Eigenvalues: Special Matrices - Linear Algebra

Let $A$ be a $4 \times 4$ upper triangular matrix with diagonal entries $2, 3, 3, 5$ . If $B = A^2 - 6A + 9I$ , what can be concluded about the eigenvalues of $B$ ?
Answer: The eigenvalues of $B$ are $1, 0, 0, 4$ but geometric multiplicities may be less than algebraic multiplicities Since $A$ is upper triangular, its eigenvalues are its diagonal entries: $2, 3, 3, 5$ . For any polynomial $p(x)$ , if $\lambda$ is an eigenvalue of $A$ , then $p(\lambda)$ is an eigenvalue of $p(A)$ . Here $p(x) = x^2 - 6x + 9 = (x-3)^2$ , so the eigenvalues of $B$ are $(2-3)^2 = 1$ , $(3-3)^2 = 0$ , $(3-3)^2 = 0$ , and $(5-3)^2 = 4$ . However, while the algebraic multiplicities are preserved, the geometric multiplicities may decrease when applying polynomial functions, so we cannot guarantee that geometric multiplicities equal algebraic multiplicities without more information about the Jordan form of $A$ .
A matrix $P$ is idempotent if $P^2=P$ . Let $M$ be a $3 \times 3$ upper triangular matrix that is idempotent. If $M$ is not the zero matrix or the identity matrix, which of the following could be the trace of $M$ ?
Answer: $2$ If $\lambda$ is an eigenvalue of an idempotent matrix $P$ , then $\lambda^2$ is an eigenvalue of $P^2$ . Since $P^2=P$ , their eigenvalues must be the same, so $\lambda^2 = \lambda$ , which gives $\lambda=0$ or $\lambda=1$ . The matrix $M$ is upper triangular, so its eigenvalues are its diagonal entries, and they must be either 0 or 1. The trace of $M$ is the sum of its eigenvalues. Since $M$ is not the zero matrix, its trace cannot be $0+0+0=0$ . Since $M$ is not the identity matrix, its trace cannot be $1+1+1=3$ . The only other possibilities for the set of eigenvalues are $\{1, 0, 0\}$ or $\{1, 1, 0\}$ . The corresponding traces are $1+0+0=1$ or $1+1+0=2$ . Of the choices given, 2 is a possible value for the trace.
Consider the block diagonal matrix $M = \begin{pmatrix} 2I_3 & 0 \\ 0 & J \end{pmatrix}$ where $I_3$ is the $3 \times 3$ identity matrix and $J = \begin{pmatrix} 4 & 1 & 0 \\ 0 & 4 & 1 \\ 0 & 0 & 4 \end{pmatrix}$ . What is the algebraic multiplicity of eigenvalue $4$ for matrix $M$ ?
Answer: The algebraic multiplicity is $3$ since $J$ is a $3 \times 3$ Jordan block with eigenvalue $4$ For block diagonal matrices, the characteristic polynomial is the product of the characteristic polynomials of each block. The $2I_3$ block contributes $(\lambda - 2)^3$ and the Jordan block $J$ contributes $(\lambda - 4)^3$ since all diagonal entries of $J$ are $4$ . Therefore, the characteristic polynomial of $M$ is $(\lambda - 2)^3(\lambda - 4)^3$ , giving eigenvalue $4$ an algebraic multiplicity of $3$ . The eigenvalue $2$ also has algebraic multiplicity $3$ .
Suppose $A$ is a $4 \times 4$ skew-symmetric matrix over the real numbers. Which of the following statements about the eigenvalues of $A^2$ must be true?
