Originally Posted by

**alice8675309** For the linear operator T on a vector space V, test for diagonalizibility and if T is diagonalizable then find a basis beta for V such that [T]_beta is a diagonal matrix.

V=P_3(R) and T is defined by T(f(x))=f'(x)+f"(x).

so far i have:

{1,x,x^2,x^3}: T(1)=0, T(x)=1, T(x^2)=2x+2, T(x^3)=3x^2+6x

and then I found [T]_beta to be the matrix

0120

0026

0003

0000

I found that the characteristic polynomial is (0-t)^4 so the eigenvalues are t=0 (Im using t instead of lamda because I always mess up the latex) so t=0 has multiplicity of 4 and since everything factors we know it splits and then it is NOT diagonalizable because n-rank(A-tI)=1 and 1 does not equal 4 or is there something else that i'm missing. I know those are the two tests we need to look for, but I found a different answer that looks weird to what I found in terms of diagonalizibilty.