Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 1 / 8E (Linear Algebra  )

The Nilpotent Jordan block $J_d$ shifts each standard basis vector one place toward the first coordinate. ThereforeThus $J_d^0=I$, for $1\le n<d$ the matrix has ones precisely on its $n$th superdiagonal, and

Since $\lambda I$ commutes with $J_d$, the binomial theorem for commuting matrices and part (i) giveEquivalently, its $k$th superdiagonal is constant with value $\left(\genfrac{}{}{0ex}{}{n}{k}\right){\lambda }^{n-k}$ for $0\le k<d$, with the convention that this value is zero when $k>n$.

If $\varphi v=\mu v$ for a nonzero eigenvector $v$, thenHence every eigenvalue satisfies $\mu =0$. Since a complex endomorphism has an eigenvalue, zero is the only possible eigenvalue.

The Jordan normal form can contain only blocks $J_r(0)$ with eigenvalue zero. Moreover,Thus the possible blocks are precisely the nilpotent Jordan blocks of sizes $1\le r\le n$.

Assume ${\varphi }^2=0$. ThenSet $W_1=\mathrm{im}\varphi$, choose a direct-sum complement $U$ such that $\ker \varphi =U\oplus W_1$, and choose a complement $W_2$ such that $V=\ker \varphi \oplus W_2$.

The restriction $\varphi {|}_{W_2}:W_2\to W_1$ is injective because $W_2\cap \ker \varphi =0$, and it is surjective because every image $\varphi (v)$ equals $\varphi (w_2)$ after decomposing $v=k+w_2$. It is therefore a isomorphism, so $\dim W_2=\dim W_1$ and $\varphi (W_2)=W_1$. Also $\varphi (U)=\varphi (W_1)=0$. Hencehas all the required properties.