# Find the signature of the quadratic form

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?

• What signature do you mean? – Jlamprong Jan 15 '14 at 12:10
• I mean $p,q,r$ where $p$ is the number of positive values of the diagonal form, $q$ is the number of negative values of the diagonal form, and $r$ is the number of $0$s in the diagonal form – Oria Gruber Jan 15 '14 at 12:15
• But, your matrix isn't diagonalizable. So, how can we define it? – Jlamprong Jan 15 '14 at 12:25
• It's not the matrix I'm talking about when I'm saying diagonal form, I'm talking about $Q(v)$. – Oria Gruber Jan 15 '14 at 12:27
• OK, if $v=(x,y,z)^T$ then $Q(v)= x(x+z)+y(x+y+z)+yz$ which is scalar. So the diagonal form of $Q(v)$ is itself – Jlamprong Jan 15 '14 at 12:32

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.

• There's actually a very general fact underlying your answer: if $A$ is a square matrix, then $$v^T A v = v^T A_{\text{sym}} v, \quad A_{\text{sym}} = \frac{1}{2}(A+A^T).$$ More abstractly, the quadratic form associated to a bilinear form only sees the symmetric part of the bilinear form. – Branimir Ćaćić Jan 15 '14 at 14:05
• That's true for all quadratic forms? – Oria Gruber Jan 15 '14 at 14:18
• Yes. If $B$ is a bilinear form, then $$B = B_s + B_a,$$ where $$B_s(x,y) = \frac{1}{2}(B(x,y) + B(y,x))$$ is the symmetric part of $B$ and $$B_a(x,y) = \frac{1}{2}(B(x,y)-B(y,x))$$ is the antisymmetric part of $B$. Hence, $$B(x,x) = B_s(x,x) + B_a(x,x) = B_s(x,x),$$ since $B_a(x,x) = 0$ by antisymmetry of $B_a$. – Branimir Ćaćić Jan 16 '14 at 15:18