# Let $A$ and $B$ be matrix symmetric matrix. Show that $AB-BA$ is a skew symmetric matrix.

I solved it by myself, but I don't know I did it right. I hope you guys to check this out.

Suppose $(AB-BA)^{T}=BA-AB$

$(AB-BA)^{T}=B^{T}A^{T}-A^{T}B^{T}$

then,

$B^{T}=B$

$A^{T}=A$

therefore,

$B^{T}A^{T}-A^{T}B^{T}=BA-AB$

It's my first time to ask question here and my English is not that good since I learned English for second language. I hope you to teach me a lot.

What you showed was that if $AB - BA$ was skew symmetric then $A,B$ are symmetric. However what they asked us to show was the opposite, if $A,B$ are symmetric then show $AB - BA$ is skew symmetric.
To prove $AB -BA$ is skew symmetric you must show that $(AB - BA)^T = BA - AB.$ $$(AB-BA)^T = (AB)^T - (BA)^T = B^TA^T - A^TB^T$$
Since $A = A^T, B= B^T$ $$(AB - BA)^T = BA - AB$$