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April 6th, 2016, 10:18 AM  #1 
Newbie Joined: Jun 2015 From: Warsaw Posts: 3 Thanks: 1  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 ? 
May 22nd, 2016, 03:08 AM  #2 
Senior Member Joined: Aug 2012 Posts: 229 Thanks: 3 
Hey hegendroffer. Hint  Use the submultiplicative norm with respect to the eigendecomposition of the matrix and consider the result of the inverse of the matrix (in terms of how the eigenvalues become reciprocated). The eigendecomposition should have three matrices of PDP_inverse and the submultiplicative law of matrix norms in combination with how eigenvalues are determined for the original matrix and the inverse relate to that result. 
June 2nd, 2016, 05:30 AM  #3 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,214 Thanks: 91 
Let ui and $\displaystyle \lambda_{i}$ be eigenvectors and eigenvalues of A x=x1u1+..xnun, x=1 A=maxAx 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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eigenvalue, matrix, norm, proof 
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