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I am really have some difficulty understanding how to do this problem. It asks to show that if T is one-to-one and onto, then T is invertible, and why T being invertible is equivalent to being one to one and onto. ( T: V $\to$ V)

Here is what I know. A linear transformation T is invertible if T $ \circ T^{-1} $= I and $T^{-1} \circ T $ = I as well.

Here is what I have, been I really am not sure. Suppose that T is one to one and onto, and let v $\in V $, because T is onto there exists a v' $\in V $ such that T(v')=v. This v' is also unique because we assumed T was also one-to-one.

Next, suppose there is some transformation $T^{-1}$: V $\to V$, and say $T^{-1}(v)$=v'

then T $\circ T^-{1}(v)$=T($T^{-1}(v$))=T(v')=v and in the other direction of composition, $T^{-1} \circ T (v')$= $T^{-1} $(T(v'))= $ T^{-1} (v) $=v' which shows they both map to the identity.

But I am not sure if this is valid, and I also don't know how to go about showing that the inverse itself is a linear transformation.

Any help, advice, hints or answers would be greatly appreciated.

Thank you.

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  • $\begingroup$ To show the inverse is a linear transformation. Use the definition of "linear transformation" ... check that all its parts are satisfied. $\endgroup$ – GEdgar Feb 14 '15 at 19:05
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Another way of stating your question is

$T$ is invertible $\Leftrightarrow$ $T$ is bijective (one to one and onto).

You wrote down some good first steps for the $\Leftarrow$ direction, that is assuming $T$ is bijective and trying to show that it is invertible. You defined a new map "$T^{-1}$" as follows. Given any $v\in V$ you know that there is some unique $v'\in V$ such that $T(v') = v$. Then you define $T^{-1}(v) = v'$. Your reasoning is fine except you still need to show that the map "$T^{-1}$" is in fact a linear transformation. So far you have defined a map of sets. To do this you'll need to go back and use the definition of linear transformation and show that $T^{-1}$ satisfies all the necessary properties. For example, one thing you will need to show is that $T^{-1}(v+w) = T^{-1}(v) + T^{-1}(w)$.

For the $\Rightarrow$ direction, you get to assume $T$ is invertible. That means there exists an inverse $T^{-1}:V\to V$ such that $T\circ T^{-1} = T^{-1}\circ T = Id_V$. Now show that because $T^{-1}\circ T$ is one-to-one, we must have that $T$ is one-to-one. It's easy to see the contrapositive. If $T(v) = T(w)$ for some $v,w,\in V$ then $T\circ T^{-1}(v) = T\circ T^{-1}(w)$. You may have to think about what this is actually saying because it is an (important) exercise in logic. Similarly you can show because $T\circ T^{-1}$ is onto, we must have that $T$ is onto.

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You are on the right track.

Hints:

  1. If $T:V\to W$ is one-to-one and onto, then, as you concluded, each $w\in W$ has a unique preimage $v\in V$, i.e. $T$ as function, has an inverse. Its linearity simply follows from linearity of $T$ (and uniqueness of the preimage).
  2. Your proof for the other direction seems incomplete. You should show that, given an invertible linear transformation $T$, it is also one-to-one and surjective.
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