An equality with matrices
An equality with matrices
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.
$$A - \left ( A^{-1} +\left ( B^{-1} - A \right )^{-1} \right )^{-1} = ABA$$
provided all the inverses appearing on the left hand side exist.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
- Tolaso J Kos
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Re: An equality with matrices
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*}
\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*}
Imagination is much more important than knowledge.
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