Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 2 / 15G (Linear Mathematics)

If $U$ is a unipotent matrix, $N=U-I$ is a nilpotent matrix. If $Uv=\lambda v$ for a nonzero eigenvector, then $N^{m}v=(\lambda -1{)}^{m}v=0$ for some $m$, forcing $\lambda =1$.

Conversely, suppose the only eigenvalue is $1$. Over $\mathbb{C}$, the characteristic polynomial is then $(t-1{)}^n$. The Cayley-Hamilton theorem gives $(U-I{)}^n=0$, so $U-I$ is nilpotent. Thus $U$ is unipotent if and only if its only eigenvalue is one.

Choose an invertible change-of-basis matrix $P$ so that $J=PT{P}^{-1}$ is the Jordan normal form of $T$. Write each block aswhere $N_{m_j}$ has ones on its first superdiagonal and all other entries zero. DefineEvery ${\lambda }_j\ne 0$ because $T$ is invertible, so $D_0$ is an invertible diagonal matrix. The matrix $N$ is strictly upper triangular and hence nilpotent. On each block $D_0$ is a scalar multiple of the identity, so it commutes with $N$ on that block and therefore on the whole space. This gives

For $D=P^{-1}{D}_{0}P$, conjugating $U=D^{-1}T$ givesBecause $D_0^{-1}$ commutes with $N$, and $N^n=0$,Thus $U-I$ is similar to a nilpotent matrix, and is itself nilpotent. Therefore $U$ is a unipotent matrix.

The definition gives $T=DU$. The matrix $D=P^{-1}{D}_{0}P$ is invertible and diagonalizable. In the same basis $U$ is $I+D_0^{-1}N$, which commutes with $D_0$ because $N$ does. Conjugating back givesThis is the multiplicative Jordan decomposition in an affine algebraic group, here applied to the general linear group. The explicit construction also handles repeated eigenvalues and nontrivial Jordan blocks.