Question

Let $A$ and $B$ be $4\times 4$ matrices with real entries satisfying $\mathrm{tr}\left(A^T+B{B}^T\right)=\mathrm{tr}\left(AB+A^T{B}^T\right)$ then (Note : $\mathrm{tr}$-stands for trace and $A^T$ stands for transpose of matrix A)

• A. For the $4\times 4$ matrix $A$ with real entries $\mathrm{tr}\left(A{A}^T\right)$ is the sum of the squares of the entries of $A$
• B. For the $4\times 4$ matrix $A$ with real entries $\mathrm{tr}\left(A{A}^T\right)$ is the sum of the cubes of the entries of $A$
• C. $A=B^T$
• D. $A+B^T=O_4$

Answer 1

Answer: (A), (C)

Using a property of trace that, for any nxn matrix $X$ with real entries $\mathrm{tr}\left(X^T\right)$ is the sum of the squares of the entries of $X$ and this number is non-negative and is equal to 0 iff $X$ is a zero matrix, we consider,
$\mathrm{tr}\left[\left(A-B^T\right){\left(A-B^T\right)}^T\right]$
$=\mathrm{tr}\left[\left(A-B^{\prime }\right)\left(A^{\prime }-B\right)\right]=\mathrm{tr}\left(A{A}^{\prime }+B{B}^{\prime }-AB-B^{\prime }{A}^{\prime }\right)$
$=\mathrm{tr}\left(A{A}^{\prime }+B^{\prime }B\right)-\mathrm{tr}\left(AB+B^{\prime }{A}^{\prime }\right)$
$=\mathrm{tr}\left(A{A}^{\prime }\right)+\mathrm{tr}\left(B^{\prime }\right)-\mathrm{tr}(AB)-\mathrm{tr}\left(B^{\prime }{A}^{\prime }\right)$
$=\mathrm{tr}\left(A{A}^{\prime }\right)+\mathrm{tr}(BB)-\mathrm{tr}(AB)-\mathrm{tr}\left(A^{\prime }{B}^{\prime }\right)[\text{ Since }\mathrm{tr}(XY)=\mathrm{tr}(YX)]$
$=O$ from given relation
Hence, $A-B^T=O_4$