Let A be a square matrix such that AA^T=I.

Question

Let $A$ be a square matrix such that ${AA}^{T}=I$. Then $\frac{1}{2}A[{(A+{A}^{T})}^{2}+{(A-{A}^{T})}^{2}]$ is equal to

Accepted Answer

$$\mathrm{AA}^{\mathrm{T}}=\mathrm{I}=\mathrm{A}^{\mathrm{T}} \mathrm{A}$$

On solving given expression, we get

$$\begin{aligned}
& \frac{1}{2} \mathrm{~A}\left[\mathrm{~A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2+2 \mathrm{~A} \mathrm{~A}^{\mathrm{T}}+\mathrm{A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2-2 \mathrm{~A} \mathrm{~A}^{\mathrm{T}}\right] \
& =\mathrm{A}\left[\mathrm{A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2\right]=\mathrm{A}^3+\mathrm{A}^{\mathrm{T}}
\end{aligned}$$

Solutions · 1