3
$\begingroup$

So we want to find the basis for the eigenspace of each eigenvalue $\lambda$ for some matrix $A$.

Through making this question, I have noticed that the basis for the eigenspace of a certain eigenvalue has some sort of connection to the eigenvector of said eigenvalue. Now I'm not sure if they actually equal each other, because I have some trouble when it comes to eigenvalues with a geometric multiplicity of two or more.

Take the following example:

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

This matrix has a characteristic polynomial $- \lambda ^3 + 6 \lambda ^2 - 12 \lambda + 8$. The root of this is $\lambda = 2$, which has an algebraic multiplicity of 3.

When I try to find the basis for the eigenspace of the eigenvalue $\lambda = 2$, I kind of get confused. Because when I solve $(A - 2I)\mathbf{v}$ I simply get $(0,0,1)$ (actually $(0,0,n)$ where $n \in \mathbb{R}$) as the answer for the basis, even though this eigenvalue has two associated linearly independent eigenvectors, namely $(0,0,1)$ and $(1,-2,0)$. This leaves me with the following questions:

  • Is it true that the "basis of the eigenspace of the eigenvalue" is simply all of the eigenvectors of a certain eigenvalue (so in our example, the basis would be $(0,0,1), (1,-2,0)$)?

  • If so, why am I not able to get both eigenvectors with my method? And how would I be able to get them both?

$\endgroup$
  • $\begingroup$ When you solve $(A-2I)v = 0$ you should get two-dimensional solution space. How did you solve it? $\endgroup$ – Abstraction Dec 7 '15 at 15:24
0
$\begingroup$

Your first question is correct, the "basis of the eigenspace of the eigenvalue" is simply all of the eigenvectors of a certain eigenvalue.

Something went wrong in calculating the basis for the eigenspace belonging to $\lambda=2$. To calculate eigenvectors, I usually inspect $(A-\lambda I)\textbf{v}=0$. In this case take $\textbf{v}=\left(\begin{array}{c}v_1\\v_2\\v_3\end{array}\right)$ to see: $$(A-\lambda I)\textbf{v}=\left(\begin{array}{ccc}-2 & -1 & 0\\4 & 2 & 0\\2 & 1 & 0\end{array}\right)\left(\begin{array}{c}v_1\\v_2\\v_3\end{array}\right)=\left(\begin{array}{c}-2v_1-v_2\\4v_1+2v_2\\2v_1+v_2\end{array}\right)=\left(\begin{array}{c}0\\0\\0\end{array}\right)$$ Note that all three equations are in fact the same (i.e. $A-\lambda I$ has three linearly dependent rows), such that the only restriction you have on $v_1,$ $v_2$ and $v_3$ is $$2v_1=-v_2$$ Three unknowns with only one restriction leaves 2 degrees of freedom, i.e. you arrive at two eigenvectors. Now it is easy to see $$\left(\begin{array}{c}1\\-2\\0\end{array}\right) \ \mbox{ and } \ \left(\begin{array}{c}0\\0\\1\end{array}\right)$$ are two of such eigenvectors.

$\endgroup$
  • $\begingroup$ Hmm, up until the $2v_1 = - v_2$ result I did exactly the same. I failed to see the 2 degrees of freedom = 2 eigenvectors. Thanks $\endgroup$ – Algebreh Dec 7 '15 at 16:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.