Let $\|\cdot\|_2·$ be the norm of $C^n$ defined by $\|x\|_2=(|x_1|^2+|x_2|^2+...+|x_n|^2)^{1/2}$.

Question

Let $\|\cdot\|_2·$ be the norm of $C^n$ defined by $\|x\|_2=\left(|x_1|^2+|x_2|^2+...+|x_n|^2\right)^{1/2}$. Prove that for the subordinate matrix norm $∥⋅∥_2$ must be $\|A\|_2=\max\{ \sqrt{|λ|}:λ \text{ is an eigenvalue of } AA^∗ \}$ where $A^*$ is the conjugate transpose of $A$.

Answer 1

Hint:

Since $AA^*$ is Hermitian, it is unitarily similar to a diagonal matrix, i.e. $$ AA^*=U\operatorname{diag}(|\lambda_1|,\cdots,|\lambda_n|)U^* $$ where $\lambda_i$ are eigenvalues of $AA^*$ and $UU^*=I$. Then $$ \|A\|_2^2=\sup_{\|x\|=1}\|Ax\|^2=\sup_{\|x\|=1}\sum_{1\leqslant i\leqslant n}|U^*x|_{i}^2|\lambda_i|=\max_{1\leqslant i\leqslant n}{(|\lambda_i|)} $$

Answer 2

$ A^*A$ is a nonnegative matrix. Then there is an unitary matrix $P$ such that $$A^*A=P^*\operatorname{diag}(\lambda_1,\ldots,\lambda_n)P$$ where the $\lambda_j$'s are the eigen values of $A^*A$ which happen to be all nonnegative. Then $$\|Ax\|^2_2=x^*A^*Ax=x^*P^*\operatorname{diag}(\lambda_1,\ldots,\lambda_n) Px$$ Since $\|A\|=\sup_{\|x\|_2}\|Ax\|_2$ $$\begin{align}\sup_{\|x\|_2=1}\|Ax\|^2=\sup_{\|x\|_2=1}\Big((Px)^*\operatorname{diag}(\lambda_1,\ldots,\lambda_n) Px\Big)\tag{1}\label{1}\end{align}$$ Since and $P$ is unitary ($P^*P=I$), $\|Px\|_2=\|x\|_2$ for all $x\in\mathbb{C}^n$. Letting $u=Px=[u_1\ldots u_n]^\intercal$ in \eqref{1} yields $$\begin{align}\sup_{\|x\|_2=1}\|Ax\|^2&=\sup_{\|u\|_2=1}\Big(u^*\operatorname{diag}(\lambda_1,\ldots,\lambda_n) u\Big)\\ &=\sup_{\|u\|_2=1}\lambda_1|u_1|^2+\ldots+\lambda_n|u_n|^2=\max_{1\leq k\leq n}\lambda_j \tag{2}\label{2}\end{align}$$ The conclusion follows by noticing that $AA^*$ and $A^*A$ have the same eigenvalues.