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

moonbeam

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Let A\displaystyle A and B\displaystyle B be n x n square matrices with A\displaystyle A being invertible. Show that if AB=BA\displaystyle AB = BA, then A1B=BA1\displaystyle A^{-1}B = BA^{-1}. Also, give an example such that AB=BA\displaystyle AB = BA, but ATBBAT\displaystyle A^T B \neq BA^T.
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Is this right?

If AB=BA\displaystyle AB = BA, then A1AB=A1BA\displaystyle A^{-1}AB = A^{-1}BA

B=A1BA\displaystyle \Longrightarrow B=A^{-1}BA

BA1=A1B\displaystyle \Longrightarrow BA^{-1} = A^{-1}B

For AB=BA\displaystyle AB=BA and ATBBAT\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 AB=AA1=A1A=BA\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 ATBBAT\displaystyle A^TB \neq BA^T.
 
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 AB1=B1A\displaystyle AB^{-1}=B^{-1}A

But that is pretty much analogous to the above.

Does that help?.
 
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