Transposes and Inverses: AB=BA, but A^T B neq. B A^T

[moonbeam]

Let \(\displaystyle A\) and \(\displaystyle B\) be n x n square matrices with \(\displaystyle A\) being invertible. Show that if \(\displaystyle AB = BA\), then \(\displaystyle A^{-1}B = BA^{-1}\). Also, give an example such that \(\displaystyle AB = BA\), but \(\displaystyle A^T B \neq BA^T\).
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Is this right?

If \(\displaystyle AB = BA\), then \(\displaystyle A^{-1}AB = A^{-1}BA\)

\(\displaystyle \Longrightarrow B=A^{-1}BA\)

\(\displaystyle \Longrightarrow BA^{-1} = A^{-1}B\)

For \(\displaystyle AB=BA\) and \(\displaystyle A^TB \neq BA^T\), let \(\displaystyle A=\left\[\begin{array}{cc} 1&2\\3&4\end{array}\right\]\) and let \(\displaystyle B=A^{-1}=\left\[\begin{array}{cc}-2&1\\\frac{3}{2}&-\frac{1}{2}\end{array}\right\]\).

Then \(\displaystyle AB=AA^{-1}=A^{-1}A=BA\).

\(\displaystyle A^TB=\left\[\begin{array}{cc}1&3\\2&4\end{array}\right\] \left\[\begin{array}{cc}-2&1\\\frac{3}{2}&-\frac{1}{2}\end{array}\right\] = \left\[\begin{array}{cc}\frac{5}{2}&-\frac{1}{2}\\2&0\end{array}\right\]\)

\(\displaystyle BA^T = \left[\begin{array}{cc}-2&1\\\frac{3}{2}&-\frac{1}{2}\end{array}\right\] \left\[\begin{array}{cc}1&3\\2&4\end{array}\right\] = \left\[\begin{array}{cc}0&-2\\\frac{1}{2}&\frac{5}{2}\end{array}\right\]\).

Therefore \(\displaystyle A^TB \neq BA^T\).

 

[galactus]

First assume that \(\displaystyle \L\\AB^{-1}=B^{-1}A\)

Then we have:

\(\displaystyle \L\\(AB^{-1})B=(B^{-1}A)B\)

or

\(\displaystyle \L\\A=B^{-1}AB\)

so that

\(\displaystyle \L\\BA=B(B^{-1}AB)=(BB^{-1})(AB)=AB\)

You still should show that if AB=BA then \(\displaystyle AB^{-1}=B^{-1}A\)

But that is pretty much analogous to the above.

Does that help?.