Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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
add comment

1 Answer

up vote 1 down vote accepted

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.

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.