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Category:Electronic design automation
Category:Electronic circuit simulatorsQ:

Absolute value of a product of matrices

Suppose $A$ and $B$ are $n\times n$ matrices and $AB$ and $BA$ are $n\times n$ matrices, does $|AB| = |BA|?$
I can get to a point of proving that $|AB| \le |BA|$ since \begin{align} |AB|^2 = AB\cdot AB &= (A\cdot B + BA) \cdot (A\cdot B + BA)\\ |BA|^2 = BA\cdot BA &= (B\cdot A + AB) \cdot (B\cdot A + AB)\\ \end{align}
I’m wondering if there’s a clean way to show $AB \le BA$ though since the above derivation doesn’t seem to be very useful.

A:

In a general Banach space the answer is yes.
If $X$ is a finite dimensional Banach space $||AB|| \le ||A||||B||$ and your claim follows.
It may not be true in infinite dimensional Banach space if $A$ is not bounded.
For example consider $X=c_0$, with the product topology. Take $A=e_1$.
Then, for $B=e_2$, $||A||=1$ and $||B||=2$ so that
||AB||=||Ae_2||=||\{1,0\}||=\sqrt 2Lovable Dog Failures

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