Transpose & Symmetric Matrices - Linear Algebra

A $3 \times 3$ matrix $A$ is constructed using the rule $a_{ij} = 2i + 2j$ for its entries. Which of the following properties does matrix $A$ have?
Answer: $A$ is symmetric. To check if $A$ is symmetric, we must verify if $a_{ij} = a_{ji}$ for all $i, j$ . According to the rule, $a_{ij} = 2i + 2j$ . For the transposed entry, we have $a_{ji} = 2j + 2i$ . Since addition is commutative, $2i + 2j = 2j + 2i$ , which means $a_{ij} = a_{ji}$ . Therefore, the matrix $A$ is symmetric. Let's check the other options. (A) For $A$ to be skew-symmetric, $a_{ij} = -a_{ji}$ , which is not true here. (D) The diagonal entries are $a_{11}=4, a_{22}=8, a_{33}=12$ , so the trace is $4+8+12=24 \neq 0$ . (B) Let's write out the matrix: $A = \begin{pmatrix} 4 & 6 & 8 \\ 6 & 8 & 10 \\ 8 & 10 & 12 \end{pmatrix}$ . The second row is the first row plus $(2, 2, 2)$ , and the third row is the second row plus $(2, 2, 2)$ . This means the rows are linearly dependent (e.g., $R_1 + R_3 = 2R_2$ ), so the determinant must be zero. Thus, $A$ is not invertible.
Let $A$ be a $3 \times 3$ matrix. The matrix $B = A + A^T$ is given by $B = \begin{pmatrix} 4 & 6 & 0 \\ 6 & -2 & 5 \\ 0 & 5 & 8 \end{pmatrix}$ . Based on this information, what are the diagonal entries of matrix $A$ ?
Answer: $2, -1, 4$ Let $A = (a_{ij})$ and $B = (b_{ij})$ . By definition, $B = A + A^T$ , so the entries of $B$ are given by $b_{ij} = a_{ij} + a_{ji}$ . For the diagonal entries, where $i=j$ , we have $b_{ii} = a_{ii} + a_{ii} = 2a_{ii}$ . Therefore, $a_{ii} = b_{ii} / 2$ . We can find the diagonal entries of $A$ by halving the diagonal entries of $B$ . The diagonal of $B$ is $(4, -2, 8)$ . Thus, the diagonal of $A$ is $(4/2, -2/2, 8/2) = (2, -1, 4)$ .
Let $A$ be a $3 \times 2$ matrix. Which of the following operations results in a $2 \times 2$ symmetric matrix?
Answer: $A^T A$ Let's analyze the dimensions of the resulting matrix for each option. The matrix $A$ has dimensions $3 \times 2$ , so its transpose $A^T$ has dimensions $2 \times 3$ . A) $A^2 = A A$ . The product is not defined because the inner dimensions do not match ( $3 \times 2$ and $3 \times 2$ ). B) $A A^T$ . The product dimensions are $(3 \times 2)(2 \times 3)$ , resulting in a $3 \times 3$ matrix. This is not a $2 \times 2$ matrix. C) $A + A^T$ . Addition is not defined because the matrices have different dimensions ( $3 \times 2$ and $2 \times 3$ ). D) $A^T A$ . The product dimensions are $(2 \times 3)(3 \times 2)$ , resulting in a $2 \times 2$ matrix. To check if it's symmetric, we take the transpose: $(A^T A)^T = A^T (A^T)^T = A^T A$ . Since the matrix is equal to its own transpose, it is symmetric. This is the correct choice.
Let $A$ be an invertible, symmetric matrix. Which of the following is equivalent to the expression $(A^{-1} A^T)^T$ ?
Answer: $I$ We are given that $A$ is invertible and symmetric, so $A^{-1}$ exists and $A^T = A$ . We can simplify the expression in two ways. Method 1: Substitute $A^T=A$ first. The expression becomes $(A^{-1} A)^T$ . Since $A^{-1}A = I$ (the identity matrix), the expression is $I^T$ . The transpose of the identity matrix is itself, so $I^T = I$ . Method 2: Use transpose properties first. $(A^{-1} A^T)^T = (A^T)^T (A^{-1})^T$ . We know $(A^T)^T = A$ and $(A^{-1})^T = (A^T)^{-1}$ . So the expression becomes $A(A^T)^{-1}$ . Now use the symmetric property $A^T=A$ . The expression is $A A^{-1}$ , which equals $I$ . Both methods yield the identity matrix $I$ .