Answer: All eigenvalues of $A^2$ are real and nonpositive, with even algebraic multiplicities for nonzero eigenvalues When you encounter questions about eigenvalues of powers of special matrices, focus on how the matrix's structure constrains its spectral properties. Skew-symmetric matrices have unique characteristics that directly impact their eigenvalues and those of their powers. For a skew-symmetric matrix $A$ (where $A^T = -A$ ), the eigenvalues come in purely imaginary conjugate pairs $\pm i\alpha$ for real $\alpha \geq 0$ , plus possibly zero eigenvalues. Since $A$ is $4 \times 4$ , we have an even number of entries, so the eigenvalues appear as conjugate pairs with matching multiplicities. When we form $A^2$ , each eigenvalue $\lambda$ of $A$ becomes $\lambda^2$ in $A^2$ . If $A$ has eigenvalues $\pm i\alpha$ , then $A^2$ has eigenvalues $(\pm i\alpha)^2 = -\alpha^2 \leq 0$ . Zero eigenvalues of $A$ remain zero in $A^2$ . Since conjugate pairs from $A$ both map to the same negative real value $-\alpha^2$ , each nonzero eigenvalue of $A^2$ has even algebraic multiplicity. Choice A incorrectly requires at least one zero eigenvalue, but this isn't guaranteed. Choice B wrongly claims eigenvalues are nonnegative—they're actually nonpositive since $-\alpha^2 \leq 0$ . Choice C misunderstands that $A^2$ always has real eigenvalues regardless of $A$ 's symmetry, since the purely imaginary eigenvalues of skew-symmetric $A$ square to real values. Remember: For skew-symmetric matrices, squaring converts purely imaginary eigenvalues to real, nonpositive ones with even multiplicities—this pairing principle is key to solving such problems.
Let $A$ and $B$ be $3 \times 3$ upper triangular matrices. Which of the following matrices is not guaranteed to be upper triangular?
Answer: $A^T$ When you encounter questions about matrix operations and special matrix types like upper triangular matrices, you need to understand how operations preserve or change structural properties. An upper triangular matrix has all entries below the main diagonal equal to zero. Let's examine why option D is correct. The transpose operation $A^T$ flips a matrix across its main diagonal, swapping rows and columns. When you transpose an upper triangular matrix, the zeros that were below the diagonal move above the diagonal, creating a lower triangular matrix instead. For example, if $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{pmatrix}$ , then $A^T = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 4 & 0 \\ 3 & 5 & 6 \end{pmatrix}$ , which is lower triangular. Now for why the other options preserve the upper triangular structure: Option A ( $A+B$ ) works because adding corresponding entries of two upper triangular matrices keeps zeros below the diagonal. Option B ( $AB$ ) preserves upper triangular form due to how matrix multiplication works—when multiplying upper triangular matrices, entries below the diagonal remain zero. Option C ( $5A$ ) simply scales every entry by 5, so zeros below the diagonal stay zero. Remember this key principle: matrix addition and multiplication preserve the upper triangular property, but transposition converts upper triangular matrices to lower triangular. When studying matrix properties, always visualize how operations affect the position of zeros in the matrix structure.
Consider the matrix $A = \begin{pmatrix} 3 & a & b \\ 0 & 3 & c \\ 0 & 0 & 5 \end{pmatrix}$ where $a, b, c$ are nonzero real numbers. If $\det(A - 3I) = 0$ , what can be concluded about the geometric multiplicity of eigenvalue $3$ ?
Answer: The geometric multiplicity is $1$ regardless of the values of $a, b, c$ 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 $a, b, c$ .
Consider the circulant matrix $C = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{pmatrix}$ . If $\omega = e^{2\pi i/3}$ is a primitive cube root of unity, what is the eigenvalue corresponding to eigenvector $\begin{pmatrix} 1 \\ \omega^2 \\ \omega^4 \end{pmatrix}$ ?
Answer: $1 + 2\omega^2 + 3\omega^4 = 1 + 2\omega^2 + 3\omega$ For a circulant matrix with first row $(c_0, c_1, \ldots, c_{n-1})$ , the eigenvalue corresponding to eigenvector $(1, \omega^k, \omega^{2k}, \ldots, \omega^{(n-1)k})^T\$ is$ \sum_{j=0}^{n-1} $c_j$ \omega^{jk} $. Here we have$ k=2 $(since the eigenvector is$ (1, \omega^2, \omega^4)^T $), and the first row is$ (1, 2, 3) $. The eigenvalue is$ 1 \cdot \omega^0 + 2 \cdot \omega^2 + 3 \cdot \omega^4 $. Since$ \omega^3 = 1 $, we have$ \omega^4 = \omega^{3+1} = \omega^3 \cdot \omega = \omega $. Therefore the eigenvalue is$ 1 + 2\omega^2 + 3\omega$$.
Let $L$ be a $3 \times 3$ lower triangular matrix with diagonal entries $d_1, d_2, d_3$ . Let $D$ be a diagonal matrix with diagonal entries $2, 1, 3$ . The product $M=LD$ is formed. What are the diagonal entries of the resulting matrix $M$ ?
