If $X$ is symmetric, show $k(X^2)$ = $k(X)^2$ Suppose we have a symmetric invertible matrix $X$. How can I find $k(X^2)$ in terms of $k(X)$?
Note that $k$ represents the condition number operation, that is $k(X) = \|X\|\,\|X^{-1}\|$.
So, after messing around in Matlab for a while, I think $$k(X^2) = k(X)^2$$ although I dont know how to prove it.
I think I need to take advantage of the fact that symmetric matrices can be diagonalized by orthogonal matrices, but I wouldn't know where to go from there. Does anybody know how to show this?

(this assumes that $X$ has real entries; otherwise we would need $X$ selfadjoint and not just symmetric)
As you say, if $X$ is symmetric and invertible, then
$$
X=UDU^*, 
$$
where $D$ is diagonal with nonzero diagonal entries $d_1,\ldots,d_n$, and $U$ is a unitary. We can have the diagonal entries of $D$ ordered so that $|d_n|=\max\{|d_1|,\ldots,|d_n|\}$ and $|d_1|=\min\{d_1,\ldots,d_n\}$.
We have
$$
\|X\|=\|UDU^*\|=\|D\|=d_n,\ \ \|X^{-1}\|=\|UD^{-1}U^*\|=\frac1{d_1}.
$$
So $k(X)=|d_n/d_1|$. Similarly, $X^2=UD^2U^*$, so $k(X^2)=d_n^2/d_1^2$. In other words,
$$
k(X^2)=k(X)^2.
$$
