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An equality with matrices

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Riemann
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An equality with matrices

#1

Post by Riemann » Sat Jan 21, 2017 9:52 pm

Let $A, B$ be $3 \times 3$ matrices with real entries. Prove that

$$A - \left ( A^{-1} +\left ( B^{-1} - A \right )^{-1} \right )^{-1} = ABA$$

provided all the inverses appearing on the left hand side exist.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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Tolaso J Kos
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Re: An equality with matrices

#2

Post by Tolaso J Kos » Fri Nov 06, 2020 6:50 am

Let $A, B$ be elements of an arbitrary associative algebra with unit. Then:

\begin{align*}
\left ( A^{-1} +\left ( B^{-1} - A \right )^{-1} \right )^{-1} &= \left ( A^{-1} \left ( B^{-1} - A \right )\left ( B^{-1} - A \right )^{-1} + A^{-1} A \left ( B^{-1} - A \right )^{-1} \right )^{-1} \\
&=\left ( A^{-1} \left ( \left ( B^{-1} - A \right ) +A \right )\left ( B^{-1} -A \right )^{-1} \right )^{-1} \\
&= \left ( A^{-1} B^{-1} \left ( B^{-1} -A \right )^{-1} \right )^{-1}\\
&= \left ( B^{-1}-A \right ) BA \\
&= A - ABA
\end{align*}
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