TITLE: Show that this matrix is not diagonalizable
QUESTION [6 upvotes]: Say I have a matrix:
$$A = \begin{bmatrix} 2 & 0 \\ -1 & 2 \end{bmatrix} $$

Is this matrix diagonalizable?
Does a 2x2 matrix always have 2 eigenvalues (multipicity counts). Why is this? I know this matrix (because it's lower triangular) has the eigenvalue of 2 with multiplicity 2... but does a matrix of this size always have 2 eigenvalues. Why is this?

Is there any way to know if the eigenvalue of 2 has two eigenvectors or not quickly? Here's the way I know to find the eigenvector:
$$\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} $$
$$ eigenvector = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ t \end{bmatrix} = t * \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$
By this theorem, it is not diagonalizable because it only has 1 eigenvector right and the matrix has 2 rows and 2 columns:

REPLY [3 votes]: The geometric multiplicity of an eigenvalue (that is, the number of independent eigenvectors corresponding to that eigenvalue), is the dimension of the null space of the matrix minus the identity matrix times that eigenvalue (that is, you take the matrix and subtract the eigenvalue from each main diagonal entry). The dimension of the null space of a $n \times n$ matrix is $n$ minus the rank of the matrix. So if $n=2$, then to have a geometric multiplicity of 2, you have to have the rank be zero. But a rank zero matrix is simply he zero matrix. So once you subtract 2 from the main diagonal and see that you don't get the zero matrix, you know that there's only one linearly independent eigenvector for that eigenvalue.