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Recently I started exploring convergence of some iterative methods and spotted the equivalent of the spectral radius and a matrix norm.

For instance, http://www.scribd.com/doc/37323755/36/Richardson-Iteration states in Example 1.28 that 2-norm of a matrix is its spectral radius. On the other hand, What is the difference between the Frobenius norm and the 2-norm of a matrix? states a difference.

What is the difference between the Frobenius and 2-norm of a matrix? Is the class of symmetric matrices for which the equality of 2-norm and the spectral radius holds?

Rodrigo de Azevedo
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user506901
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  • Frobenius norm is a matrix norm while 2-norm is a vector norm. The 2-norm of a matrix is defined in terms of vectors 2-norm (look for the definition). On the other hand matrix norms are defined (axiomatically) directly for matrices. – fmoura2005 Sep 13 '12 at 12:47
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    @fmoura2005 What is then 2-norm of a matrix $A\in\mathbb{R}^{n\times n}$? What is its relation to the spectral radius of $A$? – user506901 Sep 13 '12 at 13:00
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    $|A|_2=\max|Ax|_2$, over $|x|_2=1$. This max coincides with maximum singular value of $A$. This, by the way, answer your question about the symmetric matrices. In general, the spectral radius is less or equal than the matrix norm. – fmoura2005 Sep 13 '12 at 13:57
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    Thanks; it would be good if you could make a formal answer. – user506901 Sep 13 '12 at 15:39

1 Answers1

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OK. Regarding your first question, the difference I see is that Frobenius norm is a matrix norm while a matrix 2-norm is induced by the vector 2-norm, i.e., $\|A\|_2=\max_{\|x\|=1}\|Ax\|_2$. In fact, $\|A\|_2$ is the maximal singular value of $A$, that is, the square root of the maximal eingenvalue of $A^TA$ (this more computable). This also answer your second question. In general the spectral radius of a matrix is less or equal than the matrix norm.

paperskilltrees
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fmoura2005
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  • Could you explain why the matrix 2-norm is also equal to the square root of the maximal eigenvalue of A^TA? – Hku Sep 16 '12 at 22:38
  • @ClausW. Oh Yes. In fact, since $A^TA$ is symetric, it has an ortornormal set of eingenvectors (which is a basis for $\mathbb{R}^n$). So, you can easily deduce that $\max\frac{x^TA^TAx}{x^Tx}$ is exactly the maximum eigenvalue of $A^TA$ (this is the Rayleigh quotient); but $\max\frac{x^TA^TAx}{x^Tx}$ is $|A|^2$! – fmoura2005 Sep 24 '12 at 20:40
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    Why would $||A||_2$ be the square root of the maximum singular value of A? I thought it is directly equal to the max. singular value of A. – Ondrej Skopek Feb 09 '16 at 13:01
  • @OndrejSkopek This was clearly a mistake, which is now corrected in the edit. – paperskilltrees Feb 09 '21 at 17:21