Let $S$ be the set of all $3 \times 3$ symmetric matrices and $T$ be the set of all $3 \times 3$ skew-symmetric matrices. If $A \in S$ , $B \in T$ , and $\det(A + B) \neq 0$ , what can be said about $(A + B)^{-T}$ ?
Answer: $(A + B)^{-T}$ belongs to neither $S$ nor $T$ in general Since $A^T = A$ and $B^T = -B$ , we have $(A + B)^T = A^T + B^T = A - B$ . Therefore $(A + B)^{-T} = ((A + B)^T)^{-1} = (A - B)^{-1}$ . Taking transpose: $((A + B)^{-T})^T = ((A - B)^{-1})^T = ((A - B)^T)^{-1} = (A^T - B^T)^{-1} = (A - (-B))^{-1} = (A + B)^{-1}$ . Since $(A + B)^{-T} = (A - B)^{-1}$ and $((A + B)^{-T})^T = (A + B)^{-1}$ , we have $(A - B)^{-1} \neq (A + B)^{-1}$ in general, so $(A + B)^{-T}$ is neither symmetric nor skew-symmetric.
Let $A = \begin{pmatrix} k & 3 \\ 2k-1 & 4 \end{pmatrix}$ . The matrix $S = A + A^T$ is symmetric and has a trace of 10. What is the value of $k$ ?
Answer: $1$ For any square matrix $A$ , the matrix $S = A + A^T$ is always symmetric. The condition that $S$ is symmetric is a distractor, as it holds for any $k$ . The core of the problem is the trace. First, let's find $A+A^T$ : $A^T = \begin{pmatrix} k & 2k-1 \\ 3 & 4 \end{pmatrix}$ $S = A + A^T = \begin{pmatrix} k+k & 3+(2k-1) \\ (2k-1)+3 & 4+4 \end{pmatrix} = \begin{pmatrix} 2k & 2k+2 \\ 2k+2 & 8 \end{pmatrix}$ The trace of a matrix is the sum of its diagonal elements. So, $\mathrm{tr}(S) = 2k + 8$ . We are given that the trace is 10. $2k + 8 = 10$ $2k = 2$ $k = 1$
Let $A$ and $B$ be arbitrary $n \times n$ matrices. Which of the following statements is always true?
Answer: The diagonal entries of $A - A^T$ are always zero. Let's evaluate each statement. A) This is false. A counterexample is the zero matrix, which is symmetric ( $O^T = O$ ) but not invertible. B) The diagonal entries of any matrix $M$ are $m_{ii}$ . For the matrix $M = A - A^T$ , the diagonal entries are $m_{ii} = a_{ii} - (A^T)_{ii}$ . By definition of transpose, $(A^T)_{ii} = a_{ii}$ . So, $m_{ii} = a_{ii} - a_{ii} = 0$ . This is always true. C) This is false. The correct property is $(AB)^T = B^T A^T$ . Matrix multiplication is not generally commutative, so $A^T B^T \neq B^T A^T$ . D) This is false. As shown in another question, $(AB)^T = B^T A^T$ . If $A$ and $B$ are symmetric, this becomes $BA$ . $AB$ is only symmetric if $BA=AB$ , i.e., if $A$ and $B$ commute.
Any square matrix $A$ can be written as the sum of a symmetric matrix $S$ and a skew-symmetric matrix $K$ . If $A = \begin{pmatrix} 3 & 5 \\ -1 & 7 \end{pmatrix}$ , what is the skew-symmetric component $K$ ?
