T is a linear map from $R^3$ to $R^3 $. It is known that $T(1,1,1) = (2,2,2), T(1,1,0) = (-5,-5, 0)$ and $T(1,-1,0) = (a-4,a+6,0)$
For which values of $a$ is $T$ diagonisable? For which values of $a$ is $T$ not diagonisable?
It was easy to find $(1,-1,0)$ is an eigenvector for $a=-1$, and since it is linearly independent from the other eigenvectors, the sum of dimensions of the eigenspaces must be at least 3.
Beyond this, however, I am lost. Help would be appreciated. Are there other values of a? How do I prove when it is not diagonisable?
Since$$T(1,-1,0)=(1+a)(1,1,0)-5(1,-1,0),$$the matrix of $T$ with respect to the basis $\{(1,1,1),(1,1,0),(1,-1,0)\}$ is$$\begin{bmatrix}2&0&0\\0&-5&1+a\\0&0&-5\end{bmatrix}.$$Therefore, $T$ is diagonalizable if and only if $a=-1$.