# Lie Groups question from Brian Hall's Lie Groups, Lie Algebras and their representations.

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?

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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.
I know, but why can I plug $t=1$ to get $\exp(\log A)=A$ if it's satified only for small values of $t$? –  MathematicalPhysicist Oct 26 '12 at 3:27