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Question: Let us consider the quadratic form $q: R^3 -> R$, $$q(x,y,z) = x^2+25y^2+10xy+2xz$$Find the corresponding symmetric bilinear form $f$ and a basis $B$ such that $[f]_B$ has the real canonical form. State the signature and rank of $[f]_B$.

I thought that the best way to start this question was to find an orthogonal basis, C, using the Gram Schmitd process. I got the vectors $(1,0,0),(-5,1,1),(\frac{5}{2},-\frac{1}{2},\frac{1}{2})$ as my orthogonal basis, but I dont know how to use these to find a basis B such that $[f]_B$ has the real canonical form. Any help would be greatly appreciated. Thanks.

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A way to proceed to get the signature and the rank is to reduce $q$ thanks to the Gauss' algorithm : $$q\left(x,y,z\right)=x^2+25y^2+10xy+2xz =\left(x+5y+z\right)^2-10yz-z^2 =\left(x+5y+z\right)^2-\left(z+5y\right)^2+25y^2.$$ Let then $X=\left(x+5y+z\right)$, $Y=\left(z+5y\right)$ and $Z=5y$. Then the reduced equation of $q$ is given by $$X^2-Y^2+Z^2=0.$$ It is a cone (signature $\left(2,1\right)$, rank=$3$). Furthermore, we have $$\begin{pmatrix}X\\ Y\\ Z \end{pmatrix}=\underbrace{\begin{pmatrix}1 & 5 & 1\\ 0 & 5 & 1\\ 0 & 5 & 0 \end{pmatrix}}_{=:P}\begin{pmatrix}x\\ y\\ z \end{pmatrix} $$ and the matrix $$P^{-1}=\begin{pmatrix}1 & -1 & 0\\ 0 & 0 & \frac{1}{5}\\ 0 & 1 & -1 \end{pmatrix}$$ is the passage matrix from the canonical basis to a base where $q$ as the above reduced form.

Verification : The representative matrix in the canonical base of $\mathbb{R}^3$ of the polar bilinear form associated to $q$ is $$\begin{pmatrix}1 & 5 & 1\\ 5 & 25 & 0\\ 1 & 0 & 0 \end{pmatrix}.$$ We have : $$q\left(x,y,z\right)=\begin{pmatrix}x & y & z\end{pmatrix}\begin{pmatrix}1 & 5 & 1\\ 5 & 25 & 0\\ 1 & 0 & 0 \end{pmatrix}\begin{pmatrix}x\\ y\\ z \end{pmatrix}=\begin{pmatrix}x\\ y\\ z \end{pmatrix}^{T}\begin{pmatrix}1 & 5 & 1\\ 5 & 25 & 0\\ 1 & 0 & 0 \end{pmatrix}\begin{pmatrix}x\\ y\\ z \end{pmatrix}$$

$$=\left(P^{-1}\begin{pmatrix}X\\ Y\\ Z \end{pmatrix}\right)^{T}\begin{pmatrix}1 & 5 & 1\\ 5 & 25 & 0\\ 1 & 0 & 0 \end{pmatrix}\left(P^{-1}\begin{pmatrix}X\\ Y\\ Z \end{pmatrix}\right)=\begin{pmatrix}X\\ Y\\ Z \end{pmatrix}^{T}\left(P^{-1}\right)^{T}\begin{pmatrix}1 & 5 & 1\\ 5 & 25 & 0\\ 1 & 0 & 0 \end{pmatrix}P^{-1}\begin{pmatrix}X\\ Y\\ Z \end{pmatrix}$$ $$=\begin{pmatrix}X & Y & Z\end{pmatrix}\left(\left(P^{-1}\right)^{T}\begin{pmatrix}1 & 5 & 1\\ 5 & 25 & 0\\ 1 & 0 & 0 \end{pmatrix}P^{-1}\right)\begin{pmatrix}X\\ Y\\ Z \end{pmatrix} =\begin{pmatrix}X & Y & Z\end{pmatrix}\begin{pmatrix}1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix}X\\ Y\\ Z \end{pmatrix}=X^{2}-Y^{2}+Z^{2} $$

hence $q$ is reduced on the basis associated to the matrix $P^{-1}$ (where we read its rank and its signature).

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  • $\begingroup$ Cheers. I managed to work out the question using a much longer method - I found an orthogonal basis using Gram-Schmidt and then used Sylvesters Law to convert it into the new basis B. Thanks for this quicker method though! $\endgroup$ – Arron Apr 28 '15 at 18:19
  • $\begingroup$ Sorry I have one quick question. In order to calculate the real canonical form using the method you have supplied, do you have to calculate $(P^{-1})^TA(P^{-1})$, where $A$ is the quadratic form in matrix representation? This is what I read online, however when I tried that with the $P^{-1}$ you have supplied, I did not a get a matrix in real canonical form. $\endgroup$ – Arron Apr 28 '15 at 19:10
  • $\begingroup$ I add some comments in my answer to precise some details. $\endgroup$ – Nicolas Apr 28 '15 at 19:34
  • $\begingroup$ Brilliant thank you for all your help! $\endgroup$ – Arron Apr 28 '15 at 20:02
  • $\begingroup$ You're welcome! $\endgroup$ – Nicolas Apr 28 '15 at 20:05

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