Answer: $\begin{pmatrix} 0 & 3 \\ -3 & 0 \end{pmatrix}$ The skew-symmetric component $K$ of a matrix $A$ is given by the formula $K = \frac{1}{2}(A - A^T)$ . First, we find the transpose of $A$ : $A^T = \begin{pmatrix} 3 & -1 \\ 5 & 7 \end{pmatrix}$ . Next, we compute the difference $A - A^T$ : $A - A^T = \begin{pmatrix} 3 & 5 \\ -1 & 7 \end{pmatrix} - \begin{pmatrix} 3 & -1 \\ 5 & 7 \end{pmatrix} = \begin{pmatrix} 3-3 & 5-(-1) \\ -1-5 & 7-7 \end{pmatrix} = \begin{pmatrix} 0 & 6 \\ -6 & 0 \end{pmatrix}$ Finally, we multiply by $\frac{1}{2}$ : $K = \frac{1}{2}\begin{pmatrix} 0 & 6 \\ -6 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 3 \\ -3 & 0 \end{pmatrix}$
Let $A$ and $B$ be $n \times n$ symmetric matrices. Which of the following statements is not always true?
Answer: $AB$ is a symmetric matrix. Let's analyze the choices given that $A^T=A$ and $B^T=B$ . A) $(A+B)^T = A^T + B^T = A+B$ . This is always true. B) $(cA)^T = cA^T = cA$ . This is always true. C) $(AB)^T = B^T A^T = BA$ . For $AB$ to be symmetric, we need $(AB)^T = AB$ , which means $BA=AB$ . This is only true if $A$ and $B$ commute, which is not always the case for symmetric matrices. Thus, this statement is not always true. D) Since $A$ is symmetric, $A^T = A$ , which by definition is a symmetric matrix. This is always true.
Consider the matrix equation $X + X^T = A$ where $A$ is a given $3 \times 3$ matrix. For this equation to have a solution, which condition must $A$ satisfy?
Answer: $A$ must be symmetric but may have any trace value Taking the transpose of both sides: $(X + X^T)^T = A^T$ , which gives $X^T + X = A^T$ . For consistency with the original equation $X + X^T = A$ , we need $A = A^T$ , so $A$ must be symmetric. If $A$ is symmetric, we can always construct a solution $X = \frac{1}{2}A + S$ where $S$ is any skew-symmetric matrix. The trace, determinant sign, and eigenvalue properties of $A$ don't affect the existence of solutions.
Given that $A$ is a $3 \times 3$ symmetric matrix with eigenvalues $\lambda_1, \lambda_2, \lambda_3$ , and $B = A^2 + 2A + 3I$ , what is the trace of $B^{-1}$ in terms of the eigenvalues of $A$ ?
Answer: $\frac{1}{\lambda_1^2 + 2\lambda_1 + 3} + \frac{1}{\lambda_2^2 + 2\lambda_2 + 3} + \frac{1}{\lambda_3^2 + 2\lambda_3 + 3}$ Since $A$ is symmetric, it has an orthonormal eigenbasis. If $A$ has eigenvalues $\lambda_i$ , then $B = A^2 + 2A + 3I$ has eigenvalues $\lambda_i^2 + 2\lambda_i + 3$ . The matrix $B^{-1}$ has eigenvalues $\frac{1}{\lambda_i^2 + 2\lambda_i + 3}$ . The trace of $B^{-1}$ equals the sum of its eigenvalues. Choice B incorrectly treats the sum of eigenvalues as a single eigenvalue, C attempts to use determinant relationships incorrectly, and D uses an incorrect formula mixing trace and determinant.
If $A$ is an $n \times n$ symmetric matrix and $B = A^3 - 2A^2 + A$ , which of the following must be true?
Answer: $B$ is symmetric and commutes with $A$ Since $A$ is symmetric, $A^T = A$ . Taking the transpose: $B^T = (A^3)^T - 2(A^2)^T + A^T = (A^T)^3 - 2(A^T)^2 + A^T = A^3 - 2A^2 + A = B$ . So $B$ is symmetric. Also, $B = A(A^2 - 2A + I)$ and since $A$ commutes with any polynomial in $A$ , we have $AB = BA$ . Choice A is circular reasoning, C is incorrect since $B$ is symmetric, and D is false since symmetry is preserved regardless of whether $A$ is diagonal.
