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.
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 16:42And 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