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Given a matrix $A$ find a two dimensional subspace $V \subset\mathbb{R}^4$ for which $\forall x \in V : x^TAx=0$

$$A = \begin{pmatrix}1&2&0&1\\ 2&3&1&1\\ 0&1&0&1\\ 1&1&1&-1 \end{pmatrix}$$

I understand that I have to find a subspace whose image in relation to the matrix $A$ is orthogonal to itself (original subspace).

$$Ax = y$$ $$ x^Ty = 0 \iff x \perp y $$

The question is how to do it? I know that I could construct 4 quadratic equations directly from $x^TAx=0$ but I believe that there has to be an easier way of doing things.

EDIT: Ok, I actually can't, at best I'm able to construct one equation which doesn't help me much.

Petrroll
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    The matrix is symmetric, so it can be diagonalised with an orthogonal matrix. What can you conclude from that? – copper.hat May 26 '15 at 16:28
  • That it has 4 real eigenvalues, 4 orthogonal eigenvectors and that it is possible to create an eigendecomposition. (damn, it's hard to find translations to these math stuff :) ) – Petrroll May 26 '15 at 16:38
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    Before you spend energy there, $\det A \neq 0$ should be a warning... – copper.hat May 26 '15 at 16:39
  • Well, I know that it's kernel contains only null vector, but I just can't see how is that supposed to help me.

    I know that if the kernel was at least two dimensional I just use it as $Ax$ would always be zero and therefore even $x^TAx$...

    – Petrroll May 26 '15 at 16:42
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    BTW: It's an exam exercise so given the fact that the eigennumbers of matrix $A$ not very numerically nice (definitely not easy to be computed by hand) I'd suspect that it has to be doable without eigendecomposition. – Petrroll May 26 '15 at 16:45
  • is the set ${x:x^\top Ax = 0}$ even a subspace? – abel May 26 '15 at 16:52
  • AFAIK it doens't have to be and probably even is not. But there's (well, unless the answer to the exrercise is "It's not possible") a subset of the said set that is a two dimensional subspace. – Petrroll May 26 '15 at 16:57
  • This is Witt cancellation, it is not necessary to find eigenvectors. Reduction of the quadratic form in the general method of Hermite suffices. Why do you say this is an exam question? – Will Jagy May 26 '15 at 17:00
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    Because it is an exam question. Obviously an old one released for students to prepare for the exam.

    And it being an exam question means its solution probably doesn't require finding eigenvectors (which is really hard to do by hand for $4x4$ matrixes) which I thought was suggested.

    – Petrroll May 26 '15 at 17:03
  • And I probably see how it's supposed to be done, thanks! – Petrroll May 26 '15 at 17:16
  • This seems like a computationally tough question for an exam. – copper.hat May 26 '15 at 17:30

1 Answers1

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They may or may not have mentioned the name Hermite, or Witt, in this area. The idea is to start at the upper left corner, choose coefficients in a square, perhaps with a constant coefficient, that wipes out the first row and column. I ordered my variables $(x,y,z,w),$ the coefficient of $x^2$ is one, to take care of the top row and left column I used $(x + 2 * y + w)^2.$ After that, the variable $x$ no longer appears anywhere, and we begin the same process with the letter $y,$ where we needed a coefficient of $-1$ to continue. The quadratic form indicated by your matrix is

$$ (x + 2 y + w)^2 - ( y - z + w)^2 + z^2 - w^2. $$ This task is fairly quick, no eigenvalues... more typing in a minute

This is ad hoc, I wrote the null cone as $$ (x + 2 y + w)^2 - w^2 = ( y - z + w)^2 - z^2, $$ or factoring differences of squares, $$ ( x + 2 y) (x+2y+2w) = (y+w)(y-2z+w). $$ One way to make this true is to require $$ y+w = 0; \; \; x=0 $$ which specifies a $2$-plane in $\mathbb R^4,$ because $(x+2y+2w)$ also becomes zero.

EDIT: a few hours later. I scribbled some things. There are infinitely many 2-planes contained in this null cone. Take any four real numbers $A,B,C,D$ such that $$ A^2 + B^2 = C^2 + D^2. $$ There is a 2-plane contained in the null cone spanned by the $(x,y,z,w)$ vectors $$ (A-2B-2C+D, \; B+C-D, \; B, \; D ), $$ $$ (-2A-B+C+2D, \; A-C-D, \; A, \; C). $$

There is a differential geometry phrase that goes with this: there are infinitely many 2-planes passing through the origin and contained in the cone over the Clifford torus.

Will Jagy
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  • Wow, what I was thinking was complete overkill. – copper.hat May 26 '15 at 17:36
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    @copper.hat, right, no need for eigenvalues, the null cone of a quadratic form is not the same as the nullspace of its Gram/Hessian matrix. Now I need to think of how to extract a vector space in the null cone, never done that before. – Will Jagy May 26 '15 at 17:38
  • @copper.hat, that worked out. I should read some of my books, on Witt's Lemma, and see how the final step happens without guessing at factoring polynomials. – Will Jagy May 26 '15 at 17:54