- #1

PiAreSquared

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I have been working through Linear Algebra Done Right by Axler and I have a quick question about Linear Maps, and in particular, their inverses. My question arose while working through the following proof:

A linear map is invertible iff it is bijective.

My qualm is not with the proof itself, as it is quite straightforward. My question however is this: When assuming that a linear map T:V [itex]\rightarrow[/itex] W is bijective and showing that T must then be invertible, the proof first shows that if, for each w [itex]\in[/itex] W, Sw is defined to be the unique element of V that is mapped to w, that is T(Sw) = w, then clearly TS=I and it is shown that ST=I, where the first I is the Identity Map on W and the second is the Identity Map on V. This direction of the proof then finishes by showing that S is linear. But my question is, if S satisfies the properties: ST=I and TS=I, but S is not linear, what is S? Is it just a non-linear map that "undoes" T, because by definition, S cannot be the inverse of T?

Thanks.