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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?

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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
@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
$\|A\|_2=\max\|Ax\|_2$, 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
Thanks; it would be good if you could make a formal answer. –  user506901 Sep 13 '12 at 15:39

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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.

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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? –  ClausW 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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