# Thread: Testing for diagonalizability and then finding a basis beta for V

1. ## Testing for diagonalizability and then finding a basis beta for V

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.

2. 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.

Looks fine to me...you could also mention that as T is nilpotent then it is diagonalizable iff it is the zero matrix (why? Look at the minimal polynomial)

Tonio

3. Originally Posted by alice8675309
... or is there something else that i'm missing
You are completely right, $f$ is not diagonalizable and its canonical form is:

$J=\begin{bmatrix}{0}&{1}&{0} &{0}\\{0}&{0}&{1}&{0}\\{0}&{0}&{0}&{1}\\{0}&{0}&{0 }&{0}\end{bmatrix}$

Regards.

Fernando Revilla