The operator norm of complex matrices

Let $M_{n}$ be the algebra of $n\times n$ complex matrices. By identifying $M_{n}$ with $B(\mathbb{C^{n}})$, the set of all bounded linear maps from the n-dimensional Hilbert space $\mathbb{C^{n}}$ to $\mathbb{C^{n}}$, with operator norm, i.e. $\|x\|=sup_{\eta~\in~\mathbb{C^{n}},~\|\eta\|~\leq1}|x(\eta)|$. It is easy to see that $M_{n}$ is a Banach algebra.

But I do not know how to compute the exact value of norm of a specific $M_{n}$. For instance, if $a=\left(\begin{array}{ccc} 1/2 & 1 & 0 \\ 0 & 1/2 & 0 \\ 0 & 0 & 1/3 \end{array}\right)$ , according to the operator norm defined above, how to compute its norm $\|a\|$?

• An equivalent definition of $\|A\|$ is $\sup_{\|x\|=1}|\langle Ax,x\rangle|$. Can you work from there? – Ian Coley Sep 19 '13 at 18:58
• In general, this is not an easy problem. In this case you have to maximize a function of 3 variables subject to 1 constraint, so it can be done. Because of the direct-sum structure in your example, you can simplify further. – GEdgar Sep 19 '13 at 19:01
• @IanColey: No, that is not the norm in question. That is the numerical radius, and it can be as small as $\frac12\|A\|$ even for $2$-by-$2$ matrices. – Jonas Meyer Sep 20 '13 at 5:12
• @GEdgar I think you are right. It is not easy to compute the norm of a matrix. – Insomnia Sep 20 '13 at 17:18

Alternatively, you can use the fact that for any matrix $M$ we have $$\Vert M\Vert=\sup\{ \sqrt\lambda;\; \lambda\;\hbox{eigenvalue of}\; M^*M\}\, ,$$ where $M^*$ is the adjoint matrix.
For this particular example, it should not be too difficult to find the eigenvalues of the $3\times 3$ matrix $M^*M$; but of course this is hopeless for a "general" larger matrix.
• Could you explain to me why this equation $\|M\|=sup\{\sqrt{λ}: λ$ is eigenvalue of $M^{∗}M\}$ hold? – Insomnia Sep 20 '13 at 2:13
• A detailed answer would be a little bit too long for a "comment". Here is briefly how it goes. First, $M^*M$ is a self-adjoint nonnegative matrix, so it can be put in diagonal form in some orthonormal basis, and its eigenvalues are $\geq 0$. Next, you know that $\Vert M\Vert^2=\langle M^*Mx,x\rangle$ for any $x\in\mathbb C^n$. Using this, you can express $\Vert Mx\Vert^2$ in termes of the eigenvalues of $M^*M$ and the coordinates of $x$ (in the good basis where $M^*M$ is diagonal). then you should be able to conclude. – Etienne Sep 20 '13 at 22:23