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April 6th, 2016, 09:18 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 ? ![]() |
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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. |
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June 2nd, 2016, 04:30 AM | #3 |
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,329 Thanks: 94 |
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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eigenvalue, matrix, norm, proof |
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