Answer: $2d_1, d_2, 3d_3$ When a matrix $L$ is post-multiplied (multiplied on the right) by a diagonal matrix $D$ , the $j$ -th column of $L$ is scaled by the $j$ -th diagonal entry of $D$ . Let $L = \begin{pmatrix} d_1 & 0 & 0 \\ l_{21} & d_2 & 0 \\ l_{31} & l_{32} & d_3 \end{pmatrix}$ and $D = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ . The product is $M = LD = \begin{pmatrix} 2d_1 & 0 & 0 \\ 2l_{21} & 1d_2 & 0 \\ 2l_{31} & 1l_{32} & 3d_3 \end{pmatrix}$ . The resulting matrix $M$ is also lower triangular, and its diagonal entries are $2d_1, d_2, 3d_3$ .
Let $S$ be a real symmetric $5 \times 5$ matrix with eigenvalues $-2, -1, 0, 1, 3$ . If $T = S^4 + 2S^2$ , which statement about the definiteness of $T$ is correct?
Answer: $T$ is positive semidefinite but not positive definite since $T$ has a zero eigenvalue Since $S$ is symmetric with eigenvalues $\lambda_i \in \{-2, -1, 0, 1, 3\}$ , the eigenvalues of $T = S^4 + 2S^2$ are $\lambda_i^4 + 2\lambda_i^2$ . Computing: for $\lambda = -2$ : $16 + 8 = 24$ ; for $\lambda = -1$ : $1 + 2 = 3$ ; for $\lambda = 0$ : $0 + 0 = 0$ ; for $\lambda = 1$ : $1 + 2 = 3$ ; for $\lambda = 3$ : $81 + 18 = 99$ . The eigenvalues of $T$ are $\{24, 3, 0, 3, 99\}$ . Since all eigenvalues are nonnegative but one is zero, $T$ is positive semidefinite but not positive definite.
Consider the matrix $M = \begin{pmatrix} 4 & 0 & 0 \\ 1 & 4 & 0 \\ -2 & 3 & 4 \end{pmatrix}$ . Which statement correctly describes the eigenvalues of $M$ ?
Answer: The only eigenvalue is 4, with an algebraic multiplicity of 3. The matrix $M$ is a lower triangular matrix. The eigenvalues of any triangular matrix are its diagonal entries. The diagonal entries of $M$ are 4, 4, and 4. Therefore, the only distinct eigenvalue is 4. Since it appears three times on the diagonal, it has an algebraic multiplicity of 3.
A $4 \times 4$ matrix $A$ is nilpotent with index 3, meaning $A^3 = O$ but $A^2 \neq O$ . The matrix $A$ is similar to an upper triangular matrix $T$ . What is the value of the determinant of $T$ ?
Answer: $0$ If $\lambda$ is an eigenvalue of $A$ , then $\lambda^3$ is an eigenvalue of $A^3 = O$ . The only eigenvalue of the zero matrix is 0, so $\lambda^3 = 0$ , which implies $\lambda = 0$ . Thus, all eigenvalues of the nilpotent matrix $A$ are 0. Since $A$ is similar to $T$ , they share the same eigenvalues. Therefore, all eigenvalues of $T$ are 0. As $T$ is an upper triangular matrix, its eigenvalues are its diagonal entries, so all its diagonal entries must be 0. The determinant of a triangular matrix is the product of its diagonal entries, so $\det(T) = 0 \cdot 0 \cdot 0 \cdot 0 = 0$ .
Let $P$ be an orthogonal projection matrix onto a $2$ -dimensional subspace of $\mathbb{R}^5$ . If $Q = 3P + 2(I - P)$ , what are the possible eigenvalues of $Q$ ?
Answer: Only $2$ and $3$ with multiplicities $3$ and $2$ respectively Since $P$ is an orthogonal projection onto a 2-dimensional subspace of $\mathbb{R}^5$ , the eigenvalues of $P$ are $1$ (with multiplicity $2$ ) and $0$ (with multiplicity $3$ ). We can write $Q = 3P + 2(I - P) = 3P + 2I - 2P = P + 2I$ . If $\lambda$ is an eigenvalue of $P$ , then $\lambda + 2$ is an eigenvalue of $Q$ . Therefore, the eigenvalues of $Q$ are $1 + 2 = 3$ (with multiplicity $2$ ) and $0 + 2 = 2$ (with multiplicity $3$ ).
