Prove that any symmetric or anti-symmetric matrix is an eigenvector for the given linear map L: M_n(R) -> M_n(R) M-> 2M - 3(M)ˆtr
My attempt: Let a be an eigenvalue of M. Let x be an eigenvector corresponding to a. Hence we have M.x = a.x M = Mˆtr (symmetric matrix) M = -Mˆtr (antisymmetric matrix)
1) Proof for symmetric matrix. We replace M.x=a.x with the linear map's outpit 2M-3(M)ˆtr [2M-3(M)ˆtr].x - a.x = 0 Since M = Mˆtr => x[2(M)ˆtr-3M - a] = 0 => x(-M-a) = 0 => This is what I get at the end: -M.x = a.x
I am not sure how to approach this task.