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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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