I just computed a 15x15 matrix by hand :(
It is not upper triangular as I hoped it would be. But my computations agree with what's offered in the student solution.
My question is: the solution then says "this matrix is nilpotent, so all the eigenvalues are zero."
I get the part where the spectrum = {0} i.f.f. the operator is nilpotent, but how can I tell it actually is nilpotent, just by observing the matrix that I got?
EDIT: Here is my computation of the matrix:
The vector space is the space of polynomials in two variables x,y, of degree less than or equal to 4.
So, I let the basis for this space be the set $${ 1,x,x^2,x^3,x^4, y, y^2, y^3, y^4, xy, xy^2, x^2y, xy^3, x^2y^2, x^3y }$$
The operator is a Laplacian operator on polynomials:
f(x,y) --> f(x+1,y) + f(x-1,y) + f(x,y-1) + f(x,y+1) - 4f(x,y), for f(x,y) in V.
With the above (ordered) basis, I applied this Laplacian operator to each basis element. I wrote each result as a linear combination of the ordered basis. Now, taking the transpose of the coefficients, the matrix w.r.t. the ordered basis is:
$$ \begin{bmatrix} 0 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 6 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 12& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 6 & 0 & 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 12 & 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 6 & 0 & 6 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix} $$
And thanks to the commenters (below), I now see that my matrix is indeed nilpotent, since it is upper triangular, with zeros on the main diagonal; in my initial sketch of the matrix - I was way off. :-)
(I had thought that the 6's were below the diagonal.)
Thanks,
This is too large for a comment, but I thought it might help to shed some light on the underlying reason that the matrix should be expected to be upper-triangular, and indeed nilpotent (so that you would sooner suspect an error in formatting, or perhaps in the choice of basis order).
Your transformation (call it $L[f]$) can be broken down into $L[f] = L_x[f] + L_y[f]$, where
$$ L_x[f(x,y)] = f(x+1,y) + f(x-1,y) - 2f(x,y),$$ $$ L_y[f(x,y)] = f(x,y+1) + f(x,y-1) - 2f(x,y).$$
Now notice that $L_x$ has the effect of reducing the $x$-degree of any polynomial you put into it, without increasing the $y$-degree. That's because the coefficients $(1,1,-2)$ sum to zero so the highest $x$-degree term of $f$ gets cancelled out in computing $L_x[f]$ (and nothing happens to the $y$-coefficients attached to that term).
[In fact we could say more: $L_x$ is a second-difference operator which acts much like a second derivative, so it actually lowers the $x$-degree by $2$.]
Similarly, $L_y$ has the effect of lowering the $y$-degree without raising the $x$-degree. So together, $L_x + L_y$ has the net effect of consistently lowering the degree of $f$. Morally, this is the real reason why the matrix is nilpotent: if you keep lowering the degree enough times, eventually any polynomial you start with will go to $0$, just as in repeated differentiation.
You might have noticed that while computing the effect of the Laplacian on each basis polynomial, the terms that come out always seem to be composed of basis polynomials occurring earlier in the list. If you think about it, this is exactly what it means for your matrix to be (strictly) upper-triangular.
But for your particular choice of basis ordering, there was a bit of coincidence involved (several of the basis polynomials transform to $0$, so the ordering among those polynomials doesn't matter so much). From the above discussion, the idea that each basis polynomial breaks down into terms to the left becomes much more obvious when you sort them by total degree, for instance:
$$1,x,y,x^2,xy,y^2,x^3,x^2y,xy^2,y^3,x^4,x^3y,x^2y^2,xy^3,y^4.$$
This also highlights the fact that questions like these can be sensitive to the basis order. Choosing a different ordering (which is equivalent to swapping rows and then swapping the corresponding columns) can turn a non-triangular matrix into a triangular one. For a random nilpotent matrix this strategy isn't likely to bear fruit [which is probably why it didn't show up in any earlier comments written before you added the critical details of the matrix origin].