2
$\begingroup$

In page 60 of Hall's textbook, ex. 8 assignment (c), he asks me to prove that if $A$ is a unipotent matrix then $\exp(\log A))=A$.

In the hint he gives to show that for $A(t)=I+t(A-I)$ we get

$\exp(\log A(t)) = A(t) , \ t<<1$

I don't see how I can plug here $t=1$, this equality is true only for $t \rightarrow 0$, right?

2 Answers 2

3

Initially, we only have the equality for small $t$. But if we also know that both sides are polynomial functions of $t$, then equality for all sufficiently small $t$ implies equality for all $t$.

3

If $A$ is unipotent, then $A - I$ is nilpotent, meaning that $(A-I)^n = 0$ for all sufficiently large $n$. This will turn your power series into a polynomial.

  • 0
    I know, but why can I plug $t=1$ to get $\exp(\log A)=A$ if it's satified only for small values of $t$?2012-10-26