Q−1AQB = Q^{-1}AQ for some invertible matrix QQ. If AA has eigenvalues 2,3,52, 3, 5 and det⁡(A−4I)=−6\det(A - 4I) = -6, what is det⁡(B−4I)\det(B - 4I)?

  1. -24
  2. 6
  3. 24
  4. -6 (correct answer)
Explanation: When you encounter matrices related by similarity (B=Q−1AQB = Q^{-1}AQ), remember that similar matrices share many fundamental properties, including determinants of shifted matrices. Since B=Q−1AQB = Q^{-1}AQ, we can find det⁡(B−4I)\det(B - 4I) by substituting: B−4I=Q−1AQ−4I=Q−1AQ−4Q−1IQ=Q−1(A−4I)QB - 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)\det(XYZ) = \det(X)\det(Y)\det(Z), we get: det⁡(B−4I)=det⁡(Q−1)det⁡(A−4I)det⁡(Q)\det(B - 4I) = \det(Q^{-1})\det(A - 4I)\det(Q). Since det⁡(Q−1)=1det⁡(Q)\det(Q^{-1}) = \frac{1}{\det(Q)}, this simplifies to det⁡(B−4I)=det⁡(A−4I)=−6\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\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)\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.

Question 17

Matrices AA and BB are similar. If AA is invertible, what can be concluded about matrix BB?

  1. BB is the inverse of AA.
  2. BB is also invertible, and its inverse B−1B^{-1} is similar to A−1A^{-1}. (correct answer)
  3. BB is also invertible, but its inverse B−1B^{-1} is not necessarily similar to A−1A^{-1}.
  4. BB may or may not be invertible.
Explanation: If AA and BB are similar, then B=P−1APB = P^{-1}AP for some invertible matrix PP. The determinant is an invariant of similarity, so det⁡(B)=det⁡(P−1AP)=det⁡(P−1)det⁡(A)det⁡(P)=det⁡(A)\det(B) = \det(P^{-1}AP) = \det(P^{-1})\det(A)\det(P) = \det(A). Since AA is invertible, det⁡(A)≠0\det(A) \neq 0. Therefore, det⁡(B)≠0\det(B) \neq 0, which means BB is also invertible. To find the relationship between the inverses, we take the inverse of the similarity equation: B−1=(P−1AP)−1=P−1A−1(P−1)−1=P−1A−1PB^{-1} = (P^{-1}AP)^{-1} = P^{-1}A^{-1}(P^{-1})^{-1} = P^{-1}A^{-1}P. This equation shows, by definition, that B−1B^{-1} is similar to A−1A^{-1}.

Question 18

The fundamental reason that similar matrices share the same eigenvalues is that:

  1. the determinant of a product of matrices is the product of their determinants.
  2. they represent the same linear transformation, and eigenvalues are geometric properties of the transformation itself, independent of the coordinate system. (correct answer)
  3. the change-of-basis matrix PP is always orthogonal, so it preserves lengths and angles.
  4. the characteristic polynomial is easy to compute for any matrix, and similarity is defined by this polynomial.
Explanation: 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 λ\lambda such that for some nonzero vector v⃗\vec{v}, T(v⃗)=λv⃗T(\vec{v}) = \lambda \vec{v}. This property—that the transformation TT acts on v⃗\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, PP is not always orthogonal) or describe properties that are consequences, not the fundamental reason (A, D).

Question 19

Which of the following statements is always true for any two n×nn \times n similar matrices AA and BB?

  1. AA and BB have the same eigenvectors.
  2. The null space of AA is the same as the null space of BB.
  3. A+IA+I and B−IB-I are similar.
  4. AkA^k and BkB^k are similar for any positive integer kk. (correct answer)
Explanation: When you encounter questions about similar matrices, focus on properties that are preserved under similarity transformations. Two matrices AA and BB are similar if there exists an invertible matrix PP such that B=P−1APB = P^{-1}AP. The correct answer is D because similarity is preserved under matrix powers. If B=P−1APB = P^{-1}AP, then Bk=(P−1AP)k=P−1AkPB^k = (P^{-1}AP)^k = P^{-1}A^kP, making AkA^k and BkB^k similar. This follows from the fact that (P−1AP)k=P−1AkP(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 vv is an eigenvector of AA, then P−1vP^{-1}v (not vv) is the corresponding eigenvector of BB. Option B is wrong because similar matrices don't preserve null spaces. The null space of BB is P−1(null space of A)P^{-1}(\text{null space of } A), not the null space of AA itself. The dimensions are equal, but the actual subspaces differ. Option C fails because adding different matrices (II to one, −I-I to another) breaks the similarity relationship. If B=P−1APB = P^{-1}AP, then B−I=P−1AP−I≠P−1(A+I)PB - 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.