# Geometric multiplicity of repeated Eigenvalues

I am still finding it difficult to determine the geometric multiplicity for repeated eigenvalues and the resultant eigenspace. For example, I am not quite sure what to do with the following matrix, where repeated Eigenvalues $\lambda_1 = \lambda_2 = 5$:

$$A=\begin{bmatrix} 5 & -4 & 0\\ 1 & 0 & 2\\ 0 & 2 & 5 \end{bmatrix}, [A-\lambda I] = \begin{bmatrix} 0 & -4 & 0\\ 1 & -5 & 2\\ 0 & 2 & 0 \end{bmatrix}$$

It is not obvious how to determine the Eigenvectors from this, as there are no free variables, and moving $e_2$ for example (for the first row) such that $-4e_2 = 0 \rightarrow e_2 = 0$ shows that all other values result to zero as well (which is not a valid eigenvector). How does one go about determining the geometric multiplicity and the Eigenspace with such a matrix?

• I'm getting an eigenvector of $\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}$. Jun 12, 2013 at 0:45

We are given:

$$A=\begin{bmatrix} 5 & -4 & 0\\ 1 & 0 & 2\\0 & 2 & 5 \end{bmatrix}$$

We form and solve: $|A-\lambda I|=\begin{bmatrix}0 & -4 & 0\\ 1 & -5 & 2\\0 & 2 & 0\end{bmatrix} = 0$

This yields a characteristic polynomial and eigenvalues as:

$$-(\lambda-5)^2 \lambda = 0 ~~~\rightarrow ~~~ \lambda_1 = 0, \lambda_{2,3} = 5$$

We have multiplicities of $1$ and $2$ for those eigenvalues.

To find the eigenvectors, we generally solve $[ A - \lambda_i I]v_i = 0$, but since we have a repeated eigenvalue, we may need to change that strategy and find a generalized eigenvalue.

So, for $\lambda_1 = 0$, we have:

$[A- 0I]v_1 = \begin{bmatrix}5 & -4 & 0\\ 1 & 0 & 2\\0 & 2 & 5\end{bmatrix}v_1 = 0$

Doing row-reduced-echelon-form (RREF), yields:

$\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & \dfrac{5}{2} \\ 0 & 0 & 0\end{bmatrix}v_1 = 0$

Thus, $b = -\dfrac{5}{2}c, a = -2c \rightarrow ~~\text{let}~~ c = 2 \rightarrow b = -5, a= -4, v_1 = (-4,-5,2)$.

Repeating this same process for the second eigenvalue, $\lambda_2 = 5$, we have as RREF:

$\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}v_2 = 0$

So, $b = 0$, $a = -2c$, let $c = 1 ~~\rightarrow a = -2, v_2 = (-2,0,1)$

Unfortunately, we cannot get another linearly independent eigenvector, so need to get a generalized one, by doing $[A - \lambda_3 I]v_3 = v_2$ (this does not always work), so we have:

$\begin{bmatrix}0 & -4 & 0 \\ 1 & -5 & 2 \\0 & 2 & 0\end{bmatrix}v_3 = \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}$

After RREF, we arrive at:

$\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}v_3 = \begin{bmatrix} \dfrac{5}{2} \\ \dfrac{1}{2} \\ 0 \end{bmatrix}$

So, we have: $a = \dfrac{5}{2} -2c, b = \dfrac{1}{2} \rightarrow ~~ \text{let} ~~ c = 0 \rightarrow a = \dfrac{5}{2}, b = \dfrac{1}{2}$, thus $v_3 = (\dfrac{5}{2},\dfrac{1}{2},0)$

You should get your hands around the above regarding your algebraic versus geometric multiplicities.

Putting all of this together, we have the eigenvalue/eigenvector pairs:

• $\lambda_1 = 0, v_1 = (-4, -5, 2)$
• $\lambda_2 = 5, v_2 = (-2, 0, 1)$
• $\lambda_3 = 5, v_3 = (\dfrac{5}{2},\dfrac{1}{2},0)$
• $\checkmark +$ Nice work, and it "took", two!! - excuse the play on words ;-) Jun 12, 2013 at 2:08
• For some reason, I didn't think to row reduce. And I forgot about your method for solving for $v_3$ by setting it equal to $v_2$. Thanks, that helped a lot.
– dtg
Jun 12, 2013 at 2:25
• @Dylan: You are very welcome. Hopefully, you can now use this and review the matrix ranks and see how that gives you the free variables to work with. Regards! Jun 12, 2013 at 2:27
• I've just gotten "spoiled" or accustomed to being more productive (and successful) at night, and I sleep better...but I should at least slow down when there's a "slow-down" anyway, here on the site! Jun 12, 2013 at 4:52
• Pay attention in particular to the fact that $v_3$ is a "generalized eigenvector", and not an eigenvector in the sense that $A v_3=\lambda v_3$. The eigenvectors of a matrix $A$ do not always produce a basis of the domain; they do not in this case because $5$ has geometric multiplicity 1 but algebraic multiplicity 2. The generalized eigenvectors, however, will always form a basis of the domain. Using this fact, you can put any matrix in Jordan canonical form. Jun 12, 2013 at 13:50

The eigenspace corresponding to $\lambda_1=\lambda_2=5$ is precisely the kernel of the matrix $$[A-5I]= \begin{bmatrix} 0 & -4 & 0\\ 1 & -5 & 2\\ 0 & 2 & 0 \end{bmatrix}$$ That is, we are looking for the solution set for $$\begin{bmatrix} 0 & -4 & 0\\ 1 & -5 & 2\\ 0 & 2 & 0 \end{bmatrix}\,\, \vec v=\vec0$$ We can solve this by finding the kernel of the row reduced version of this matrix, which is $$\begin{bmatrix} 1 & 0 & 2\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{bmatrix}$$ As you can see, the third entry is our free variable. The vector $$\vec v_5= \begin{bmatrix} 2 \\ 0 \\ -1 \end{bmatrix}$$ forms a basis of the kernel of this matrix, which tells us that $\vec v_5$ is a basis of our eigenspace. That is, the geometric multiplicity of this eigenvalue is $1$.

The usual procedure to find the eigenspace and GM(geometric multiplicity) for a particular eigenvector of a matrix A is as follows:

1).Solve A's characteristic polynomial |A-lamda*I|, which yields an n-th polynomial in regard to lamda.

2)If you have obtained a group real solutions, say, lamda 1,lamda 2,lamda 3...lamda r, from the previous step, for each of the lamda s, put it back into the homogeneous equation (A-lamda*I)x= 0 in regard to x. If you find the equation has only trivial solution, then this lamda is not an eigenvalue, so you can just quit and go to the next lamda. Otherwise, this lamda is indeed an eigenvalue. Solve the equation to obtain the kernel of (A- lamdaI), whose basic vectors are just the eigenvectors corresponding to A and whose dimension is just the GM for this lamda.

And note that for any eigenvalues GM<=AM(algebraic multiplicity) always holds.

First note that the matrix $A-5I$ has rank $2$ so $\dim(\ker(A-5I)) = 1$. This means that the geometric multiplicity of the eigenvalue $\lambda=5$ is 1. To get a nontrivial vector in the kernel of $A-5I$, try row reducing since this amounts to multiplying on the left by a non-singular matrix. Such an operation does not change the kernel of the linear transformation. Once you get this eigenvector, you will have determined the eigenspace completely.

Row-reducing gives: $$\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 0\\ 0 & 0 & 0 \end{bmatrix}$$

so you can multiply by an arbitrary vector $[x\; y\; z]^t$, set the result equal to $[0\; 0\;0]^t$ and solve for $x,y,z$.