Prove Corollary of Rank-Nullity theorem

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SUMMARY

The discussion centers on proving the corollary of the Rank-Nullity theorem, specifically that a linear transformation τ from vector space V to W is injective if and only if it is surjective, given that dim(V) = dim(W) < infinity. Key points include the relationship between the kernel and image of τ, where the dimension of the kernel plus the dimension of the image equals the dimension of V. The conclusion drawn is that if τ is injective, then the image of τ must equal W, confirming surjectivity.

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Homework Statement


Prove:
Let \tau \in L(V,W), where dim(V) = dim(W) < infinity. Then \tau is injective iff it is surjective.


Homework Equations


L(V,W) is the set of all linear transformations from V to W.

1. Any complement of ker(t) is isomorphic to im(t)
2. dim(ker(t)) + dim(im(t)) = dim(V)


The Attempt at a Solution



I'm pretty lost in starting this.
I know it is surjective iff im(t) = W
I know it is injective iff ker(t) = {0}

Should I assume its injective but not surjective (to move towards a contradiction)?

And maybe I don't understand the concept of an isomorphism but if:
im(\tau) = W and
ker(\tau)^{c} \approx im(\tau)
then how does the ker(t)^c relate to W?
 
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You can prove this by only looking at the formula dim(ker(t))+dim(im(t))=dim(V).

Assume that t is injective. As you stated, this implies that ker(t)={0}. Thus dim(ker(t))=0.
So, what does this imply in the above formula?
 
that implies dim(im(t)) = dim(V) = dim(W)

that doesn't necessarily imply though that im(t) = W

am I just not thinking about it enough?
 
That's correct. But since im(t)\subseteq W and dim(im(t))=dim(W) (as proven), then this DOES imply that im(t)=W.

In general, if you have two finite-dimensional spaces V and W such that V\subseteq W and dim(V)=dim(W), then V=W!
 
Ah ! okay, that makes perfect sense

thanks for the help
 

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