Given Ker(T) and Ker(T-2I), is T diagonalizable?

Question

I'm stuck on this problem that I found on my book of linear algebra.
Be $T:\mathbb{R}ˆ3 \rightarrow \mathbb{R}ˆ3$ a linear map such that $$Ker(T-2I) = \{(x,y,z) | x+y=0\} $$ and $$Ker(T) = \{ (x,y,z)|x=y-z=0\} $$

1. Is T diagonalizable?
2. Write the characteristic polynomial of T
3. Write the matrix that represents T in the standard basis

I tried solving 1. and 2. by writing first the matrix, but I'm not sure that's the correct approach to the problem. I basically solved a linear system with the equations given by the problem and I found T, which I think is $T \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2x \\ -2x \\ -2x -2y + 2z \end{bmatrix}$. In this way T satisfies the rules given by the text of the problem. So the matrix associated to T in the standard basis is $\begin{bmatrix} 2&0&0\\-2&0&0\\-2&-2&2\\ \end{bmatrix}$. From here I can see that $L_T$ is diagonalizable and I can find the characteristic polynomial.

What I don't get is how to answer questions number 1 and 2 without first finding the matrix. I've thought about finding a basis for $Ker(T)$ and $Ker(T-2I)$ in order to solve the first question, but after writing the basis I don't know what else to do. Thank you

Answer 1

Your answer to question 3 (matrix of $T$ in the standard basis) is correct.

The simplest way to answer questions 1 and 2 is to use that:

1. an endomorphism $T$ of a space $E$ of dimension $n$ is diagonalizable iff the sum of the dimensions of its eigenspaces is $n$ (or equivalently: is $\ge n$).
2. the characteristic polynomial of $T$ is $\det(xI_n-M_B(T))$ where $M_B(T)$ is the matrix of $T$ in any basis $B$ of $E,$ and this determinant is easy to calculate if that matrix is diagonal.