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.
Best Answer
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.