I'm reading Hoffman and Kunze's linear algebra book and on page 112 the authors state the following theorem as a consequence of previous commentaries:

I didn't understand why this transformation is unique.


Suppose $T^t_1$ and $T^t_2$ are two linear transformations with the property above. Then for each $g$, we have $T^t_1(g)=T^t_2(g)$. Why? Well, the property above says exactly "$T^t(g)$ is the map which sends $\alpha$ to $g(T\alpha)$".

What's really going on here is the following: the rule $$T^tg(\alpha)=g(T\alpha),$$ or better yet $$T^t(g)={}"\alpha\mapsto g(T\alpha)"$$ describes a unique function from $W^*$ to $V^*$; it then turns out that this map is actually a linear transformation.

  • $\begingroup$ ha ok! I think I misunderstood what the authors say by "unique" $\endgroup$ – user42912 May 13 '16 at 18:48
  • $\begingroup$ @user42912 They just mean that there is only one linear transformation with that property. $\endgroup$ – Noah Schweber May 13 '16 at 19:04

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