Spivak Change of Variable

Question

I was a bit confused about the following statement. On page 69, Spivak says "in fact, if T is the linear transformation $Dg(a)$, then $(T^{-1}\circ g)'(a)=I $. I may be applying the chain rule incorrectly, but I do not immediately see why the result should be the identity matrix.

Answer 1

Because $(T^{-1}\circ g)'(a)=(T^{-1})'\bigl(g(a)\bigr)\circ g'(a)=T^{-1}\circ Dg(a)=\operatorname{Id}$. Note that the equality $(T^{-1})'\bigl(g(a)\bigr)=T^{-1}$ comes from the fact that $T$ is linear.