Let $A$ be a $3 \times 3$ matrix such that $A^T = -A$ . If $\text{tr}(A^TA) = 18$ , what is $\text{tr}(A)$ ?
Answer: $0$ Since $A^T = -A$ , matrix $A$ is skew-symmetric. For any skew-symmetric matrix, all diagonal entries must be zero (since $a_{ii} = -a_{ii}$ implies $a_{ii} = 0$ ). Therefore, $\text{tr}(A) = 0$ . The condition $\text{tr}(A^TA) = 18$ provides information about the sum of squares of all entries but doesn't affect the trace of $A$ .
Let $P$ be a $4 \times 4$ matrix such that $P^T P = I$ . If $A$ is symmetric and $B = P^T A P$ , which statement about the relationship between $A$ and $B$ is most precise?
Answer: $B$ is symmetric, $\text{tr}(A) = \text{tr}(B)$ , and $\det(A) = \det(B)$ Since $P^TP = I$ , matrix $P$ is orthogonal, so $P^{-1} = P^T$ . We have $B^T = (P^TAP)^T = P^TA^T(P^T)^T = P^TAP = B$ since $A^T = A$ . So $B$ is symmetric. For trace: $\text{tr}(B) = \text{tr}(P^TAP) = \text{tr}(APP^T) = \text{tr}(A)$ since $PP^T = I$ . For determinant: $\det(B) = \det(P^T)\det(A)\det(P) = \det(P)^2\det(A) = \det(A)$ since $\det(P) = \pm 1$ for orthogonal matrices.
Let matrices $A$ and $B$ be defined as $A = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$ and $B^T = \begin{pmatrix} -1 & 0 \\ 4 & 2 \end{pmatrix}$ . What is the matrix $(AB)^T$ ?
Answer: $\begin{pmatrix} -1 & 0 \\ 8 & 6 \end{pmatrix}$ The property of transposes states that $(AB)^T = B^T A^T$ . We are given $B^T$ and can find $A^T$ . The transpose of $A$ is $A^T = \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix}$ . Now, we compute the product $B^T A^T$ : $B^T A^T = \begin{pmatrix} -1 & 0 \\ 4 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} (-1)(1)+(0)(2) & (-1)(0)+(0)(3) \\ (4)(1)+(2)(2) & (4)(0)+(2)(3) \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 8 & 6 \end{pmatrix}$
Consider the block matrix $M = \begin{pmatrix} A & B \\ B^T & C \end{pmatrix}$ where $A$ and $C$ are symmetric matrices. If $M^2 = M$ , what must be true about the relationship between $A$ , $B$ , and $C$ ?
Answer: $A^2 + BB^T = A$ , $AB + BC = B$ , and $C^2 + B^TB = C$ When you encounter a block matrix equation like $M^2 = M$ , you're dealing with an idempotent matrix. The key is to compute the matrix multiplication carefully using block matrix rules and then equate corresponding blocks. To find $M^2$ , multiply the block matrix by itself: $M^2 = \begin{pmatrix} A & B \\ B^T & C \end{pmatrix} \begin{pmatrix} A & B \\ B^T & C \end{pmatrix} = \begin{pmatrix} A^2 + BB^T & AB + BC \\ B^TA + CB^T & B^TB + C^2 \end{pmatrix}$ Since $M^2 = M$ , each block must equal its corresponding block in the original matrix. This gives us: - Top-left: $A^2 + BB^T = A$ - Top-right: $AB + BC = B$ - Bottom-left: $B^TA + CB^T = B^T$ (which is equivalent to the top-right when transposed, since $A$ and $C$ are symmetric) - Bottom-right: $B^TB + C^2 = C$ This matches option C exactly. Option A assumes $B = 0$ , which is unnecessarily restrictive. While this would satisfy the idempotent condition, it's not required—non-zero $B$ can work if the other conditions hold. Option B incorrectly uses $BC^T$ instead of $BC$ . Since $C$ is symmetric ( $C = C^T$ ), this seems minor but represents a computational error in the block multiplication. Option D introduces unrelated conditions about the sum and skew-symmetry that don't follow from the idempotent property. Study tip: When working with block matrices, always multiply block-by-block systematically, and remember that symmetry conditions (like $A = A^T$ ) often simplify your final expressions.
