How to diagonalize matrices with repeated eigenvalues?

Question

Consider the matrix $$A=\begin{pmatrix}q & p & p\\p & q & p\\p & p & q\end{pmatrix}$$ with $p,q\neq 0$. Its eigenvalues are $\lambda_{1,2}=q-p$ and $\lambda_3=q+2p$ where one eigenvalue is repeated. I'm having trouble diagonalizing such matrices. The eigenvectors $X_1$ and $X_2$ corresponding to the eigenvalue $(q-p)$ have to be chosen in a way so that they are linearly independent. Otherwise the diagonalizing matrix $S$ becomes non-invertible. What is the systematic way to find normalized linearly independent eigenvectors in this situation?

Answer 1

The sum of each row of the matrix is $q+2p$ and therefore $(1,1,1)$ is an eigenvector corresponding to the eigenvalue $q+2p$. Now to compute the remaining eigenvectors, look for a basis of the null space of$$A-(q-p)\operatorname{Id}=\begin{pmatrix}p&p&p\\p&p&p\\p&p&p\end{pmatrix}.$$You can take $(1,-1,0)$ and $(0,1,-1)$.

Answer 2

In linear algebra, there is a distinction is made between algebraic and geometric multiplicity. The algebraic multiplicity of an eigenvalue is its multiplicity in the characteristic polynomial, and the geometric multiplicity is the dimension of its associated eigenspace. If the geometric multiplicity of the eigenvalues matches their algebraic multiplicities, then we can diagonalize.

To find eigenvalues, one can take A-$\lambda$I and row reduce it, which gives somewhat arbitrary solutions. For instance, here you can first try to get an eigenvector using just the "first" two dimensions, and get (1,-1,0), and then get an eigenvector of the remaining eigenspace. Or you could take (1,-.5,-.5) as your first eigenvector, and end up with a different basis.

If there is an eigenvalue with algebraic multiplicity larger than its geometric multiplicity, then you cannot diagonalize the matrix, but there is a more generalized concept called Jordan Canonical Form that applies to every matrix.