Let ${a}_{1},{a}_{2}\dots $ 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 $A(A+I{)}^{-1}$?

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Paige Paul

Answered 2022-10-03
Author has **11** answers

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

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

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.

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

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.

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