$2\times 2$ matrix with $2$ Distinct Eigenvalues is Diagonalizable

Question

In a class of mine, my teacher said that any $2\times 2$ matrix with $2$ distinct eigenvalues is diagonalizable. I've been trying to find a proof of this or a theorem to support it but cannot find anything. Is this statement true?

Answer 1

$2\times 2$ matrix has only $2$ eigen values...Also they are distinct. So minimal poly. contains distinct roots and hence diagonalisable.

Answer 2

We can use the fact that eigenvectors corresponding to distinct eigenvalues are linearly independent. Now recall that an $n\times n$ matrix is diagonalizable if and only if it has $n$ linearly independent eigenvectors.

Answer 3

Not only that, but any $n\times n$ matrix with $n$ distinct eigenvalues is diagonalizable.

The trick with diagonalizability is that you have to have $n$ independent eigenvectors. Then the diagonalization is three matrices: the first transforms the eigenvectors to your basis vectors, the second, a diagonal matrix, scales along the basis vectors, and the third transforms the basis vectors back to the original eigenvectors.

Each eigenvalue demands the existence of an eigenspace of dimension at least $1$ and at most $k$ associated with it, where $k$ is the multiplicity of the eigenvalue. Since a matrix with all distinct eigenvalues has multiplicity of 1 for each eigenvalue, you have $n$ eigenspaces of dimension 1, and thus $n$ eigenvectors, and thus the matrix is diagonalizable.