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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 ?
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 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 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,548 Thanks: 111 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/|$\displaystyle \lambda$|

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