# Multiple eigenvectors for an eigenvalue and how to know

We have the following matrix:

$$\begin{pmatrix} 0 & -1 & 0 \\ 4 & 4 & 0 \\ 2 & 1 & 2 \end{pmatrix}$$

This matrix has characteristic equation $- \lambda ^3 + 6 \lambda ^2 - 12 \lambda + 8$, which gives us the eigenvalue $\lambda = 2$. However, this eigenvalue has two eigenvectors, namely $(1,-2,0)$ and $(0,0,1)$.

How would you mathematically be able to know how many eigenvectors are associated with a single eigenvector? Is there a way to find out?

• Are you familiar with the concept of a Null Space? – Xoque55 Dec 5 '15 at 16:09
• Find all solutions to $(A-2I)v=0$. – user137731 Dec 5 '15 at 16:10
• To clarify, any eigenvalue has infinitely many eigenvectors. What you are asking about is the maximum number of linearly independent eigenvectors it can have. – Santiago Canez Dec 5 '15 at 16:18

The number of linearly independent eigenvectors that are associated with an eigenvalue, is called the geometric multiplicity of the eigenvalue. It can be found by solving the system $$\dim \ker(λΙ-Α)$$ In your case $λ=2$. To avoid confusion, the algebraic multiplicity of the eigenvalue is the number of times that this eigenvalue is a root of the characteristic polynomial. In your case $$p_A(λ)=-(λ-2)^3$$ so the algebraic multiplicity is equal to $3$. It is true that the geometric multiplicity is less or equal to the algebraic multiplicity.
Yes let $v=(x,y,z)^T$ an eigenvector for the given matrix $A$ associated to $2$ then we have $$Av=2v$$ and if we solve the system of equations we find that $v$ is a linear combination of the two vectors given in your question.