I am reading Linear Algebra: An Introductory Approach by Curtis.
This is what I want to prove:
Let $V$ be a finite dimensional vector space over an algebraically
closed field $F$. Let $T$ be an invertible linear transformation on
$V$. Then $T=DU$ where $D$ is a diagonalizable and $U$ is a unipotent
linear transformation.
Note that a unipotent linear transformation $U$ is by definition such that $1-U$ is nilpotent.
I am familiar with the Primary Decomposition Theorem, Triangular Form Theorem and the Jordan Decomposition Theorem.
I have not been able to prove this but here my attempt:
Let $T$ be a linear transformation on $V$. Since the field $F$ over which $V$ lies is algebraically closed, we can write $T=D+N$ where $D=f(T)$ and $N=g(T)$ for some polynomials $f,g\in F[x]$ and are diagonalisable and nilpotent respectively by the Jordan Decomposition Theorem.
Can some clever algebra do the trick here on the linear transformations? Hints will be appreciated.
Best Answer
There's a detail about the additive Jordan normal form $T = D+N$ you haven't used: $D$ and $N$ commute with each other. Therefore $D^{-1}N$ remains nilpotent, and hence $U = 1+D^{-1}N$ is unipotent.