link to image of matrix A:
http://s8.postimg.org/iel25yp9h/EIGENVECTORS.jpg
[1 -1 0]
[-1 2 -1]
[ 0 -1 1]
Find the eigenvectors

(1−λ)((2−λ)(1−λ)−1)+(−1)(1−λ)=
−λ3+4λ2−3λ=
−(λ−3)(λ−1)λ=0
eigenvalues found:
λ1=0, λ2=1, λ3=3

Thanks in advance...

What step of the process (see for example, Eigenvalues and Eigenvectors - HMC Calculus Tutorial (http://www.math.hmc.edu/calculus/tutorials/eigenstuff/) ) do you get stuck at?

What step of the process (see for example, Eigenvalues and Eigenvectors - HMC Calculus Tutorial (http://www.math.hmc.edu/calculus/tutorials/eigenstuff/) ) do you get stuck at?

For λ = 0, we solve (A - 0I)v = 0:
[1 -1 0|0]
[-1 2 -1|0]
[0 -1 1|0], which reduces to

[1 0 -1|0]
[0 1 -1|0]
[0 0 0|0], yielding eigenvector v = (1, 1, 1).
--------------
For λ = 1, we solve (A - 1I)v = 0:
[0 -1 0|0]
[-1 1 -1|0]
[0 -1 0|0], which reduces to

[1 0 1|0]
[0 1 0|0]
[0 0 0|0], yielding eigenvector v = (-1, 0, 1).
--------------
For λ = 3, we solve (A - 3I)v = 0:
[-2 -1 0|0]
[-1 -1 -1|0]
[0 -1 -2|0], which reduces to

[1 0 -1|0]
[0 1 2|0]
[0 0 0|0], yielding eigenvector v = (1, -2, 1).

is it OK?
Thanks in advance...

Yes - you have the correct values for eigenvalues and eigenvectors. You can check your results by seeing if the original matrix times an eigenvector equals the eigenvalue multiple of the eigenvector. For example:


\left| \begin{matrix} \ 1 & -1 & \ 0 \\ -1 & \ 2 & -1 \\ \ 0 & -1 & \ 1 \end{matrix} \right| \left| \begin{matrix} \ 1 \\ -2 \\ \ 1 \end{matrix} \right| = \left| \begin{matrix} \ 3 \\ -6 \\ \ 3 \end{matrix} \right| = 3 \left| \begin{matrix} \ 1 \\ -2 \\ \ 1 \end{matrix} \right|