What are the eigenvalues of $A(A+I)^{-1}$ in terms of the eigenvalues of SPSD matrix $A$?

Solution 1:

Let $A = Q\Lambda Q^T$ be the spectral decomposition. Then $A + I = Q\Lambda Q^T + QQ^T = Q(\Lambda + I)Q^T$. Thus, when we compute the inverse, we get that

$$A(A+I)^{-1} = Q\Lambda Q^T (Q(\Lambda+1)Q^T)^{-1} = Q \Lambda Q^T Q (\Lambda+I)^{-1} Q^T = Q \Lambda(\Lambda+I)^{-1} Q^T.$$

Thus is $\lambda_i$ is an eigenvalue of $A$, we have that $\frac{\lambda_i}{\lambda_i+1}$ is an eigenvalue of $A(A+I)^{-1}$.

In general if all matrices are simultaneously diagonalizable then, we can just assume that our matrices are diagonal and derive the result for this case to get the general result.