# 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?

## 1 Answer

(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, $$k(X^2)=k(X)^2.$$

• Very late, but this was wonderful, thank you! Very clear explanation. Oct 2, 2016 at 22:05
• Glad I could help. Oct 2, 2016 at 22:08