Question: If $\text{A}$and$B$are symmetric matrices, show that their commutator is antisymmetric [see equation 6.3].

A symmetric matrix is a square matrix that is identical to its transpose in linear algebra.

An antisymmetric matrix is a square matrix whose negative transpose equals its positive transpose.

The commutator of matrices and can be mathematically presented as$\left[A,B\right]$

Therefore,$[A,B]=AB-BA$, which is a commutator of and .

Now, transpose of commutator of matrices and is mathematically presented as,

$$\begin{gathered} [A,B{]}^T=(AB-BA{)}^T \\ =(AB{)}^T-(BA{)}^T\dots \dots .(\text{Using transpose property)} \\ =B^T{A}^T-A^T{B}^T\dots \dots \text{(1)} \end{gathered}$$

Since, then matrices and are symmetric, which means$A=A^T$and $B=B^T$.

Therefore, plugging in (1),

$=-(AB-BA)\$$

$=-[A,B]\$$

Hence, it is proven that$[A,B{]}^T=-[A,B]$, which means that the transpose of commutator of two matrices $A$and B , which are symmetric, is equal to the negative of the commutator of matrices A and B. Therefore, if A and B are symmetric matrices, then their commutator is antisymmetric.