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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. Tags invertible, matrix, proof, symmetric Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Magnesium Linear Algebra 2 December 11th, 2013 02:09 AM shine123 Linear Algebra 1 September 21st, 2012 08:47 AM frankpupu Linear Algebra 3 March 6th, 2012 11:45 AM problem Linear Algebra 3 August 31st, 2011 05:30 AM 450081592 Linear Algebra 1 June 24th, 2010 07:13 AM

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