# Let a_1,a_2… be the eigenvalues of symmetric positive semidefinite matrix A. How to find the eigenvalues of matrix A(A+I)^(-1)?

kasibug1v 2022-10-02 Answered
Let ${a}_{1},{a}_{2}\dots$ be the eigenvalues of symmetric positive semidefinite matrix A. How to find the eigenvalues of matrix $A\left(A+I{\right)}^{-1}$?
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## Answers (1)

Paige Paul
Answered 2022-10-03 Author has 11 answers
Let $A=Q\mathrm{\Lambda }{Q}^{T}$ be the spectral decomposition. Then $A+I=Q\mathrm{\Lambda }{Q}^{T}+Q{Q}^{T}=Q\left(\mathrm{\Lambda }+I\right){Q}^{T}$. Thus, when we compute the inverse, we get that
$A\left(A+I{\right)}^{-1}=Q\mathrm{\Lambda }{Q}^{T}\left(Q\left(\mathrm{\Lambda }+1\right){Q}^{T}{\right)}^{-1}=Q\mathrm{\Lambda }{Q}^{T}Q\left(\mathrm{\Lambda }+I{\right)}^{-1}{Q}^{T}=Q\mathrm{\Lambda }\left(\mathrm{\Lambda }+I{\right)}^{-1}{Q}^{T}.$
Thus is ${\lambda }_{i}$ is an eigenvalue of A, we have that $\frac{{\lambda }_{i}}{{\lambda }_{i}+1}$ is an eigenvalue of $A\left(A+I{\right)}^{-1}$
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.
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