We are given:
$$A=\begin{bmatrix}
5 & -4 & 0\\ 1 & 0 & 2\\0 & 2 & 5
\end{bmatrix}$$
We form and solve: $|A-\lambda I|=\begin{bmatrix}0 & -4 & 0\\ 1 & -5 & 2\\0 & 2 & 0\end{bmatrix} = 0$
This yields a characteristic polynomial and eigenvalues as:
$$-(\lambda-5)^2 \lambda = 0 ~~~\rightarrow ~~~ \lambda_1 = 0, \lambda_{2,3} = 5$$
We have multiplicities of $1$ and $2$ for those eigenvalues.
To find the eigenvectors, we generally solve $[ A - \lambda_i I]v_i = 0$, but since we have a repeated eigenvalue, we may need to change that strategy and find a generalized eigenvalue.
So, for $\lambda_1 = 0$, we have:
$[A- 0I]v_1 = \begin{bmatrix}5 & -4 & 0\\ 1 & 0 & 2\\0 & 2 & 5\end{bmatrix}v_1 = 0$
Doing row-reduced-echelon-form (RREF), yields:
$\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & \dfrac{5}{2} \\ 0 & 0 & 0\end{bmatrix}v_1 = 0$
Thus, $b = -\dfrac{5}{2}c, a = -2c \rightarrow ~~\text{let}~~ c = 2 \rightarrow b = -5, a= -4, v_1 = (-4,-5,2)$.
Repeating this same process for the second eigenvalue, $\lambda_2 = 5$, we have as RREF:
$\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}v_2 = 0$
So, $b = 0$, $a = -2c$, let $c = 1 ~~\rightarrow a = -2, v_2 = (-2,0,1)$
Unfortunately, we cannot get another linearly independent eigenvector, so need to get a generalized one, by doing $[A - \lambda_3 I]v_3 = v_2$ (this does not always work), so we have:
$\begin{bmatrix}0 & -4 & 0 \\ 1 & -5 & 2 \\0 & 2 & 0\end{bmatrix}v_3 = \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}$
After RREF, we arrive at:
$\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}v_3 = \begin{bmatrix} \dfrac{5}{2} \\ \dfrac{1}{2} \\ 0 \end{bmatrix}$
So, we have: $a = \dfrac{5}{2} -2c, b = \dfrac{1}{2} \rightarrow ~~ \text{let} ~~ c = 0 \rightarrow a = \dfrac{5}{2}, b = \dfrac{1}{2}$, thus $v_3 = (\dfrac{5}{2},\dfrac{1}{2},0)$
You should get your hands around the above regarding your algebraic versus geometric multiplicities.
Putting all of this together, we have the eigenvalue/eigenvector pairs:
- $\lambda_1 = 0, v_1 = (-4, -5, 2)$
- $\lambda_2 = 5, v_2 = (-2, 0, 1)$
- $\lambda_3 = 5, v_3 = (\dfrac{5}{2},\dfrac{1}{2},0)$
All you need is just to solve the system. The system of equations
$$
\begin{cases}
2x+y-z=1,\\
\qquad y-z=1
\end{cases}
$$
has the solution
$$
\begin{bmatrix}
x\\y\\z
\end{bmatrix}=\begin{bmatrix}
0\\1+t\\t
\end{bmatrix},\quad t\in\Bbb R.
$$
Now take $t=0$.
The second vector is obtained similarly: solve
$$
\begin{cases}
2x+y-z=1,\\
\qquad y-z=0
\end{cases}
$$
to get
$$
\begin{bmatrix}
x\\y\\z
\end{bmatrix}=\begin{bmatrix}
1/2\\-t\\t
\end{bmatrix},\quad t\in\Bbb R
$$
and set $t=0$.
P.S. Those other two vectors are not eigenvectors. They are called generalized eigenvectors.
Best Answer
Your construction of the eigenvector assumes $c$ is non-zero. A matrix with full geometric multiplicity is diagonalizable, so the only such matrix is $\lambda I$.