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In relation to the Euclidean norm...

What are the conditions for when this occurs? Is it only real symmetric matrices?

When is this not the case?

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    Which norm do you mean?2012-09-19
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    Euclidean norm... edited not. Thx.2012-09-19
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    @DirkCalloway Surely you mean the operator norm of the matrix.2012-09-19
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    So the matrix norm subordinated to the euclidian norm, namely $\sup_{x\neq 0}\frac{|Ax|}{|x|}$, where $|\cdot|$ is the euclidian norm?2012-09-19
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    @DavideGiraudo I think so... I've been taking the norms of square matrices using the Euclidean norm command in Maple and getting this result over and over and was wondering the conditions for it...2012-09-19
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    And do you mean the largest in modulus? Because nothing _a priori_ gives that the eigenvalues are non negative, which can of course happen even when the matrix is symmetric. And when you don't assume this, the eigenvalues can be complex and not real.2012-09-19
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    Okay, how about if the matrix is SPD with all real eigenvalues?2012-09-19
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    In this case, the matrix is symmetric and we can work with spectral theorem.2012-09-19
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    @DirkCalloway: If the matrix is diagonalizable, you have the largest absolute value of the eigenvalues. Since a positive semidefinite matrix (I assume that's what you meant with SPD) has only positive eigenvalues, for this it is the same as the largest eigenvalue. I don't know what happens for non-diagonalizable matrices, though.2012-09-19
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    @celtschk Yeah, that's what I've been seeing. The absolute value of the largest eigenvalue keeps coming up as the norm. I guess this is related to the spectral theorem which I haven't gotten into yet?By SPD I meant symmetric postive definite...2012-09-19

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