glamrockqueen7

2021-10-28

If $\alpha \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\beta$ are transformations on a set S, prove that ${\alpha }^{-1}\circ {\beta }^{-1}$ and ${\beta }^{-1}\circ {\alpha }^{-1}$ are transformations. Which is the inverse of $\alpha \circ \beta$ ? Prove your answer.

Bentley Leach

Step 1
Linear transformation are represented as the matrix .
${\left(AB\right)}^{-1}\left({B}^{-1}{A}^{-1}\right)=I$
If A and B are two matrix then ${\left(AB\right)}^{-1}={B}^{-1}{A}^{-1}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}},\left({B}^{-1}{A}^{-1}\right){\left(AB\right)}^{-1}=I$
So,${\left(\alpha \circ \beta \right)}^{-1}={\beta }^{-1}\circ {\alpha }^{-1}$
Let us prove the ${\left(\alpha \circ \beta \right)}^{-1}={\beta }^{-1}\circ {\alpha }^{-1}$
Let us consider $s\in S$
Show that
${\left(\alpha \circ \beta \right)}^{-1}\circ \left({\beta }^{-1}\circ {\alpha }^{-1}\right)\left(s\right)={I}_{s}\left(s\right)$
$\left({\beta }^{-1}\circ {\alpha }^{-1}\right)\circ {\left(\alpha \circ \beta \right)}^{-1}\left(s\right)={I}_{s}\left(s\right)$
${\left(\alpha \circ \beta \right)}^{-1}\circ \left({\beta }^{-1}\circ {\alpha }^{-1}\right)\left(s\right)={\alpha }^{\beta \left({\beta }^{-1}{\alpha }^{-1}\left(s\right)\right)}$
$=\alpha \left({I}_{s}\left({\alpha }^{-1}\left(s\right)\right)\right)$
$=\alpha \left({\alpha }^{-1}\left(s\right)\right)$
$={I}_{s}\left(s\right)$
Step 2
$\left({\beta }^{-1}\circ {\alpha }^{-1}\right)\circ {\left(\alpha \circ \beta \right)}^{-\left\{1\right\}}\left(s\right)={\beta }^{-1}\left({\alpha }^{-1}\alpha \left(\beta \left(s\right)\right)\right)$
$=\beta \left({I}_{s}\left({\beta }^{-1}\left(s\right)\right)\right)$
$=\beta \left({\beta }^{-1}\left(s\right)\right)$
$={I}_{s}\left(s\right)$
So,${\left(\alpha \circ \beta \right)}^{-1}={\beta }^{-1}\circ {\alpha }^{-1}$
Hence proved

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