# Determine the eigenvector and eigenspace and the basis of the eigenspace

The yellow marked area is correct, so don't check for accuracy :)

$A=\begin{pmatrix} 0 & -1 & 0\\ 4 & 4 & 0\\ 2 & 1 & 2 \end{pmatrix}$ is the matrix.

Characteristic polynomial is $-\lambda^{3}+6\lambda^{2}-12\lambda+8=0$

The (tripple) eigenvalue is $\lambda=2$.

Calculate the eigenvectors now:

$\begin{pmatrix} -2 & -1 & 0\\ 4 & 2 & 0\\ 2 & 1 & 0 \end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix}= \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$

We get the equations:

$I: -2x-y=0 \Leftrightarrow y = -2x$

$II: 4x+2y=0$

$III: 2x+y=0 \Leftrightarrow 2x-2x=0 \Leftrightarrow 0=0$

We see that in every eequation $z$ is unknown, so we can choose an arbitrary $z$.

$x\begin{pmatrix} 1\\ -2\\ z \end{pmatrix}$ and this is the eigenspace...?

And what is the basis of this eigenspace? Can I just set $x=1$ and some value for $z$? So this would be a correct basis of the eigenspace: $\begin{pmatrix} 1\\ -2\\ 3 \end{pmatrix}$?

Now we need three linearly independent eigenvectors but I couldn't find them as I always got linearly dependent vectors...

I need a detailled, not too complicated answer that explains it well and I will give that answer a nice bounty (up to 200 rep) because I couldn't find another site explaining this correctly to me and I'm really in need of it.

• "Now we need three linear independent eigenvectors"...no, not really – imranfat Mar 8 '17 at 16:55

The $x$ shouldn't be outside the vector. The solution to equations I,II, and III is

\begin{pmatrix} x\\ -2x\\ z \end{pmatrix}

where $x$ and $z$ are arbitrary. Every vector of this form is an eigenvector for $A$. You can write each such vector as a linear combination of two vectors $e_1$ and $e_2$ defined by

$$e_1:= \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}$$ and $$e_2:= \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}.$$

More concretely, we have $$\begin{pmatrix} x \\ -2x \\ z \end{pmatrix} = \begin{pmatrix} x \\ -2x \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ z \end{pmatrix} = x\begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix} + z\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = xe_1 + ze_2$$

A basis for the eigenspace is the two vectors $e_1$ and $e_2$, since every vector in the eigenspace can be written uniquely as a linear combination of those two vectors.

There's no reason it should have 3 linearly independent eigenvectors.

• So let's say the task says "Determine the eigenvectors", what is the best way to write them for this example? Is it $v_{1}=x\begin{pmatrix} 1\\ -2\\ 0 \end{pmatrix}+z\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}$? Or rather $\begin{pmatrix} 1\\ -2\\ 0 \end{pmatrix},\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}$? – tenepolis Mar 8 '17 at 19:23
• Technically the eigenvectors are that first one. Though depending on the class, I've seen people refer to (1,-2,0) and (0,0,1) as the eigenvectors. – Nathan H. Mar 9 '17 at 1:58

You must solve $(A - 2I)\vec{x} = \vec{0}$, i.e. $$\begin{pmatrix} -2 & -1 & 0\\ 4 & 2 & 0\\ 2 & 1 & 0 \end{pmatrix} \begin{pmatrix}x\\ y\\ z\end{pmatrix} =\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}$$ Note that all equations are multiple of each other, so let's leave the last one to use and eliminate all others. You get the constraint $2x+y = 0$. So Your solutions will have to have $y = -2x$, in other words, $$\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}x\\ -2x\\ z\end{pmatrix} = x \begin{pmatrix}1\\ -2\\ 0\end{pmatrix} + z \begin{pmatrix}0\\ 0\\ 1\end{pmatrix}$$ and so you see that there are 2 eigenvectors that can be used as a basis.

As for the number of independent eigenvectors in the basis not equal the multiplicity of the root of the characteristic equation, this has to do with the fact that for your matrix, the geometric and algebraic multiplicities don't match for the eigenvalue $\lambda = 2$. You can read more about it here.

From your calculation you have that all eigenvectors are of the form $\begin{pmatrix} x\\ -2x\\ z \end{pmatrix} = x\begin{pmatrix} 1\\ -2\\ 0 \end{pmatrix} + z\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}$

If $\lambda$ is an eigenvalue of $A$, then its corresponding eigenvectors are vectors $\mathbf v$ that satisfy $A\mathbf v=\lambda\mathbf v$, or $(A-\lambda I)\mathbf v=0$. That is, the eigenspace of $\lambda$ is the null space of the matrix $A-\lambda I$. This answer describes how to read a basis for the null space directly from the row-reduced echelon form of the matrix. Remember that the dimension of the eigenspace—the geometric multiplicity of $\lambda$—will be at most the algebraic multiplicity of $\lambda$, which is 3 in this case, but it might be less than that.