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April 6th, 2016, 09:18 AM   #1
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Proof concerning eigenvalue and matrix norm.

Hi, I am struggling with the following proof. I think I know how to prove the right hand side inequality |lambda| <= ||A||, but I still don't know how to prove the left hand side inequality.

Could you please give me some hint ? Attached Images Proof.jpg (13.7 KB, 23 views) May 22nd, 2016, 02:08 AM #2 Senior Member   Joined: Aug 2012 Posts: 229 Thanks: 3 Hey hegendroffer. Hint - Use the sub-multiplicative norm with respect to the eigen-decomposition of the matrix and consider the result of the inverse of the matrix (in terms of how the eigen-values become reciprocated). The eigen-decomposition should have three matrices of PDP_inverse and the sub-multiplicative law of matrix norms in combination with how eigen-values are determined for the original matrix and the inverse relate to that result. June 2nd, 2016, 04:30 AM #3 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Let ui and $\displaystyle \lambda_{i}$ be eigenvectors and eigenvalues of A x=x1u1+..xnun, ||x||=1 ||A||=max||Ax|| ||Ax||=||$\displaystyle \lambda_{1}$x1u1+...$\displaystyle \lambda_{n}$xnun|| $\displaystyle \geq$ |$\displaystyle \lambda_{1}$|,..,|$\displaystyle \lambda_{n}$| ||A|| $\displaystyle \geq |\lambda|$ Au=$\displaystyle \lambda$ u A$\displaystyle ^{-1}$u=1/$\displaystyle \lambda$ u ||A$\displaystyle ^{-1}$||$\displaystyle \geq$ 1/|$\displaystyle \lambda$| Tags eigenvalue, matrix, norm, proof 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 mophiejoe Linear Algebra 1 April 1st, 2015 06:34 PM calypso Linear Algebra 3 February 4th, 2015 02:16 PM johnnyb Linear Algebra 3 September 17th, 2013 11:42 AM onako Linear Algebra 0 February 1st, 2012 12:59 AM ricconna Linear Algebra 0 August 23rd, 2011 08:55 PM

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