Diagonlisation of certain matrices

Question

Why is it that the matrix $\begin{pmatrix} -1 & -3 & -1 \\ -3 & 5 & -1 \\ -3 & 3 & 1 \end{pmatrix}$ is diagonalisable, even though it has eigenvalues 1, 2, 2 (which are not all distinct) but $\begin{pmatrix} 1 & 1 & 0 \\ -1 & 3 & 0 \\ 0 & 1 & 3 \end{pmatrix}$ is not, despite the fact that it too has two eigenvalues which are not distinct (in fact exactly the same eigenvalues: 1, 2, 2)?

Answer 1

A $n \times n$ matrix (over $\mathbb{R}$ or any other field) is diagonalizable if the set of his eigenvector is a basis for $\mathbb{R}^n$. So the matrix must have $n$ linearly independent eigenvector.

For $n=3$, if the matrix has an eigenvalue of multiplicity $2$ (as in your case), then we have two possibilities:

1) this ''double'' eigenvalue has two linearly independent eigenvector, and in this case we say that it has algebraic multiplicity $2$ and geometric multiplicity $2$ ( this means that the eigenvalue ''span'' a plane in $R^3$).

2) The ''double'' eigenvalue has only one eigenvector, and his geometric multiplicity is $1$ ( the eigenvalue ''span'' a straight line).

So, your first matrix has two distinct eigenvalues, one of them ( $\lambda=2$) of algebraic multiplicity 2, but has three linearly independent eigenvectors since $\lambda=2$ has geometric multiplicity $2$. So this matrix is diagonalisable. For the second matrix you can see that the geometric multiplicity of the eigenvalue $\lambda=2$ is $1$, so that there are not three linearly independent eigenvectors.