Let $U$ be an $n \times n$ upper triangular matrix with eigenvalues $\lambda_1, \dots, \lambda_n$ . Let $A = U - 3I$ , where $I$ is the $n \times n$ identity matrix. Which of the following represents the set of eigenvalues of matrix $A$ ?
Answer: The eigenvalues of $U$ are each shifted by -3, $\lbrace \lambda_1 - 3, \dots, \lambda_n - 3 \rbrace$ . If a matrix $U$ has an eigenvalue $\lambda$ with corresponding eigenvector $\mathbf{v}$ , so that $U\mathbf{v} = \lambda\mathbf{v}$ , then the matrix $A = U - 3I$ has an eigenvalue $\lambda - 3$ . This can be shown by $A\mathbf{v} = (U - 3I)\mathbf{v} = U\mathbf{v} - 3I\mathbf{v} = \lambda\mathbf{v} - 3\mathbf{v} = (\lambda - 3)\mathbf{v}$ . Thus, each eigenvalue of $U$ is shifted by -3 to become an eigenvalue of $A$ .
The matrix $A = \begin{pmatrix} 2 & 3 & 4 \\ 0 & x & 5 \\ 0 & 0 & y \end{pmatrix}$ is similar to the diagonal matrix $D = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -3 \end{pmatrix}$ . What is the value of $x+y$ ?
Answer: $-2$ Similar matrices have the same set of eigenvalues. The eigenvalues of the upper triangular matrix $A$ are its diagonal entries: $\{2, x, y\}$ . The eigenvalues of the diagonal matrix $D$ are its diagonal entries: $\{2, 1, -3\}$ . Since the sets of eigenvalues must be identical, the set $\{x, y\}$ must be equal to the set $\{1, -3\}$ . Therefore, the sum is $x+y = 1 + (-3) = -2$ . Alternatively, since the trace is the sum of the eigenvalues and is invariant under similarity, $\text{tr}(A) = \text{tr}(D)$ . So, $2+x+y = 2+1+(-3) = 0$ . Solving for $x+y$ gives $x+y = -2$ .
Consider the matrix $A = \begin{pmatrix} 3 & 0 & 0 \\ -1 & 2 & 0 \\ 4 & 1 & k \end{pmatrix}$ . The sum of the eigenvalues of the matrix $A^2$ is 29. What is the value of $k^2$ ?
Answer: $16$ When you encounter questions about eigenvalues and matrix powers, remember two key relationships: the trace of a matrix equals the sum of its eigenvalues, and if λ is an eigenvalue of matrix A, then λ² is an eigenvalue of A². Since A is upper triangular, its eigenvalues are simply the diagonal entries: 3, 2, and k. Therefore, the eigenvalues of A² are 3² = 9, 2² = 4, and k². Given that the sum of eigenvalues of A² equals 29, we have: $9 + 4 + k^2 = 29$ $13 + k^2 = 29$ $k^2 = 16$ Let's examine why the other answers are incorrect. Choice A (4) would mean k² = 4, giving us k = ±2. This would make the sum of A²'s eigenvalues equal 9 + 4 + 4 = 17, not 29. Choice B (13) represents a common error where students might subtract incorrectly and think k² = 29 - 13 - 3 = 13, forgetting that the eigenvalues of A² are the squares of A's eigenvalues. Choice C (24) might result from the misconception that k² = 29 - 2 - 3 = 24, using the original eigenvalues instead of their squares. Study tip: For triangular matrices, eigenvalues are always the diagonal entries. When dealing with matrix powers, remember that eigenvalues get raised to that same power. Always double-check your arithmetic, especially when squaring negative values.
Let $T$ be a $3 \times 3$ upper triangular matrix with a trace of 6 and a determinant of -12. If two of its eigenvalues are 4 and -1, what is the third eigenvalue?
Answer: $3$ For any matrix, the trace is the sum of its eigenvalues, and the determinant is the product of its eigenvalues. Since $T$ is upper triangular, its diagonal entries are its eigenvalues. Let the eigenvalues be $\lambda_1, \lambda_2, \lambda_3$ . We are given $\lambda_1=4$ and $\lambda_2=-1$ . Using the trace property: $\lambda_1 + \lambda_2 + \lambda_3 = \text{tr}(T) \implies 4 + (-1) + \lambda_3 = 6 \implies 3 + \lambda_3 = 6 \implies \lambda_3 = 3$ . We can verify this with the determinant: $\lambda_1 \cdot \lambda_2 \cdot \lambda_3 = 4 \cdot (-1) \cdot 3 = -12$ , which matches the given determinant.
