If $A$ and $B$ are both positive semi-definite matrices, is it possible to show that

$$\left\Vert \left(I+AB\right)^{-1}\right\Vert _{2}\leq1$$

where $\left\Vert \cdot\right\Vert _{2}$ is the operator norm?


Let $A = \begin{bmatrix}1&2\\2&5\end{bmatrix}$ and $B = \begin{bmatrix}1&-1\\-1&2\end{bmatrix}$.

Clearly, both $A$ and $B$ are symmetric. Also, you can check that the eigenvalues of $A$ are $3\pm\sqrt{2} > 0$ and the eigenvalues of $B$ are $\dfrac{3\pm\sqrt{5}}{2} > 0$. Hence, $A$ and $B$ are both PSD matrices.

But, $(I+AB)^{-1} = \begin{bmatrix}1&-1/3\\1/3&0\end{bmatrix}$, which has norm $\|(I+AB)^{-1}\|_2 = \dfrac{1}{2}+\dfrac{\sqrt{13}}{6} > 1.$

Thanks to Robert Israel's answer here for an example of two PSD matrices whose product isn't PSD.


If $A$ and $B$ commute, i.e., $AB=BA$, then it is true, where the proof is rather straightforward. If $A$ and $B$ commute, then the answer is negative since $AB$ is not necessary PSD.

Example: $$A=\begin{bmatrix}10 & -3\\-3 & 3\end{bmatrix}, \quad B = \begin{bmatrix}2 & -3\\-3 & 4.5\end{bmatrix},$$ where $A$ is PD and $B$ is PSD. Then (with Matlab)

A=[10 -3;-3 3];
B=[2 -3; -3 4];
Z = eye(2)+A*B;

yeilds 1.14

  • $\begingroup$ In this example B is not PSD (in either text or code), but it probably still works if the last digit of B is changed to 5. $\endgroup$ – struggling_economist Jun 20 at 7:49
  • $\begingroup$ Sorry, I update the answer. Now $A$ is PD and $B$ is PSD. $\endgroup$ – Arastas Jun 20 at 8:16

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