Diagonalisation of a linear operator Given the following linear operator
$$T(x,y,z)=(2x+z,-3y+z,-3z)$$
with $T\colon\mathbb{R}^{3}\to\mathbb{R}^{3}$, I found that its matrix in the canonical basis is
$$[T]=\begin{bmatrix} 2 & 0 & 1 \\ 0 & -3 & 1 \\ 0 & 0 & -3 \end{bmatrix}$$
such that its eigenvalues and eigenvectors are:
$$\lambda_{1}=2,\vec{v_{1}}=(1,0,0),\quad\lambda_{2}=-3,\vec{v_{2}}=(0,1,0)$$
How do I conclude that $T$ is not diagonalisable?
 A: If $T$ is diagonalizable, then the sum of the dimension of the nullspace of $T - \lambda I$ over all eigenvalues $\lambda$ of $T$ must be equal to the dimension of the space, which is in this case $3$.
However, for this case, we find that $\dim \ker(T - 2 I) = \dim \ker (T + 3I) = 1$, which means that the sum $1 + 1 = 2$ is too small.
A: Let's compute eigenvalues of $T.$ They are $2$ and $-3.$
Notice that the geometric multiplicity of $-3$ is $1,$ but the algebraic multiplicity is $2$. This shows that $T$ is Not diagonalizable.
A: I am sorry for my late reply. Ben Grossman already answered your question. Now, you know the definition of geometric multiplicity. I just want to add something with his answer. In general, the geometric multiplicity/dimension of eigenspace of an eigenvalue is less than or equal to the algebraic multiplicity of that eigenvalue. If the equality holds for every eigenvalue, then the operator is diagonalizable. Converse is also true. The proof is elementary. You should find it in any standard Linear Algebra text.
