Let A be an n times n-matrix. Assume that A can be diagonalized, with A = SAS^-1

Question

Let A be an n times n-matrix. Assume that A can be diagonalized, with A = SAS^-1 for an invertible n times n-matrix S and a diagonal n times n-matrix A. Diagonalize A^2. Diagonalize A^-1, if A is invertible. (You can use the fact that for an invertible A, the diagonal entries of A are nonzero, and so A^-1 is a diagonal matrix again.) Diagonalize A^T (the transpose of A). (The answers should be in terms of S and A. For example, A + I_n can be diagonalized as follows: A + I_n = S(A + I_n)S^-1. Indeed, S is an invertible matrix, A + I_n is a diagonal matrix (being the sum of the two diagonal matrices A and I_n), and we have A(A + I_n) S^-1 = SAS^-1 + SI_n S^-1 = A + SS^-1 = A + I_n.

Explanation / Answer

We have A = SDS-1 so that A2 = SD2S-1 . (Since D is a diagonal matrix, D2 is also a diagonal matrix). We have A-1 = (SDS-1)-1 = (S-1)-1 (SD)-1 = SD-1S-1 [( PQ)-1 = Q-1 P-1; Also inverse of the diagonal matrix D is a diagonal matrix]. We have AT = (SDS-1)T = (S-1)T (SD)T = ( ST)-1 DT ST = ( ST)-1D ST [ (PQ)T = QT PT; Also (P-1)T = (PT)-1 ; Transpose of a diagonal matrix D is D itself].

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