# Bounding the size of the inverse of $I+AB$ when $A$ and $B$ are both PSD

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;
disp(norm(inv(Z)));


yeilds 1.14

• 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. – struggling_economist Jun 20 at 7:49
• Sorry, I update the answer. Now $A$ is PD and $B$ is PSD. – Arastas Jun 20 at 8:16