Suppose that we are dealing with positive semidefinite $n\times n$
symmetric real matrices.The induced matrix norm of $A$ is defined as $$
|A|=\max\lambda(A)=\max_{x\in\mathbb{R^n},|x|=1}x'Ax. $$ Here,
$\lambda(A)$ denotes the set of eigenvalues of $A$. Is it true that
this matrix norm is continuous: if $A_n\to A$ (component-wise),
then $|A_n|\to |A|$?
I vaguely imagine that since eigenvalues are solutions to the characteristic polynomial, which are built using entries of the matrix, continuity is likely to hold. But I would love to see a rigorous proof (or a counter example).
Thank you very much. I apologize ahead if this is a duplicate. (I did search.)
Best Answer
It is maybe a too big cannon to use (or prove) the continuity of eigenvalues. It is well known that the (matrix) norms are continuous. This follows from the triangle inequality: $$ \|A\|\leq\|B\|+\|A-B\|, \quad \|B\|\leq\|A\|+\|A-B\|\quad\Rightarrow \quad |\|A\|-\|B\||\leq\|A-B\|. $$ Therefore, if $A$ and $B$ are close then $\|A\|$ and $\|B\|$ are close as well.
Then use the fact that $\|A\|=\rho(A)$ when $\|\cdot\|:=\|\cdot\|_2$ is the spectral norm and $A$ is SPD.
You can relate the componentwise closeness to the normwise closeness by realizing that $$ \|A\|_2=\max_{\|x\|_2=1}\|Ax\|_2\leq n\max_{1\leq i,j\leq n}|a_{ij}|=:n\|A\|_{\max} $$ (or by simply showing that $\|A\|_{\max}$ is a norm and use the fact that norms on a finite-dimensional space are equivalent). Therefore, if $|a_{ij}-b_{ij}|<\epsilon$ for all $i,j=1,\ldots,n$ and some $\epsilon>0$, we have $|\|A\|_2-\|B\|_2|<\delta$ with $\delta:=\epsilon/n$.