is diagonalizable has linearly independent eigenvectors.
i.e.
where vectors are linearly independent eigenvectors, and are the corresponding eigenvalues, in order.
Distinct Eigenvalues
If and has distinct eigenvalues, then is diagonalizable
Non-distinct Eigenvalues
You check that the sum of the geometric multiplicities is equal to the size of the matrix.
e.g. for
Find the eigenvalues:
We know that geomult ⇐ algmult. Therefore has 1 distinct eigenvector.
This means has to have 2 distinct eigenvectors to form a basis, so if it doesn’t then the matrix is not diagonalizable.
There is only one free columns here. Therefore, the dimension of the Nullspace is one, not two, which means the matrix is not diagonalizable.