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- #1

- Thread starter Yankel
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- Thread starter
- #1

- Feb 15, 2012

- 1,967

so:

$A^* = (A^*(A^*)^{-1}A)^* = A^*((A^*)^{-1})^*A$

therefore:

$A^*A^{-1} = A^*((A^*)^{-1})^*$

and multiplying on the left by $(A^*)^{-1}$ we get:

$A^{-1} = ((A^*)^{-1})^*$

so

$(A^{-1})^* = ((A^*)^{-1})^{**} = (A^*)^{-1}$

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