# Norm of AB+BA when A, B are well understood

Here is the situation:

I have the operator/matrix expression AB+BA and want to know the norm.

1.) B is compact, so I believe AB+BA is compact (as are AB and BA, so any finite-dimensional matrix answer should be fine, or at least informative)

2.) A is both unitary and self-adjoint (and therefore A^2 = I)

3.) AB+BA itself is self-adjoint, because I also know that AB and BA are self-adjoint.

4.) I know the norm, eigenvalues, basically everything about both A and B (and A* and B*), AND I know all of this about AB and BA (and their adjoints) as well. An expression in terms of the norms/eigenvalues of those four operators would give me everything I want.

5.) A sharp inequality would be okay but really I need to find the exact expression for the norm or the max eigenvalue.

## 1 Answer

In finite dimension $$n$$.

According to 2), $$A$$ is an orthogonal symmetry and is orthogonally similar to $$diag(I_p,-I_{n-p})$$.

We assume that $$A=diag(I,-I)$$ and $$B=\begin{pmatrix}P&Q\\R&S\end{pmatrix}$$.

Then $$AB+BA=2diag(P,-S)$$.

Since $$AB$$ is self adjoint, $$P=P^*,S=S^*,Q^*=-R$$.

We obtain, $$||AB+BA||_2=\rho(AB+BA)=2\max(\rho(P),\rho(S))$$.

EDIT 1. Let $$E_1=\ker(A-I_n),E_{-1}=\ker(A+I_n)$$.

Clearly, $$M_1=\max_{x\in E_1,||x||_2=1}|x^*Bx|=\rho(P)$$, $$M_{-1}=\max_{x\in E_{-1},||x||_2=1}|x^*Bx|=\rho(S)$$.

Finally, $$||AB+BA||_2= 2\max(M_1,M_{-1})$$.

EDIT 2. We can also use $$\rho(P)=1/2\rho((A+I)B)$$ and $$\rho(S)=1/2\rho((A-I)B)$$.

• Thanks for the start here, this is helpful! Is $p$ here just a positive integer $\leq n$? Also, is $\rho$ here meant to indicate spectral radius, i.e. max eigenvalue? – Derek Thompson Mar 5 at 22:08
• Of course, yes and yes. – loup blanc Mar 5 at 22:18
• Is there possibly a way to compute it via $|x^*Ax|$, where $x$ comes from the eigenspaces of $B$? – Derek Thompson Mar 6 at 11:36