Commit 2e5136d7 by 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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