Stuck on finding eigenvalues and eigenvectors of 3x3 matrix Im trying to calculate the eigenvalues and eigenvectors of the following matrix:
$\begin{bmatrix}1 & 1 & 0\\1& 1 & 1\\0 &1 &1\end{bmatrix}$
so far I worked out:
$A-λI=\begin{bmatrix}1-λ & 1 & 0\\1& 1-λ & 1\\0 &1 &1-λ\end{bmatrix}$
I know you have to calculate the determinant but im not exactly sure how to do it.
It is also asking me to Show that if the eigenvectors of A, {u1,u2,u3}, are normalized to unit length then they form an orthonormal set, i.e. uTi uj = δij where δij = 1 if i = j and δij = 0 if i ̸= j. Hence write A in the form A = UΛUT where Λ is a diagonal matrix and UT U = I3.
Please help.
 A: You need to solve:
$\det\left(A-λI\right)=\det \left(\begin{bmatrix}1-λ & 1 & 0\\1& 1-λ & 1\\0 &1 &1-λ\end{bmatrix}\right) \overset{!}{=} 0$
In order to calculate the determinant, I suggest you use Sarrus's rule Sarrus's Rule.
The resulting polynomial 
$(1-\lambda)^3 -2\,(1-\lambda) = (1-\lambda)\,(\lambda^2-2\lambda +1)\overset{!}{=} 0 $.
You should be able to take it from here on.
You will end up with three Eigenvalues. The corresponding eigenvectors can be obtained, finding a solution to
$ (A-\lambda_i)\, v_i \overset{!}{=} 0 $, for $i = 1,2,3$.
Note that your matrix $A$ is a symmetric matrix.
Let $v_1,v_2$ be eigenvectors to a symmetric matrix $A$ with non-ientical eigenvalues $\lambda_1,\lambda_2$ Than:
$\lambda_1\,v_1^T\,v_2 = (A\,v_1)^T\,v_2 = v_1^T\,A^T\,v_2  \overset{A=A^T}{=} v_1^T(A\,v_2) = \lambda_2 \, v_1^T\, v_2$
and thus
$(\lambda_1 -\lambda_2)v_1^T\, v_2 = 0$.
Given $\lambda_1 \neq \lambda_2$, follows that $v_1^T\,v_2 = 0$, which is the definition of orthogonality.
Summing up: 
A symmetric, real matrix $A$


*

*(has only real eigenvalues)

*It's eigenvectors to corresponding, non-identical eigenvalues, are orthogonal 

A: Hint:
You can calculate the determinant developing by the first row:
$$
\det
\begin {bmatrix}
1-\lambda&1&0\\
1&1-\lambda&1\\
0&1&1-\lambda
\end{bmatrix}=
(1-\lambda)\left(\det\begin {bmatrix}
1-\lambda&1\\
1&1-\lambda
\end{bmatrix}\right)-1\left( \det
\begin {bmatrix}
1&1\\
0&1-\lambda
\end{bmatrix}\right)=(1-\lambda)\left[(1-\lambda)^2-1\right]-(1-\lambda)=
(1-\lambda)(\lambda^2-2\lambda-1)
$$
so you find the eigenvalues:
$$
\lambda_1=1 \qquad \lambda_2=1+\sqrt{2} \qquad \lambda_3=1-\sqrt{2 }
$$
Now, solving the linear equations:
$$
\begin {bmatrix}
1&1&0\\
1&1&1\\
0&1&1
\end{bmatrix}
\begin {bmatrix}
x\\y\\z
\end {bmatrix}=\lambda_i\begin {bmatrix}
x\\y\\z
\end {bmatrix}
$$
you can find the eigenvectors corresponding to each eigenvalue $\lambda_i$.
E.g., for $\lambda_1=1$ you have:
$$
\begin{cases}
x+y=x\\
x+y+z=y\\
y+z=z
\end{cases}
$$
So the eigenspace is given by the vectors with $y=0$ and $x=-z$ and you can chose as an eigenvector $u_1=[-1,0,1]^T$.
In the some way you can find the eigenvectors for $\lambda_2$ and $\lambda_3$ and find: $u_2=[1,\sqrt{2},1]^T$,$u_3=[1,-\sqrt{2},1]^T$.
Since they are eigenvectors of different eigenvalues they are are orthogonal (you can verify this calculating the inner product) and you can normalize dividing by the modulus.
Finally: the matrix $\Lambda$ has as diagonal elements the eigenvalues and the matrix $U$ has as columns the corresponding eigenvectors.
You can see here for an other example.
