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A normal matrix defined over a complex vector space has the property, that $\|A\|_2$ is its largest eigenvalue and now I was wondering whether this is also true for matrices defined over the real numbers, just with the difference, that now we are talking about the absolute value instead of the eigenvalues itself, as there is no unitary diagonalization possible? Wikipedia suggests this in the article "normal matrix", but they do not make any difference between the real and the complex numbers. And further, I was wondering whether the condition $\|A\|_2=|\lambda_{\max}|$ is somehow related to $\max\limits_{\|v\|=1} \langle Av,v \rangle = |\lambda_{\max}|$?

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The operator 2-norm is defined by $$ \|A\|_2 = \max_{\|v\|_2\neq0}\frac{\|Av\|_2}{\|v\|_2} = \max_{\|v\|_2=1}\|Av\|_2 = \sigma_1(A), $$ where $\sigma_1(A)$ denotes the largest singular value of $A$. This definition works for all (rectangular) matrices over both $\mathbb{R}$ and $\mathbb{C}$.

When $A$ is a normal matrix over $\mathbb{R}$ or $\mathbb{C}$, its singular value is equal to the largest modulus of the its eigenvalues (over $\mathbb{C}$), hence $\|A\|_2=|\lambda|_\max(A)=\rho(A)$, the spectral radius of $A$.

When $A$ is a normal matrix and the underlying field is $\mathbb{C}$, $\|A\|_2$ is also equal to $\max_{\|v\|_2=1}|\langle Av,v\rangle|$, but this is in not true over $\mathbb{R}$ (for a counterexample, consider a $2\times2$ rotation matrix for an angle $\neq0,\pi$), unless $A$ is Hermitian. However, even if $A$ is Hermitian, the following three quantities in general do not coincide:

  • $|\lambda|_\max(A)$ (the maximum of the absolute values of eigenvalues of $A$),
  • $|\lambda_\max(A)|$ (the absolute value of the maximum eigenvalue of $A$), and
  • $\lambda_\max(A)$ (the maximum eigenvalue of $A$).

So, it is inaccurate to say that "$\|A\|_2$ is the largest eigenvalue of $A$" (as you did in your question). If $A$ is positive semidefinite (over $\mathbb{R}$ or $\mathbb{C}$), the three quantities coincide.

user1551
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  • but is $||A||=\sigma_1(A)$ true only if the regard the 2-norm or for each operator norm? –  Jun 03 '13 at 12:54
  • @Lipschitz This, of course, is not true for all operator norms, otherwise all operator norms would be identical, which is obviously wrong. – user1551 Jun 03 '13 at 12:59
  • yeah, but we are particularily looking at normal matrices. and i thought that i read such a theorem, that actually says so. are you able to construct a normal matrix so that this is obviously wrong? –  Jun 03 '13 at 13:02
  • maybe you have a look at http://en.wikipedia.org/wiki/Normal_matrix#Equivalent_definitions the last sentence there. –  Jun 03 '13 at 13:02
  • there they suggest that it does NOT depend on which operator norm you take that $||A||=\sigma_1(A)$, but you argued that this would be wrong in general. –  Jun 03 '13 at 13:26
  • @Lipschitz No. The operator norm in the Wikipedia article means the 2-norm. When the $p$ in "operator $p$-norm" is unspecified, it is usually understood as $2$. The operator $p$-norms for normal matrices are not identical to each other. For instance, when $A=nI_n-J$ where $J$ is the all-one matrix, $|A|_2=n\neq|A|_1=2(n-1)$. – user1551 Jun 03 '13 at 13:50
  • I do not get this. your example suggests that e.g. for n=2, this is the matrix with the first row being (1,-1) and the second one (-1,1). This matrix has the largest eigenvalue 2 and the 1-operatornorm is the largest absolute value of the columns, so |1|+|-1|=2. I do not understand your example, sorry. –  Jun 03 '13 at 14:05
  • @Lipschitz They happen to be equal when $n=2$. In fact, $n=2(n-1)$ if and only if $n=2$. Take any $n>2$ and you get a counterexample. – user1551 Jun 03 '13 at 14:12