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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 toa 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 posted by The K9Kotter Crew - Dec 15, 2018 Are you a dog lover who supports rescues? Of course you do! No one wants to see a dog "Die" - beit at a shelter, in a home, or at a farm - without knowing that a rescue was involved. Supporting rescues is important, and if you
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