# Generalised eigenvectors

Find the general solution of the homogeneous ordinary differential equation $\dot{\mathbf{x}}=A\mathbf{x}$.

$$A=\begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}$$

I know how to answer these kind of questions, but the specific example above I am struggling with because I'm not sure I'm finding the correct geometric multiplicity for the repeated eigenvalue. I have the eigenvalues $\lambda=4,1,1$. Hence the eigenvalue $\lambda=1$ has algebraic multiplicity of 2, so I need to find the geometric multiplicity.

I have

$$A-\lambda I=\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$$

Now, normally I would spot the number of independent rows/columns and conclude the geometric multiplicity from this. But it seems this matrix has 0 independent rows/columns, which would make the geometric multiplicity 3, but this violates the fact that the geometric multiplicity can not be greater than the algebraic multiplicity. Where is my thinking going wrong?

$$A-\lambda I=\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}$$
Why do you think this matrix has $0$ independent rows? Clearly the first row "on its own" is independent... Subtract the first row from the second and third to get:
$$\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix} \sim \begin{pmatrix} 1 & 1 & 1 \\ 0&0 & 0 \\ 0 & 0& 0 \end{pmatrix}$$ So you're looking for solutions to the homogeneous system with (only) the equation $x+y+z=0$. The geometric multiplicity is $2$ as well and two linearly independent eigenvectors are, for example, $(1,-1,0)$ and $(1,0,-1)$.
• One non-zero vector alone, is always linearly independent. Take for example $(1,2,3)$ and $(2,4,6)$. These vectors are clearly linearly dependent because they are multiples of one another. But that doesn't change the fact that either of those vectors alone is linearly independent. – StackTD Jan 11 '17 at 10:04