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 October 25th, 2014, 01:45 PM #1 Newbie   Joined: Oct 2014 From: Edinburgh, Scotland Posts: 3 Thanks: 0 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 01:50 PM.
October 25th, 2014, 02:01 PM   #2
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 Originally Posted by iluvmafs 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.

October 25th, 2014, 02:38 PM   #3
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 Originally Posted by Pero 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

 October 25th, 2014, 02:51 PM #4 Newbie   Joined: Oct 2014 From: Edinburgh, Scotland Posts: 3 Thanks: 0 So to prove this, do I have to show that $\displaystyle A^{-1}={(A^{-1})}^T$? Last edited by iluvmafs; October 25th, 2014 at 02:57 PM.

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