2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ What signature do you mean? $\endgroup$ – Jlamprong Jan 15 '14 at 12:10
  • $\begingroup$ 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 $\endgroup$ – Oria Gruber Jan 15 '14 at 12:15
  • $\begingroup$ But, your matrix isn't diagonalizable. So, how can we define it? $\endgroup$ – Jlamprong Jan 15 '14 at 12:25
  • $\begingroup$ It's not the matrix I'm talking about when I'm saying diagonal form, I'm talking about $Q(v)$. $\endgroup$ – Oria Gruber Jan 15 '14 at 12:27
  • $\begingroup$ 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 $\endgroup$ – Jlamprong Jan 15 '14 at 12:32
3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Branimir Ćaćić Jan 15 '14 at 14:05
  • $\begingroup$ That's true for all quadratic forms? $\endgroup$ – Oria Gruber Jan 15 '14 at 14:18
  • 1
    $\begingroup$ 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 $B_a$. $\endgroup$ – Branimir Ćaćić Jan 16 '14 at 15:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.