It can be easily checked that A, z-A, (z-A)-1 commute and thus are diagonalizable simultaneously. Furthermore, it can be easily be checked directly that if λ is an eigenvalue of A with eigenvector V, and (z-λ)-1 is an eigenvalue corresponding also to v. Therefore, A, z-A and (z-A)-1 have the same spectral projector Pλ of A= the spectral projector P(z-λ)-1of (z-A)-1, and, therefore, the spectral decomposition of (z-A)-1 is thus.1c.) Given a square matrix M its resolvent is the matrix-valued function of a square matrix A its resolvent is the matrix-valued function RA(z)=(zI-A)-1, defined for all z ∈ C and I is a n*n identity matrix.In infinite dimensions the resolvent is also called the Green’s function. Since the resolvent RA(z)is nothing else but f(A) for f(t)=(z-t)-1=1/z-t its spectral decomposition is exactly what is expected.The diagonals entries ∑i,j of ∑ are the singular values of A. The m columns of U and the N columns of V are the left-singular and right-singular vectors of A. One application that uses SVD is the pseudoinverse.A+=V∑+U*, where ∑+ is the pseudoinverse of ∑, which is formed by replacing every non-zero diagonal entry by its reciprocal and getting the transpose of the resulting matrix. It is also possible to use SVD of A to determine the orthogonal matrix R closest to the range of A. The closeness of fit is measured by the Frobenius norm of R-A. The solution is the product UV*. the orthogonal matrix would have the decomposition UIV* where I is the identity matrix, so that if