Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 3 / 17C (Linear Mathematics)

Work in the finite-dimensional vector space $V$ implicit in the requested matrix representation and $n=\dim V$. Put $K_j=\ker {\alpha }^j$, with $K_0=0$. These form an increasing sequence ending in $K_m=V$, and $\alpha (K_j)\subseteq K_{j-1}$. Choose a basis of $K_1$, extend it to a basis of $K_2$, and continue until a basis of $V$ is obtained. Each basis vector introduced at stage $j$ maps into the span of vectors chosen at earlier stages. Its matrix column can therefore have nonzero entries only in earlier rows. This proves that the matrix is strictly upper triangular.

For a strictly upper-triangular $n\times n$ matrix $N$, a nonzero contribution to $(N^r{)}_{ij}$ requires an index chain $i<i_1<\cdots <i_{r-1}<j$. No such chain has $n$ links among $n$ indices. Consequently $\boxed{{\alpha }^n=0}$. An example with exact index four is the Nilpotent Jordan blockHere $M^3$ has entry one in position $(1,4)$ and all others zero, while $M^4=0$.

In the adapted basis, $I+A$ is upper triangular with diagonal entries one, so all its eigenvalues are one. This does not persist for the product of two such matrices with unrelated adapted bases. TakeBoth square to zero, buthas characteristic polynomial $t^2-3t+1$ and eigenvalues $(3\pm \sqrt{5})/2$, neither equal to one. The obstruction is that the two nilpotent matrices need not admit a common triangularizing basis.