Question
Prove that $A+A^{\top}$ is a symmetric and $A-A^{\top}$ is a skew symmetric matrix, where
$A=\left[\begin{array}{ccc}5 & 2 & -4 \\ 3 & -7 & 2 \\ 4 & -5 & -3\end{array}\right]$
$A=\left[\begin{array}{ccc}5 & 2 & -4 \\ 3 & -7 & 2 \\ 4 & -5 & -3\end{array}\right]$
$\therefore \quad A^T=\left[\begin{array}{ccc}5 & 3 & 4 \\ 2 & -7 & -5 \\ -4 & 2 & -3\end{array}\right]$
$\begin{aligned} \therefore \quad \mathrm{A}+\mathrm{A}^{\mathrm{T}} & =\left[\begin{array}{ccc}5 & 2 & -4 \\ 3 & -7 & 2 \\ 4 & -5 & -3\end{array}\right]+\left[\begin{array}{ccc}5 & 3 & 4 \\ 2 & -7 & -5 \\ -4 & 2 & -3\end{array}\right] \\ & =\left[\begin{array}{ccc}5+5 & 2+3 & -4+4 \\ 3+2 & -7-7 & 2-5 \\ 4-4 & -5+2 & -3-3\end{array}\right]\end{aligned}$
$\therefore \quad A+A^T=\left[\begin{array}{ccc}10 & 5 & 0 \\ 5 & -14 & -3 \\ 0 & -3 & -6\end{array}\right]$
$\therefore \quad\left(A+A^T\right)^T=\left[\begin{array}{ccc}10 & 5 & 0 \\ 5 & -14 & -3 \\ 0 & -3 & -6\end{array}\right]$
$\therefore\left(A+A^{\top}\right)^{\top}=A+A^{\top}$, i.e., $A+A^{\top}=\left(A+A^{\top}\right)^{\top}$
$\therefore \mathrm{A}+\mathrm{A}^{\top}$ is a symmetric matrix.
$A-A^T=\left[\begin{array}{ccc}5 & 2 & -4 \\ 3 & -7 & 2 \\ 4 & -5 & -3\end{array}\right]-\left[\begin{array}{ccc}5 & 3 & 4 \\ 2 & -7 & -5 \\ -4 & 2 & -3\end{array}\right]$
$\begin{aligned} & =\left[\begin{array}{ccc}5-5 & 2-3 & -4-4 \\ 3-2 & -7+7 & 2+5 \\ 4+4 & -5-2 & -3+3\end{array}\right] \\ & =\left[\begin{array}{ccc}0 & -1 & -8 \\ 1 & 0 & 7 \\ 8 & -7 & 0\end{array}\right]\end{aligned}$
$\therefore \quad\left(A-A^T\right)^T=\left[\begin{array}{ccc}0 & 1 & 8 \\ -1 & 0 & -7 \\ -8 & 7 & 0\end{array}\right]=-\left[\begin{array}{ccc}0 & -1 & -8 \\ 1 & 0 & 7 \\ 8 & -7 & 0\end{array}\right]$
$ \therefore\left(A-A^{\top}\right)^{\top}=-\left(A-A^{\top}\right) $ i.e. $A-A^{\top}=-\left(A-A^{\top}\right)^{\top}$
$\therefore \mathrm{A}-\mathrm{A}^{\top}$ is skew symmetric matrix.
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