Operator norm of square root of matrix vs original

Question

If I have a nonsymmetric matrix whose operator norm is $\leq 1$ and square root it, does its operator norm remain below $1$?

More formally, I want to know whether there is always at least one square root for which this is the case even if it isn't true for all of them and under reasonable assumptions that guarantee a square root exists.

Specifically, suppose I have a square nonsymmetric matrix $A$ with $\|A\|_2 \leq 1$ where $\|\cdot\|_2$ denotes the operator norm (max singular value). Does there always exist a square matrix $B$ such that $B^2=A$ and $\|B\|_2 \leq 1$?

I am willing to assume $A$ is diagonalizable, although not unitarily diagonalizable. I am also willing to assume $A$ is of the form $A' + \Delta$ for some small random perturbation $\Delta$ and matrix $A'$. Without at least one of these assumptions, the square root might not exist at all.

I would be interested in a proof / counterexample under any nonempty subset of these assumptions. For example, a counterexample showing that the statement can be false when we assume $A$ is diagonalizable would be useful to me.

Answer 1

This is not true. For example, if

$$ A = \frac{2}{1 + \sqrt{5}}\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix} $$

then $\|A\| = 1$. Furthermore, $A$ has exactly $2$ square roots, which are

$$ B_{\pm} = \frac{\pm\sqrt{2}}{\sqrt{1 + \sqrt{5}}}\begin{bmatrix} 1 & 1/2 \\ 0 & 1\end{bmatrix}. $$

It is straightforward to check that $\|B_{\pm}\| \approx 1.0068841364 > 1$.

The matrix $A$ here is not diagonalizable, but if we add $\varepsilon > 0$ to its bottom-right entry then it becomes diagonalizable. It will then have four square roots, two of which are continuous in $\varepsilon$ (and thus still have norm $> 1$ when $\varepsilon$ is small), and the other two of which have a huge top-right entry (and thus have norm $> 1$ when $\varepsilon$ is small).