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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?

glS
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user66566
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  • Do you know about eigenvectors? – J126 Mar 13 '13 at 11:04
  • May be helpful: http://en.wikipedia.org/wiki/Spectral_theorem –  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. –  Mar 13 '13 at 23:18
  • @user66661: you need to register an account and merge your existing accounts. – robjohn Mar 14 '13 at 06:50

1 Answers1

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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 $P=E_1^TE_2^T$.

user1551
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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
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    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 http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr http://math.stackexchange.com/questions/395634/given-a-4-times-4-symmetric-matrix-is-there-an-efficient-way-to-find-its-eige/1392600#1392600 http://math.stackexchange.com/questions/1388281/find-the-transitional-matrix-that-would-transform-this-form-to-a-diagonal-form/1391117#1391117 – Will Jagy Aug 22 '15 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: http://i.stack.imgur.com/xMwJ7.png I will check after saving the comment... Yes, worked just fine. – Will Jagy Aug 22 '15 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 '15 at 14:08
  • user, all a bit mysterious. My (many) quadratic forms books do this in Hermite's direction, until recently i always answered that way. Last night I saw this, solved it by following Horn and Johnson, but left that as a reference in hopes the kid would get his hands dirty and learn something: http://math.stackexchange.com/questions/1406359/simultaneously-diagonalise-two-real-quadratic-forms I've never before had reason to simultaneously make two q. forms diagonal by congruence, but Horn and Johnson say it's important sometimes, some physics – Will Jagy Aug 23 '15 at 16:49
  • Oh, you answered the same thing http://math.stackexchange.com/questions/88022/congruence-and-diagonalizations – Will Jagy Aug 23 '15 at 16:53
  • in the final 2 by 2 step, one may use rational or integer matrix, I left you a comment there – Will Jagy Aug 23 '15 at 17:05
  • @WillJagy If you google "simultaneous diagonalization by congruence" (with double quotes), a few applications show up. Apparently some are related to weather prediction, some to signal processing and some to eigenvalue problems. – user1551 Aug 26 '15 at 21:01
  • Thanks. I requested a book from the public library that had some as exercises, by T. W. Korner. I think he always has a positive definite one that is to be sent to the identity, so he will generally be expecting square roots in the matrix $P$ with $P^T A P = I, P^T B P = D.$ Interesting; but it appears we are over the short time when a few people were asking about these topics. – Will Jagy Aug 26 '15 at 21:09