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$A\in \mathbb{C}^{n\times n}$. Set: $$ r(A) = \max_{||u||_2=1}|u^* Au| $$

Prove the following statements:

(1) $||A||_2 \leq 2 r(A)$

(2)if $A^*A=AA^*$ then $r(A)=||A||_2$

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The way this is phrased makes it sound like these are exercises from a book, or a course. If so, then ideally this should be acknowledged (at least in my personal opinion). – user16299 Mar 1 '12 at 4:08

Write $A=\frac{A+A^*}2+\frac{A-A^*}2$. Let $H_1:=\frac{A+A^*}2$, then $H_1$ is hermitian and is therefore diagonalizable. Thanks to that we can see that $||H_1||=r(H_1)$. Since $H_2:=\frac{A-A^*}2$ is anti-hermitian it's in particular a normal matrix, so $H_2$ is unitary diagonalizable and $||H_2||=r(H_2)$. Since $r(H_1)\leq r(A)$ and $r(H_2)\leq r(A)$ we have $$||A||_2\leq ||H_1||_2+||H_2||_2=r(H_1)+r(H_2)\leq 2r(A).$$

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could you please help me again? The problem is now reedited. – BerSerK Feb 26 '12 at 3:16
"Since $H_2$ is anti-hermitian it's in particular an unitary matrix", are you saying every anti-hermitian matrix is a unitary matrix? I don't believe that is true. – Calle Feb 26 '12 at 3:28
These are certainly not all unitary (only those with eigenvalues $\pm i$) but that is also not required. $H_2$ is diagonalizable with respect to an orthonormal basis and that suffices. (To see it note that $iH_2$ is hermitian or, more general, $H_2$ is normal.) – WimC Feb 26 '12 at 7:47
I probably made a confusion, thinking that if $A$ is anti-hermitian, $A^*A=-AA=A(-A)=AA^*$, but I still don't know why I wrote unitary instead of normal. – Davide Giraudo Feb 26 '12 at 9:55

For part (2) note that $A$ is normal in this case and hence is diagonalizable with respect to an orthonormal basis. (See here for more about normal operators.) As noted above, this implies that $||A||_2 = r(A)$: If $||v||_2=1$ and $v = a_1e_1 + \dotsc + a_ne_n$ in this basis then

$$ |v^{\ast}Av| = |\langle Av, v \rangle| = \left|\lambda_1 |a_1|^2 + \dotsc + \lambda_n |a_n|^2 \right| \leq \max_k(|\lambda_k|) \cdot (|a_1|^2 + \dotsc + |a_n|^2) = \max_k(|\lambda_k|) $$

and this maximum is attained for some $e_k$ and so $r(A) = \max_k(|\lambda_k|)$. A similar computation shows that this indeed equals $||A||_2$.

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