Let $\varphi, \phi$ be quadratic forms on $V$ and suppose $\varphi$ is positive definite.

I want to find a basis for V such that $\varphi$ and $\phi$ are both represented by diagonal matrices.

My idea is to define an inner product $<,>: V\times V \rightarrow F$ where $<v,w> = \varphi(v,w)$.

I know that if I can find a basis that is orthonormal w.r.t. this inner product that diagonalises $\phi$, then I am done, since $\varphi$ will be represented by the identity with respect to this basis.

I know that I can use Gram-Schmidt to get an orthonormal basis for $V$, but I don't understand how to choose a basis that diagonalises $\phi$.

I realised I didn't understand what I was doing when I tried the example where $\phi$ is the symmetric bilinear form associated to $2x^2 + 3y^2 +3z^2 - 2yz = (\sqrt{3}z - \frac{1}{\sqrt{3}}y)^2 + 2x^2 + \frac{8}{3}y^2$ which is positive definite, and wish to simultaneously diagonalise this and $\phi$ which is the symmetric bilinear form associated to the quadratic form $3x^2 + 3y^2 + z^2 +2xy - 3xz + 3yz$.

Help in general, or with relevance to this particular example, gratefully received.

  • $\begingroup$ Well, seeing as this is mostly a question about bilinear forms, the natural notion of diagonalisation is invertible $P, P^T AP$. I'm not so much after a quick way to do the computation, more an explanation of what is going on when diagonalising $\phi$. I don't have access to the book, unfortunately. $\endgroup$ – probablystuck May 10 '17 at 14:28
  • 1
    $\begingroup$ Notably, $\varphi$ and $\phi$ are both "phi". $\endgroup$ – Omnomnomnom May 10 '17 at 14:41

Let $\varphi$ denote the positive definite form. Let $\psi$ denote the second form. I'll assume you're working over $F = \Bbb R$; for $F = \Bbb C$ you'd have to clarify whether we're considering a "sesquilinear" quadratic form.

If $A$ is the matrix of $\psi$ relative to a $\varphi$-orthonormal basis, you question amounts to asking whether there exists an orthogonal matrix $U$ such that $U^TAU$ is diagonal. By the spectral theorem, this is possible whenever $A$ is symmetric. Since $A$ is the matrix associated with a quadratic form, it is always possible; $A$ is necessarily symmetric.

In order to find such a basis, it suffices to find an orthonormal eigenbasis of $A$. That is: from our bilinear form $\psi$, we get a linear transformation $T_\psi$. Once we select mutually orthogonal eigenvectors of $T_\psi$, we have the columns of our desired $U$. To that end, it suffices to find any basis eigenvectors, then apply the Gram-Schmidt process.

  • $\begingroup$ My issue was, I think, that I was making errors during the computation whilst trying to implement the process you describe, and the fact that I couldn't get it to fall out made me assume it was wrong. Thanks - your answer meant I kept at it a while longer until it became clear where the problems lay. $\endgroup$ – probablystuck May 10 '17 at 15:48

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.