I just computed a 15×15 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,
Best Answer
Your matrix is a strictly upper triangular matrix. Upper triangular matrices $n\times n$ are all nilpotent since their characteristic polynomials, $\det (A-XI_n)$, is equal to $(-1)^nX^n$. According to Cayley-Hamilton's Theorem, $(-1)^nA^n=0\Rightarrow A^n=0$.