Very simple question but something doesn't make sense to me.
We are given a quadratic form (bilinear map but on the same vector twice):
$Q(v) = v^t *\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 0\end{pmatrix}*v$
and we are asked to find the signature....But this matrix isn't symmetric. Not only is it not symmetric, it's not diagonlizable. Is the signature defined well in this case?
Think I got it.
Suppose $u=\begin{pmatrix} x \\ y \\ z\end{pmatrix}$.
$Q(u)=\begin{pmatrix} x & y & z\end{pmatrix} \begin{pmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\1 & 1 & 0\end{pmatrix} \begin{pmatrix} x\\y\\z\end{pmatrix} = \begin{pmatrix}x+z \\ x+y+z \\y\end{pmatrix} \begin{pmatrix} x \\y\\z\end{pmatrix}=x^2+xz+xy+y^2+2yz$ which corresponds to the matrix:
$\begin{pmatrix}1 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2}& 1 & 1\\ \frac{1}{2} &1 &0\end{pmatrix}$
Where the first column is x and row is x, second column and row is y, third column and row is z. And the signature of that matrix is (2,1,0) and that's the answer.