How to interpret relation of variables as eigenvector? I am trying to calculate the eigenvectors of a square matrix $A \in \mathbb{R}^{4x4}$.
$$A = 
\begin{pmatrix}
a & 1 & 0 & 0 \\
0 & a & 1 & 0 \\
0 & 0 & a & 0 \\
0 & 0 & 0 & b
\end{pmatrix}
$$
To receive the eigenvalues I did:
$$\chi_A = \begin{vmatrix}
a-\lambda & 1 & 0 & 0 \\
0 & a-\lambda & 1 & 0 \\
0 & 0 & a-\lambda & 0 \\
0 & 0 & 0 & b-\lambda
\end{vmatrix} = (\lambda-a)^3(\lambda-b)$$
This means we have $\lambda_1 = a$ with multiplicity $m=3$ and $\lambda_2 = b$ with multiplicity $m=1$.
Now I try to calculate the eigenvectors first for $\lambda_1 = a$
$$\begin{bmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & b-a
\end{bmatrix} = \begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}$$
which gives us $x_1 \in \mathbb{R}$, $x_2 = 0$, $x_3 = 0$ and $(b-a)x_4 = 0$.
When I try to convert this result to the eigenvector for eigenvalue $a$ I am stuck with
$$Eig(\lambda_1) = <\begin{pmatrix} 1 \\ 0 \\ 0 \\ ?\end{pmatrix}>$$
I do not know what value (or variables) I could use for the fourth entry as their value depends on the statement of $(b-a)x_4 = 0$ being true.
 A: There are two cases to separate:


*

*If $a = b$, then you 'only' get one eigenvalue $a$ with multiplicity 4. In this case 
$$\begin{bmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & b-a
\end{bmatrix} = \begin{bmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix} = \begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}$$


implies $x_2 = x_3 = 0$ and $x_1,x_4 \in \mathbb{R}$ arbitrary. Hence,
$$Eig(\lambda_1 = a) = < \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} >$$
2. If $a \neq b$, then $b-a \neq 0$ and we can divide by it. Hence, it follows that $x_4 = 0$ and we have 
$$Eig(\lambda_1 = a) = < \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} >$$
A: $(b-a)x_4=0$ is true only if either $a=b$ or $x_a=0$.
So here you have two cases :


*

*$b=a$ : and $e_1$ and $e_4$ are eigenvectors with eigenvalue $a$

*$b\neq a$ : then $x_4=0$, and $e_1$ is an eigenvector  with eigenvalue $a$, but $e_4$ is not ($e_4$ is actually an eigenvector with eigenvalue $b$).


In short, if $a=b$, you have one eigenspace of dimension 2 for eigenvalue $a$ spanned by $(e_1,e_4)$, if $a \neq b$ you have two eigenspaces of dimension 1, one for $a$ spanned by $(e_1)$, and one for $b$ spanned by $(e_4)$.
note : $e_i$ design the vectors of the canonical base (ie vectors with a 1 at i-th entry, and zeros everywhere else)
