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.

  • $\begingroup$ 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

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.


You are on the right track.


  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.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.