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October 25th, 2014, 02:45 PM   #1
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Question Proof with symmetric and invertible matrix

Qu. Prove that if a symmetric matrix is invertible, then its inverse is symmetric also.

So far I have this.. is it correct?

$\displaystyle
A=A^T \\

A^{-1}A=A^{-1}A^T\\

\mathbb{I}=A^{-1}A^T\\

\mathbb{I}{(A^T)}^{-1}=A^{-1}A^T{(A^T)}^{-1}\\

{(A^T)}^{-1}=A^{-1}\mathbb{I}\\

{(A^T)}^{-1}=A^{-1}\\

$

Last edited by iluvmafs; October 25th, 2014 at 02:50 PM.
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October 25th, 2014, 03:01 PM   #2
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Quote:
Originally Posted by iluvmafs View Post
Qu. Prove that if a symmetric matrix is invertible, then its inverse is symmetric also.

So far I have this.. is it correct?

$\displaystyle
A=A^T \\

A^{-1}A=A^{-1}A^T\\

\mathbb{I}=A^{-1}A^T\\

\mathbb{I}{(A^T)}^{-1}=A^{-1}A^T{(A^T)}^{-1}\\

{(A^T)}^{-1}=A^{-1}\mathbb{I}\\

{(A^T)}^{-1}=A^{-1}\\

$
What does that prove? As $\displaystyle A = A^T$ you can write down $\displaystyle A^{-1} = (A^T)^{-1}$ immediately.
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October 25th, 2014, 03:38 PM   #3
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Quote:
Originally Posted by Pero View Post
What does that prove? As $\displaystyle A = A^T$ you can write down $\displaystyle A^{-1} = (A^T)^{-1}$ immediately.
Oh yes, how did I miss that.. Thanks for pointing that out
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October 25th, 2014, 03:51 PM   #4
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So to prove this, do I have to show that $\displaystyle A^{-1}={(A^{-1})}^T
$?

Last edited by iluvmafs; October 25th, 2014 at 03:57 PM.
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