Prove that $T$ is not diagonizable

Question

I'm having difficulties with this exercise, can anyone give me a hand?

Let $T:R^3 \rightarrow R^3$ be a linear transformation. It's know that $(1,1,0), (1,1,1)$ are eigenvectors of $T$ and: $T(1,0,1) = (1,2,1)$, $T(2,2,1) = (1,1,-1)$ Prove that $T$ is not diagonizable.

Thanks

Answer 1

Let A be the standard matrix for T, so

$A\begin{bmatrix}1&1&2\\1&0&2\\0&1&1\end{bmatrix}=\begin{bmatrix}\lambda&1&1\\\lambda&2&1\\0&1&-1\end{bmatrix}\implies A=\begin{bmatrix}\lambda&1&1\\\lambda&2&1\\0&1&-1\end{bmatrix}\begin{bmatrix}1&1&2\\1&0&2\\0&1&1\end{bmatrix}^{-1}$

$\implies A=\begin{bmatrix}\lambda&1&1\\\lambda&2&1\\0&1&-1\end{bmatrix}\begin{bmatrix}2&-1&-2\\1&-1&0\\-1&1&1\end{bmatrix}=\begin{bmatrix}2\lambda&-\lambda&-2\lambda+1\\2\lambda+1&-\lambda-1&-2\lambda+1\\2&-2&-1\end{bmatrix}$.

Then $A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}-\lambda+1\\-\lambda+1\\-1\end{bmatrix}$, so since $\begin{bmatrix}1\\1\\1\end{bmatrix}$ is an eigenvector for A,

$-\lambda+1=-1\implies\lambda=2\implies A=\begin{bmatrix}4&-2&-3\\5&-3&-3\\2&-2&-1\end{bmatrix}$.

Then $\det(A-\lambda I)=(1+\lambda)^2(2-\lambda)$, but the eigenspace for $\lambda=-1$ is only 1-dimensional since

solving $(A+I)x=0$ gives $x=\begin{bmatrix}t\\t\\t\end{bmatrix}$. Therefore A is not diagonalizable.