# Checking if a vector space has a basis consisting of eigenvectors of an operator?

Let $T \in \mathcal{L}(\text{Mat}(2,2,\mathbb{C}))$ be defined as $T \left(\begin{array} \\ a & b \\ c & d \\ \end{array}\right)= \left(\begin{array} \\ 2a-ib & ia+3c \\ 3b-2c & 0 \\ \end{array}\right)$

Question: Does $\text{Mat}(2,2,\mathbb{C})$ have a basis consisting of eigenvectors of $T$?

Attempt: I am not even sure if this is true or not.

Note that the above statement is equivalent to showing either of the following:-

$(1)$ $T$ is diagonalizable.

$(2)$ There exist $1$-dimensional subspaces $U_1, \ldots, U_n$ of $V$ each invariant under $T$ such that $V=U_1 \bigoplus \ldots \bigoplus U_n$.

$(3)$ $\text{dim} V= \text{dim} E(\lambda_1,T)+ \ldots +\text{dim} E(\lambda_m,T)$.

It clear that $0$ is an eigenvalue of $T$. But, finding other eigenvalues is not possible by hand. I have this feeling that we the only way to do this is use $(2)$ somehow.

Any hint in the right direction is highly appreciated.

Hint: Use the basis of $4$ elementary $2\times 2$ matrices to represent $T$ as a $4\times 4$ matrix.
• Yes, I know using determinants, but that is not allowed. The characteristic polynomial which we get in that scenario is $\lambda^2(\lambda-2)(\lambda+2)=0$ since the last column contains $0$ for the first three entries. But, now my problem is that $2$ and $-2$ are not eigenvalues of $T$. Am I doing a calculation mistake? – user330477 Apr 25 '18 at 1:07
• The matrix for $T$ is $\left( \begin{array}\\ 2 & -i & 0 & 0 \\ i & 0 & 3 & 0 \\ 0 & 3 & -2 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)$ – user330477 Apr 25 '18 at 1:16