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12. Suppose A is an n x n matrix with the property that A2 = 1. a. Show that if X is an eigenvalue of A, then l = 1 or 2 = -1

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I have done all parts for you. Kindly go through. I have explained in detail the steps and in b part I have given the reason of how we reached in to the conclusion of the eigen vectors of 1 and -1. The third part is easy since we have a theorem which says that a square matrix A is diagonalisable iff the direct sum of eigen spaces of A is the whole vector space. So here we proved that \mathbf{R^n} is a direct sum of the eigen spaces of A. Hence A is diagonalisable.

an ие AYE em ectors: 12. Given A²= I. Suppose a be an eigen value (a). Of A So Ax = 120 for some x+0. >> A[Ax) = A (am) Á meVem Just to get it in the form givers in the question, we did like this Eigens vectors corresfording to 1==1 : Axe = -12 (A+I1 Thu R. E(1) + (-). ( ELDO EH) » and E(1)+E6-1) =R) Thus we have A is diagonalizable since - We could express Rh as a dire

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