GATE Overflow Test Series | Linear Algebra | Test 1 | Question: 9

Question

Which of the following is/are correct? (Mark all the appropriate choices)

• A. Addition of two symmetric matrices results in a symmetric matrix.
• B. If $A$ and $B$ are two symmetric commutative matrices, i.e. $AB =BA,$ then the product of $A$ and $B$ is symmetric.
• C. If matrix $A$ is symmetric then $A^{n}$ is also symmetric, where $n$ is any positive integer.
• D. If $A$ is an invertible symmetric matrix then $A^{-1}$ is symmetric.

Answer 1

If matrix $'A'$ is the symmetric matrix, then $A^{T} = A.$

For option $(A):(A + B)^{T} = B^{T} + A^{T} = B + A = A + B.$

For option $(B): (AB)^{T} = B^{T} A^{T} = BA = AB.$

For option $(C): (A^{n})^{T} = {\underbrace{(A\cdot A \cdot A \dots A)}_{ \text{n times}}}^{T} = {\underbrace{(A^{T} \cdot A^{T} \cdot A^{T} \dots A^{T}}_ {\text{n times}})}$ $(\because (AB)^T = B^TA^T)$

$={\underbrace{(A\cdot A \cdot A \dots A)}_{ \text{n times}}} \left(\because A = A^T\right) =A^n$

$\implies \left(A^{T}\right)^{n} = A^{n}.$

Ref: https://math.stackexchange.com/questions/3208939/transpose-of-product-of-matrices

For option $(D):$ Let $A$ be an invertible symmetric matrix.

•  $\mid A \mid \neq 0 \implies \mid A^{T} \mid \neq 0 \implies (A^{T})^{-1}$ exists.  
•  $A^{T} = A$

We know that, $A^{-1}=\dfrac{\text{Adj}(A)}{\mid A \mid}$

 For the transpose matrix $A^T,$ we get

Taking transpose on both sides, we get

$(A^{-1})^T = \left(\dfrac{\text{Adj}(A)}{\mid A \mid}\right)^T$

$\implies (A^{-1})^T = \dfrac{\left(\text{Adj}(A)\right)^{T}}{\left(\mid A \mid \right)} \quad \left(|A|\;\text{is a constant}\right)$

$\implies (A^{-1})^T = \dfrac{\text{Adj}(A^{T})}{\mid A \mid } = \dfrac{\text{Adj}(A)}{\mid A \mid } \quad \left(\text{Since}\; A = A^T \right)$
$= A^{-1}$

$\implies A^{-1}$ is symmetric.

$\textbf{(OR)}$

We know that $AA^{-1} = I$

Taking the transpose on both the sides, we get

$\left(AA^{-1}\right)^{T} = (I)^{T}$

$\implies (A^{-1})^{T} A^{T} = I \quad \left(\because (AB)^{T} = B^{T}A^{T}\right)$

Multiply both sides by $(A^{T})^{-1},$ we get

$(A^{-1})^{T} A^{T} (A^{T})^{-1} = I (A^{T})^{-1}$

$\implies (A^{-1})^{T} = (A^{T})^{-1} \quad \left(\because A^{T} (A^{T})^{-1} = I \right) $

$\implies (A^{-1})^{T} = A^{-1} \quad \left(\because A^{T} = A \right)$

$\therefore$ $A^{-1}$ is a symmetric matrix.

So, the correct answer is $A; B; C;D.$

Ref: https://math.vanderbilt.edu/sapirmv/msapir/prsymmetric.html