# Bilinear Form Diagonalisation

We have the following symmetric bilinear form on $\\Q^4$ (vector space over rational numbers) with respect to standard basis $\{e_1,e_2,e_3,e_4\}$ $$g(v,w)=v^tAw$$ $$A=\begin{bmatrix}1 & 2 &3&4\\2 & 3 & 4 & 5\\3&4&5&6\\4&5&6&7\end{bmatrix}$$

How to find a basis on which g is diagonal. I suppose using Gram-Schmidt algorithm we find orthogonal basis, but could someone explain this explicitly, please?

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Do you know about eigenvectors? – Joe Johnson 126 Mar 13 '13 at 11:04
May be helpful: en.wikipedia.org/wiki/Spectral_theorem – user38268 Mar 13 '13 at 11:23
You don't really need Gram-Schmidt (since you are not looking for an orthonormal basis, and you are working on $Q^4$). If you do the naive thing you'd find that with $v_1 = (1,2,3,4)$ and $v_2 = (0,1,2,3)$. Your $A$ is equal to $v_1 v_1^T - v_2 v_2^T$. It remains to complete this to a basis and finding its dual. – Willie Wong Mar 13 '13 at 11:32
Sorry I need to find orthogonal basis on which g is diagonal. P.S there is some problem in the system and thats why I cant login and posting it as a guest. – user66661 Mar 13 '13 at 23:18
@user66661: you need to register an account and merge your existing accounts. – robjohn Mar 14 '13 at 6:50

There is no need to look for an orthogonal basis. Note that our goal is not to diagonalize $A$ using similarity transform, but to diagonalize $A$ using congruence. That is, what we are looking for is not an invertible matrix $P$ such that $P^{-1}AP$ is diagonal, but an invertible $P$ such that $P^TAP$ is diagonal.
(When $P$ is a real orthogonal matrix, $P^{-1}AP$ and $P^TAP$ coincide, but it is not always possible to find such an orthogonal $P$ over the rationals and this is not our goal anyway.)
You may apply simultaneous elementary row and column reductions to obtain $P$. More specificically, let $$E_1=\begin{pmatrix}1&0&0&0\\-2&1&0&0\\-3&0&1&0\\-4&0&0&1\end{pmatrix}.$$ Then $$B=E_1AE_1^T=\begin{pmatrix}1&0&0&0\\0&-1&-2&-3\\0&-2&-4&-6\\0&-3&-6&-9\end{pmatrix}.$$ Now let $$E_2=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&-2&1&0\\0&-3&0&1\end{pmatrix}.$$ Then $$D=E_2BE_2^T=\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}.$$ Now $D=(E_2E_1)A(E_1^TE_2^T)$. So you may take the basis vectors as the column vectors of $P=E_1^TE_2^T$.
How did you come to choose$\;E_1,E_2\;$ the way you did? – DonAntonio Oct 24 '13 at 14:07