You get the eigenvectors just the same as you would normally, except here we need to split into cases where some of the eigenvalues may be equal.
Suppose first that none of the eigenvalues are equal. We need to find vectors $v=(a,b,c)$ such that $Av=\lambda v$. For $\lambda_1=\alpha$ for example, we have the system of equations
$$a\alpha=a\alpha$$
$$a(\beta-\alpha)+b\beta= b\alpha $$
$$a(1-\beta)+b(1-\beta)+c= c\alpha$$
The first equation is trivially true, and simplifying the second equation gives $a=\dfrac{\alpha-\beta}{\beta-\alpha}b\hspace{3mm}\implies\hspace{3mm}a=-b$, which we can include with the third equation to get $c=c\alpha\hspace{3mm}\implies\hspace{3mm}c=0$. Hence setting $a=t$, we have the eigenvectors for $\lambda_1=\alpha$ are $\left[\begin{matrix}t\\-t\\0\end{matrix}\right]$ for $t\in\mathbb{R}$.
You can apply the same process to find the eigenvectors for the other eigenvalues.
Now we have to deal with other cases: where $\alpha=\beta\ne1$, where $\alpha=1\ne\beta$, where $\beta=1\ne\alpha$, and finally where $\alpha=\beta=1$.
I will demonstrate the case $\alpha=\beta\ne1$ and leave the other cases to you.
In the $\alpha=\beta\ne1$ case, the equations for the eigenvalue $\lambda_1=\alpha$ become:
$$a\alpha=a\alpha$$
$$b\alpha= b\alpha $$
$$a(1-\alpha)+b(1-\alpha)+c= c\alpha$$
Note I just used the equations from before replaced all occurrences of $\beta$ with $\alpha$ since they are the same.
Now the first two equations are trivial and the last equation gives us $(a+b)(1-\alpha)=c(\alpha-1)\hspace{3mm}\implies\hspace{3mm} a+b=-c$. Hence setting $a=t$ and $b=s$ we get $\left[\begin{matrix}t\\s\\-(s+t)\end{matrix}\right]$ where $t,s\in\mathbb{R}$ as our eigenvectors for $\lambda_1=\lambda_2=\alpha$.
For $\lambda_3=1$, we get the system of equations
$$a\alpha=a$$
$$b\alpha= b$$
$$a(1-\alpha)+b(1-\alpha)+c= c$$
and simplification gives us $a=b=0$ and $c=c$. Hence setting $c=t$ gives us the eigenvectors for this eigenvalue as $\left[\begin{matrix}0\\0\\t\end{matrix}\right]$ for $t\in\mathbb{R}$.
I leave the rest up to you.
It is clear that the eigenvector for the eiganvalue $2$ is $(0,0,1)$. The remaining two are more interesting.
Notice that the eigenvalues are roots of the polynomial $x^2-6x+1$. (This is what you get from the characteristic polynomial. I assume this is how you get them.)
In particular, any eigenvalue fulfills $\lambda^2=6\lambda-1$.
Now to find an eigenvector for the eignevalue $\lambda$ you solve the linear system (I have omitted the third variable, since we know it must be zero):
$$
\begin{pmatrix}
1-\lambda & -2 \\
-2 & 5-\lambda
\end{pmatrix}\overset{(1)}\sim
\begin{pmatrix}
1-\lambda & -2 \\
0 & 0
\end{pmatrix}
$$
(Since we know that $\lambda$ is an eigenvalue, the above matrix cannot have full rank. That's why we could replace the second row immediately by zeroes in the step (1).
But you can also check that the second row is indeed a multiple of the first or. Just use the fact that $(\lambda-1)(\lambda-5)=4$ and multiply the first row by the number $\frac{\lambda-5}2=\frac2{\lambda-1}$.)
Clearly, the vector $\vec x=(2,1-\lambda,0)$ is a solution.
We want a normalized vector, so we calculate the norm of this vector: $|\vec x|^2 = 4+(1-\lambda^2)=\lambda^2-2\lambda+5=6\lambda-1-2\lambda+5=4(\lambda-1)$.
So the normalized eigenvectors have the form
$$\frac{\vec x}{|\vec x|}=\left(\frac1{2\sqrt{\lambda-1}},-\frac{\sqrt{\lambda-1}}2,0\right).$$
Remark. I have tried to solve the system in a way in which we get a solution for both eigenvalues at the same time. (To get two eigenvectors for the price of one.) Moreover, in this way I avoided calculations with square roots.
Of course, if you prefer, you can find the solutions for $\lambda_1=3-2\sqrt2$ and for $\lambda_2=3+2\sqrt2$ separately.
In response to the following comment I will add more details:
I don't understand how the last row elements were made $0$? Is it like doing row echelon form? Why is it important to do this? From then how do you say $\vec x =(2,1−\lambda)$ is a solution
Yes, a standard way to solve a system of linear equations is get the matrix to row echelon form. In our case we have a homogeneous system represented by the matrix:
$$\begin{pmatrix}
1-\lambda & -2 \\
-2 & 5-\lambda
\end{pmatrix}$$
As I have mentioned before, we already know that the matrix does not have full rank. Therefore second row is a multiple of the first row, so by doing one simple row operation we get zeroes in the second row.
However, we might want to go a bit slower and perform the actual operations. (What I wrote above is only true if we calculated the eigenvalues correctly. So in this way we doublecheck that they are indeed eigenvalues.)
So we want add a $c$-multiple of the first row to the second row. We want to choose $c$ in such way that we get zero in the first place of the second row. So we need $c=\frac{2}{1-\lambda}$.
What do we get in the second place of the second row? Well, let's calculate:
$$5-\lambda-2c = 5-\lambda - \frac{4}{1-\lambda} = \frac{(5-\lambda)(1-\lambda)-4}{1-\lambda} = \frac{\lambda^2-6\lambda+1}{1-\lambda} = 0$$
So we see that after this row operation we have the new matrix:
$$
\begin{pmatrix}
1-\lambda & -2 \\
0 & 0
\end{pmatrix}
$$
This matrix corresponds to the linear system which has only one equation
$$(1-\lambda)x_1 - 2x_2=0.$$
How can we find a solution? For any value we choose for $x_2$, the above equation uniquely determines $x_1=\frac{2x_2}{1-\lambda}$. We can choose arbitrary non-zero value for $x_2$, So I chose ${1-\lambda}$, as to avoid fractions.
Best Answer
We are given:
$$A = \begin{bmatrix}1 & 2 & 0\\2 & 1 & 0\\0 & 0 & 1\\\end{bmatrix}$$
The eigenvalues are found by solving $|A - \lambda I\ | = 0$, which yields:
$$-\lambda^3 + 3 \lambda^2 + \lambda -3 = -(\lambda-3) (\lambda-1) (\lambda+1)$$
So, we have three distinct eigenvalues, which means we can diagonalize this system.
To find the eigenvectors, we solve $[A - \lambda_i I]v_i = 0$, yielding (we just RREF this system):
To diagonalize the matrix, we have:
$$J = P^{-1} A P$$
$P$ is formed using the column vectors $[v_1 | v_2 | v_3]$.
You should get:
$$J = \begin{bmatrix}-1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 3\\\end{bmatrix}$$
I purposely left the details out so you can work through them.
Notice that since we had three distinct eigenvalues, we could have written the diagonal matrix straight off.