LA - Identity Maps and Injectivity

[MushroomPirat]

Homework Statement

Suppose that $W$ is finite dimensional and $T:V\rightarrow W$. Prove that $T$ is injective if and only if there exists $S:W\rightarrow V$ such that $ST$ is the identity map on $V$.

Homework Equations

The Attempt at a Solution

First, suppose that $T$ is injective and let $Tu=Tv$ for $u,v\in V$. Clearly, $Tu-Tv=0$ and thus $S(Tu-Tv)=0$. From this, we can see that $STu=STv$. However, since $T$ is injective, then $u=v$. Therefore, there exists an $S$ such that $ST$ is the identity map on $V$. In the other direction, suppose $ST$ is the identity map on $V$, and let $Tu=Tv$. From the previous argument, we can see that $STu=STv$, and thus $u=v$, so $T$ is injective.


I think the second part of my proof is right, going from identity map to injectivity, but I'm just not sure about my argument for the first half.

Thanks for your help!

 

[micromass]

First, suppose that $T$ is injective and let $Tu=Tv$ for $u,v\in V$. Clearly, $Tu-Tv=0$ and thus $S(Tu-Tv)=0$. From this, we can see that $STu=STv$. However, since $T$ is injective, then $u=v$. Therefore, there exists an $S$ such that $ST$ is the identity map on $V$.

I can't see how you concluded that. The only thing you can conclude is that STu=STu. The thing you must show is that STu=u!
In fact, you haven't even defined a suitable S! You must first define a candidate S, and only then can you show that ST is the identity. So, how would you define such an S?

In the other direction, suppose $ST$ is the identity map on $V$, and let $Tu=Tv$. From the previous argument, we can see that $STu=STv$, and thus $u=v$, so $T$ is injective.

This is ok.

 

[MushroomPirat]

Thanks for your help micromass. I thought about your advice and came up with a different proof:

Suppose T is injective and let $(v_1,...,v_n)$ be a basis of V. Because this basis is linearly independent and T is injective, then $(Tv_1,...,Tv_n)$ is also linearly independent (this was a result from a previous problem). Because T is injective, then dim V = dim rangeT $\leq$ dimW. Extend the previous list to a basis $(Tv_1,...,Tv_n,w_1,...,w_m)$ of W. Now, define $S: W\rightarrow V$ such that $STv_i=v_i$ for all i=1,...,n, and $Sw_j=0$ for all j=1,...,m. Thus, if u is an arbitrary element of V, written as $u=a_1v_1+...+a_nv_n$ then $STu=ST(a_1v_1+...+a_nv_n)=a_1v_1+...+a_nv_n$ so ST must be the identity map on V.

 

[Dick]

Thanks for your help micromass. I thought about your advice and came up with a different proof:

Suppose T is injective and let $(v_1,...,v_n)$ be a basis of V. Because this basis is linearly independent and T is injective, then $(Tv_1,...,Tv_n)$ is also linearly independent (this was a result from a previous problem). Because T is injective, then dim V = dim rangeT $\leq$ dimW. Extend the previous list to a basis $(Tv_1,...,Tv_n,w_1,...,w_m)$ of W. Now, define $S: W\rightarrow V$ such that $STv_i=v_i$ for all i=1,...,n, and $Sw_j=0$ for all j=1,...,m. Thus, if u is an arbitrary element of V, written as $u=a_1v_1+...+a_nv_n$ then $STu=ST(a_1v_1+...+a_nv_n)=a_1v_1+...+a_nv_n$ so ST must be the identity map on V.

Now that looks good to me.