Can en eigen value be a vector? Given 

$$b=\sqrt \frac{3}{22}\begin{pmatrix}2\\2\\-1
\end{pmatrix}\\
B=\frac 1 9 \begin{pmatrix}1 &amp; 1 &amp; 0\\
1 &amp; 1 &amp; 0\\
0 &amp; 0 &amp; 0
\end{pmatrix}$$

How can I deduce the eigenvalue $\lambda$ associated to $b$ from $b'Bb$? 

Because 

\begin{align}
b'Bb = \lambda b \Leftrightarrow \overbrace{\frac{3}{22}\begin{pmatrix}2 &amp; 2 &amp; -1\end{pmatrix}\frac 1 9 \begin{pmatrix}1 &amp; 1 &amp; 0\\
1 &amp; 1 &amp; 0\\
0 &amp; 0 &amp; 0
\end{pmatrix}}^{\mbox{isn't a scalar !!}}b=\lambda b\\
\end{align}
Presumably you are using $b'$ to be the transpose of $b$.  It should be $b'Bb=b'(Bb)=b'\lambda b=\lambda b'b$ which is a scalar.  But in your example $b$ is not an eigenvalue of $B$.  Leaving aside the constants we have $$Bb= \begin{pmatrix}1 &amp; 1 &amp; 0\\
1 &amp; 1 &amp; 0\\
0 &amp; 0 &amp; 0
\end{pmatrix}\begin{pmatrix}2\\2\\-1
\end{pmatrix} =\begin{pmatrix}4\\4\\0
\end{pmatrix}$$
which is not proportional to $b$
I believe you have the wrong formula. You want

$$b'Bb = \lambda b'b$$

This comes from,

$$Bb = \lambda b$$
$$b'Bb = b'\lambda b$$
$$b'Bb = \lambda b'b$$

and both sides of that equation are scalar.