Suppose that nu1rightarrow nunrightarrow is a basis of Rop

Suppose that {nu_1^rightarrow, .. ., nu_n^rightarrow} is a basis of Ropf^n which are eigenvectors of both A and B. That is, say (lambda i, nu_1^rightarrow), .. .(lambda_n, nu_n^rightarrow) are eigenpairs of A (rho_i, nu_1^rightarrow), .. .(rho_n, nu_n^rightarrow) are eigenpairs of B. Show that AB = BA.

Solution

Let P be the matrix with v₁ , v₂, …,v_n as its columns and let D and S be diagonal matrices with the corresponding eigenvalues of A and B respectively as the entries on the respective leading diagonals. Then D and S are diagonal matrices of the same size (nxn) so that DS = SD.

Further, since A = PDP^-1 and B = PSP^-1, we have AB = (PDP^-1)( PSP^-1) = PD(P^-1P)SP^-1 =PDI_n SP^-1 =PDSP^-1.

Also, BA = ( PSP^-1) (PDP^-1) = PS(P^-1P)DP^-1 = PSI_nDP^-1 = PSDP^-1.

Now, since DS = SD, we have AB = BA.