Let $A$ be an $n \times n$ symmetric matrix and $B$ be an $n \times n$ skew-symmetric matrix. Which of the following expressions simplifies to the matrix $A$ ?
Answer: The symmetric part of $(A+B)$ We are given $A^T = A$ and $B^T = -B$ . The symmetric part of any matrix $M$ is given by $\frac{1}{2}(M + M^T)$ . Let's find the symmetric part of the matrix $M = A+B$ . First, find $M^T$ : $M^T = (A+B)^T = A^T + B^T$ . Using the given properties, $A^T=A$ and $B^T=-B$ , so: $M^T = A - B$ . Now, substitute $M$ and $M^T$ into the formula for the symmetric part: $\frac{1}{2}(M + M^T) = \frac{1}{2}((A+B) + (A-B)) = \frac{1}{2}(2A) = A$ . Therefore, the symmetric part of $(A+B)$ is equal to $A$ . Choice C, the skew-symmetric part, would be $\frac{1}{2}(M - M^T) = \frac{1}{2}((A+B) - (A-B)) = \frac{1}{2}(2B) = B$ .
Given the matrix equation $X^T A X = B$ where $A$ and $B$ are $n \times n$ symmetric matrices and $X$ is $n \times n$ and invertible, which condition is necessary and sufficient for a solution to exist?
Answer: $A$ and $B$ must have the same number of positive and negative eigenvalues The transformation $X^TAX$ is a congruence transformation. Two symmetric matrices are congruent if and only if they have the same inertia (same number of positive, negative, and zero eigenvalues) by Sylvester's law of inertia. This is equivalent to having the same signature. Choice A is incorrect since congruence doesn't preserve determinant or trace in general. Choice B mentions rank (preserved) but 'signature' usually refers to the difference between positive and negative eigenvalues, not their individual counts. Choice C is too strong - eigenvalues need not be identical, only their signs matter.
Let $A$ and $P$ be $n \times n$ matrices. If $A$ is symmetric and $P$ is orthogonal (meaning $P^T P = I$ ), what property must the matrix $S = P A P^T$ have?
Answer: $S$ must be symmetric. To determine the property of $S = P A P^T$ , we should examine its transpose, $S^T$ . Using the property that $(XYZ)^T = Z^T Y^T X^T$ , we have: $S^T = (P A P^T)^T = (P^T)^T A^T P^T$ Since $(P^T)^T = P$ , this simplifies to: $S^T = P A^T P^T$ We are given that $A$ is a symmetric matrix, which means $A^T = A$ . Substituting this into our expression for $S^T$ : $S^T = P A P^T$ This is the original definition of $S$ . Since $S^T = S$ , the matrix $S$ must be symmetric.
For any $n \times n$ matrix $A$ , which of the following expressions is guaranteed to result in a skew-symmetric matrix?
Answer: $A^2 - (A^T)^2$ A matrix $X$ is skew-symmetric if $X^T = -X$ . Let's test each option. Let $X$ be the expression in each choice. A) Let $S = A+A^T$ , which is symmetric. $X = S^2$ . $X^T = (S^2)^T = (S^T)^2 = S^2 = X$ . So $(A+A^T)^2$ is symmetric. B) $X^T = (A^T A + A A^T)^T = (A^T A)^T + (A A^T)^T = A^T(A^T)^T + (A^T)^T A^T = A^T A + A A^T = X$ . So it is symmetric. C) $X^T = (A^2 - (A^T)^2)^T = (A^2)^T - ((A^T)^2)^T = (A^T)^2 - A^2 = -(A^2 - (A^T)^2) = -X$ . This is skew-symmetric. D) $X^T = (A^T A - A A^T)^T = (A^T A)^T - (A A^T)^T = A^T A - A A^T = X$ . So it is symmetric.

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