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# If A and B are symmetric matrices, prove that AB - BA is a skew symmetric matrix

Toolbox:
• A square matrix $A = [a_{ij}]$ is said to be skew symmetric if $A' = -A$ that is $[a_{ij} ] = - [a_{ji} ]$ for all possible value of i and j.
• A square matrix $A = [a_{aij}]$ is said to be symmetric if $A' = A$ that is $[a_{ij}] = [a_{ji}]$ for all possible value of i and j.
Step 1: Given
$A \to symmetric \;matrix$
$B \to symmetric\; matrix$
$\implies A = A$
$\implies B = B$
From the property of transpose of matrices. we have
$(AB) = BA$
Step 2: Now consider AB - BA and by taking transpose of it, we get
\begin{align*}(AB - BA) & = (AB) - (BA) \\ & = B'A' - A'B' \end{align*}
Replace $A' = A \; and B' = B$
$\; \; \; \; \; \; = BA -AB$
$\; \; \; \; \; \; \; = -(AB - BA )$ (By taking negative common)
we know that a matrix is said to b skew symmetric matrix if $A = -A$
Hence $AB - BA$ is skew symmetric matrices