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.


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 $\rho(S)=1/2\rho((A-I)B)$.

  • $\begingroup$ 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? $\endgroup$ – Derek Thompson Mar 5 at 22:08
  • $\begingroup$ Of course, yes and yes. $\endgroup$ – loup blanc Mar 5 at 22:18
  • $\begingroup$ Is there possibly a way to compute it via $|x^*Ax|$, where $x$ comes from the eigenspaces of $B$? $\endgroup$ – Derek Thompson Mar 6 at 11:36

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