Question

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

• A. For the $4 \times 4$ matrix $A$ with real entries $\operatorname{tr}\left( AA ^{ T }\right)$ is the sum of the squares of the entries of $A$
• B. For the $4 \times 4$ matrix $A$ with real entries $\operatorname{tr}\left( AA ^{ T }\right)$ 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 $\operatorname{tr}\left(X^T\right)$ 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,
$ \operatorname{tr}\left[\left(A-B^T\right)\left(A-B^T\right)^T\right] $
$= \operatorname{tr}\left[\left(A-B^{\prime}\right)\left(A^{\prime}-B\right)\right]=\operatorname{tr}\left(A A^{\prime}+ BB ^{\prime}-A B- B ^{\prime} A ^{\prime}\right) $
$= \operatorname{tr}\left( AA ^{\prime}+ B ^{\prime} B \right)-\operatorname{tr}\left( AB + B ^{\prime} A ^{\prime}\right) $
$= \operatorname{tr}\left( AA ^{\prime}\right)+\operatorname{tr}\left( B ^{\prime}\right)-\operatorname{tr}(A B)-\operatorname{tr}\left( B ^{\prime} A ^{\prime}\right) $
$= \operatorname{tr}\left(A A^{\prime}\right)+\operatorname{tr}( B B)-\operatorname{tr}( AB )-\operatorname{tr}\left( A ^{\prime} B ^{\prime}\right)[\text { Since } \operatorname{tr}( XY )=\operatorname{tr}( YX )]$
$= O$ from given relation
Hence, $A - B ^{ T }= O _4$