Let $A$ be an invertible $n \times n$ lower triangular matrix. Which of the following statements about $A^T$ , the transpose of $A$ , is always true?
Answer: The determinant of $A^T$ is non-zero. A matrix is invertible if and only if its determinant is non-zero. A property of determinants is that $\det(A) = \det(A^T)$ . Since $A$ is invertible, $\det(A) \neq 0$ . Therefore, $\det(A^T)$ must also be non-zero. A matrix with a non-zero determinant is invertible (non-singular). Distractor A is false because the off-diagonal elements of $A$ become the off-diagonal elements of $A^T$ . Distractor B is false because $A^T$ is invertible. Distractor C describes the eigenvalues of $A^{-1}$ , not $A^T$ . A matrix and its transpose always have the same eigenvalues.
Let matrix $A$ be defined as $A = \begin{pmatrix} 2 & -1 & 4 \\ 0 & k & 5 \\ 0 & 0 & 3 \end{pmatrix}$ . For which value of $k$ is matrix $A$ singular?
Answer: $k = 0$ A matrix is singular if and only if 0 is one of its eigenvalues. For a triangular matrix, the eigenvalues are the entries on its main diagonal. The eigenvalues of matrix $A$ are 2, $k$ , and 3. For $A$ to be singular, one of these eigenvalues must be 0. Since 2 and 3 are non-zero, $k$ must be 0.
Consider the lower triangular matrix $L = \begin{pmatrix} 1 & 0 & 0 \\ 4 & -2 & 0 \\ 5 & 6 & 3 \end{pmatrix}$ . What are the eigenvalues of the matrix $L^2$ ?
Answer: $\lbrace 1, 4, 9 \rbrace$ The eigenvalues of a triangular matrix are its diagonal entries. So, the eigenvalues of $L$ are $\lambda_1 = 1$ , $\lambda_2 = -2$ , and $\lambda_3 = 3$ . If $\lambda$ is an eigenvalue of a matrix $L$ , then $\lambda^k$ is an eigenvalue of the matrix $L^k$ . Therefore, the eigenvalues of $L^2$ are the squares of the eigenvalues of $L$ . The set of eigenvalues for $L^2$ is $\lbrace 1^2, (-2)^2, 3^2 \rbrace = \lbrace 1, 4, 9 \rbrace$ .
Consider the Toeplitz matrix $T = \begin{pmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{pmatrix}$ . Given that this matrix can be written as $T = J - I$ where $J$ is the all-ones matrix and $I$ is the identity, what is the second-largest eigenvalue of $T$ ?
Answer: $-1$ since most eigenvalues of the all-ones matrix become $-1$ after subtracting $I$ When you encounter a matrix that can be expressed as a combination of simpler matrices like $T = J - I$ , the key insight is to find the eigenvalues of each component separately, then use linearity properties. The all-ones matrix $J$ has a straightforward eigenvalue structure. Its largest eigenvalue is $n = 4$ (the matrix size) with eigenvector $(1,1,1,1)^T$ , since $J$ maps this vector to $4(1,1,1,1)^T$ . The remaining three eigenvalues are all $0$ , corresponding to any vector orthogonal to $(1,1,1,1)^T$ , such as $(1,-1,0,0)^T$ , $(1,0,-1,0)^T$ , and $(1,0,0,-1)^T$ . Since $T = J - I$ , and eigenvalues behave linearly under such operations, we subtract $1$ from each eigenvalue of $J$ . This gives us eigenvalues: $4-1=3$ , $0-1=-1$ , $0-1=-1$ , and $0-1=-1$ . Therefore, the eigenvalues of $T$ are $3, -1, -1, -1$ , making the second-largest eigenvalue $-1$ . Choice A incorrectly assumes eigenvalues follow a simple arithmetic sequence. Choice B makes the error of thinking we subtract $1$ from the $4$ eigenvalue only, missing that we subtract from all eigenvalues. Choice C misapplies the trace condition—while the trace is indeed $0$ , eigenvalues need not be symmetric about zero. Study tip: When a matrix is expressed as a linear combination of simpler matrices, find the eigenvalues of each component first, then apply the linear transformation to those eigenvalues directly.

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