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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

1 Answer 1

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 $P=E_1^TE_2^T$.

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How did you come to choose$\;E_1,E_2\;$ the way you did? –  DonAntonio Oct 24 '13 at 14:07
    
It's just Gaussian elimination, isn't it? –  user1551 Oct 30 '13 at 19:02
    
Very nice. I have been seeing that people do not differentiate well between orthogonal diagonalization and, well, just diagonalization. There is a published algorithm; please see math.stackexchange.com/questions/1388421/… math.stackexchange.com/questions/395634/… math.stackexchange.com/questions/1388281/… –  Will Jagy Aug 22 at 19:21
    
I think there is a good chance you will be able to see el.Salvador's selection from the Lipschutz and Lipson Schaum's outline by clicking on this image link from his answer: i.stack.imgur.com/xMwJ7.png I will check after saving the comment... Yes, worked just fine. –  Will Jagy Aug 22 at 20:31
    
@WillJagy That's really a surprise. I used to think that Schaum's outlines are written for exam preps. –  user1551 Aug 23 at 14:08

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