# Geometric multiplicity > 1

1. May 15, 2013

### njl86

after finding out what geometric multiplicity was, I was surprised to notice that in every question I'd done it was always 1.
So I'm trying to prove an example with g.m. > 1 to see why it works.
I've found a matrix which definitely has an eigenvalue with g.m. = 2. I've checked everything with WolframAlpha, so the following is correct:

Matrix A =
$$\left( \begin{array}{ccc} 5 & 4 & 2 \\ 4 & 5 & 2 \\ 2 & 2 & 2 \end{array} \right)$$
Determinant = 10

Characteristic polynomial = $$-((x-10) (x-1)^2)$$

So eigenvalues =
10
1 < -- with a.m. = 2, and g.m. = 2

(A - 1 * I ) v = 0, the matrix being:
$$\left( \begin{array}{ccc} 4 & 4 & 2 \\ 4 & 4 & 2 \\ 2 & 2 & 1 \end{array} \right)$$

Last edited: May 15, 2013
2. May 15, 2013

### WannabeNewton

It is certainly not true that the geometric multiplicity is always 1 for each eigenvalue. All you know is that the geometric multiplicity is always less than or equal to the algebraic multiplicity; there is nothing that says, however, that the geometric multiplicity must be 1. Recall that if $A$ is a matrix and $\lambda$ is an eigenvalue of $A$ then the geometric multiplicity of $\lambda$ is $\dim E_{\lambda}$ where $E_{\lambda}$ is the associated eigenspace.

As a simple counter example, consider $A = I$. The only eigenvalue of $I$ is $\lambda = 1$ and every $v\in \mathbb{R}^{n}$ is an eigenvector of $I$ so $E_{\lambda = 1} = \mathbb{R}^{n}$ which has a geometric multiplicity of $n$ which is not 1 in general.

3. May 15, 2013

### njl86

I know, I was just looking for a normal example that showed otherwise.
Thank you for your counter example

How do I continue with my method?

4. May 15, 2013

### WannabeNewton

Well your example is also quite easy in fact. Taking $B = \begin{pmatrix} 4 & 4 &2 \\ 4& 4 &2 \\ 2 & 2 & 1 \end{pmatrix}$
we can very easily put this in reduced row echelon form as $C = \begin{pmatrix} 0 & 0 &0 \\ 0& 0 &0 \\ 2 & 2 & 1 \end{pmatrix}$. Thus, the solutions to $Cv = 0$ are given by $v_{1},v_{2} = \text{arbitrary}$ and $v_{3} = -2v_1 -2v_2$ where $v = (v_1,v_2,v_3)^{T}$. Therefore, setting $v_1 = t,v_2 = s$, we can write any $v\in \text{Null}C$ as $v = t(1,0,-2)^{T} + s(0,1,-2)^{T}$ i.e. $E_{\lambda = 1} = \text{Null}C = \text{Span}\{(1,0,-2)^{T},(0,1,-2)^{T}\}$ which of course has a geometric multiplicity of 2.

Last edited: May 15, 2013
5. May 15, 2013

### njl86

I see how it works now, thank you