Commit 2e5136d7 authored by Jim Hefferon's avatar Jim Hefferon

done map4

parent 80e0323c
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......@@ -1875,7 +1875,7 @@ are equivalent
statements about a linear map \( \map{h}{V}{W} \).
\begin{tfae}
\item \( h \) is one-to-one
\item \( h \) has an inverse, from its range to its domain, that is linear
\item \( h \) has an inverse from its range to its domain that is linear
\item \( \nullspace{h}=\set{\zero\,} \), that is, \( \nullity(h)=0 \)
\item \( \rank (h)=n \)
\item if \( \sequence{\vec{\beta}_1,\dots,\vec{\beta}_n} \)
......@@ -1894,9 +1894,10 @@ We will then show that
\).
For \( \text{(1)} \Longrightarrow \text{(2)} \),
suppose that the linear map $h$ is one-to-one and so has an inverse.
suppose that the linear map $h$ is one-to-one and so has an inverse
$\map{h^{-1}}{\rangespace{h}}{V}$.
The domain of that inverse is the range of $h$ and thus a linear combination
of two members of that domain has the form $c_1h(\vec{v}_1)+c_2h(\vec{v}_2)$.
of two members of it has the form $c_1h(\vec{v}_1)+c_2h(\vec{v}_2)$.
On that combination, the inverse \( h^{-1} \) gives this.
\begin{align*}
h^{-1}(c_1h(\vec{v}_1)+c_2h(\vec{v}_2))
......@@ -1907,7 +1908,7 @@ On that combination, the inverse \( h^{-1} \) gives this.
% +c_2\composed{h^{-1}}{h}\,(\vec{v}_2) \\
&=c_1\cdot h^{-1}(h(\vec{v}_1))+c_2\cdot h^{-1}(h(\vec{v}_2))
\end{align*}
Thus if a linear map is one-to-one, that is, if it has an inverse, then
Thus if a linear map has an inverse, then
the inverse must be linear.
But this also gives the \( \text{(2)} \Longrightarrow \text{(1)} \)
implication, because the inverse itself must be one-to-one.
......
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