Do almost commuting matrices almost commute with the square root?

Question

If $A$ and $B$ are $n\times n$ self-adjoint matrices with $B$ positive semidefinite and $\|B\|_F\leq1$ such that $$\|AB-BA\|_F\leq \epsilon,$$ where $\|A\|^2_F:=tr(A^*A)$ is the Frobenius (or Hilbert-Schmidt) norm. I'd like to know if there is a constant $C>0$ such that $\|A\sqrt{B}-\sqrt{B}A\|_F<C\epsilon$ is independent of $n$.

Using the solution to this question one can use the fact that there is a polynomial of degree at most $n$ such that $p(B)=\sqrt{B}$ and it follows that there is a constant $K>0$ such that for any polynomial $p(B)$ we see that $$\|Ap(B)-p(B)A\|_F\leq Kn\epsilon,$$ just using the submutliplicativity of $\|\cdot\|_F$ and the triangle inequality. However, in the case where $p(B)=\sqrt{B}$, I would like to know if there is a bound that depends on $\epsilon$ but not $n$.

Edit: in the comments of the linked question, the author of the answer has noted that the degree of the polynomial is at least the number of distinct eigenvalues of $B$, which makes sense if you are using the polynomial that interpolates the eigenvalues.

I guess that $| \cdot |_F$ denotes the "Frobenius norm"? I would tend to call it the Hilbert-Schmidt norm or simply the entrywise $L^2$-norm. Anyway, whatever the norm is called, you should explain what norm you're referring to in the the post. — Mike F, Apr 16 '21 at 21:58

score 2 · Accepted Answer · answered Apr 20 '21 at 13:50

2

No. Pick a self-adjoint matrix $X$ and an orthogonal projection $Y\ne I$ such that $\|XY-YX\|_F=\epsilon>0$. Let $t>0,\ A=\frac1tX$ and $B=tY$. Since $Y$ is an orthogonal projection, we have $\sqrt{Y}=Y$ and $\sqrt{B}=\sqrt{t}\,Y$. Therefore $\|AB-BA\|_F=\epsilon$ but $\|A\sqrt{B}-\sqrt{B}A\|_F=\frac{1}{\sqrt{t}}\epsilon$. When $t\to0$, we have $\|B\|_F\le1$ but $\|A\sqrt{B}-\sqrt{B}A\|_F\to\infty$.

answered Apr 20 '21 at 13:50

user1551

139,064

2

@Zerox Aren't orthogonal projections precisely those which are self-adjoint? – Rammus Apr 20 '21 at 14:03
@Rammus Although orthogonal projections are self-adjoint, where I made a mistake, not every self-adjoint operater is a pure orthogonal projection, such as $2 \cdot \operatorname{Id}$. – Zerox Apr 20 '21 at 14:16
@Zerox Yes, I meant a projection is orthogonal iff it is self-adjoint. – Rammus Apr 20 '21 at 14:19
I wonder if the statement holds with the additional assumption that $A$ is an orthogonal projection? – Condo Apr 27 '21 at 16:22
@Condo Wait. I forgot that you need $C$ to be independent of $n$. Now I am not sure... – user1551 Apr 27 '21 at 16:42
@user1551 but when $t\rightarrow \infty$ wont $|B|$ have unbounded operator norm? I want $|B|_F\leq 1$. – Condo May 21 '21 at 19:26
1

@Condo I considered $t\to0$ in my answer. Not $t\to\infty$. – user1551 May 21 '21 at 19:39
@user1551 yes, you are correct! my bad. – Condo May 21 '21 at 19:44

Do almost commuting matrices almost commute with the square root?

1 Answers1