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: