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We are given: $T: V \rightarrow W$ is a linear transformation and $U$ is a subspace of $V$.

$T(U) = \{T(u) | u \in U\}.$

We are also given $S: U \rightarrow T(U)$ is a linear transformation by $S(u) = T(u)$ for all $u \in U$.

I have already proved that $\ker(S) = U \cap \ker(T)$.

I am having a hard time proving: $\dim(T(U)) = \dim(U) - \dim(U \cap \ker(T))$.

Informally, I have said that by removing all the elements that map the linear transformation $T(u)$ to $\vec{0}$ will give you the dimension of $T(U)$ but I am having a hard time showing this mathematically.

Can someone give me some hints or point me in the right direction to go about doing this?

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  • $\begingroup$ @YadatiKiran: $T(U) \subset W$ and $U \subset V$, why would the intersection be relevant? $\endgroup$ – copper.hat Mar 10 at 5:42
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$dim(T(U)) = dim(S(U))$, so re-write the equation in terms of $S$ as $dim(S(U)) = dim(U) - dim(ker(S))$ and apply the rank-nullity theorem.

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  • $\begingroup$ Can you elaborate a little more? $\endgroup$ – Yadati Kiran Mar 10 at 17:02
  • $\begingroup$ The statement of the rank-nullity theorem is that given a linear map A: V \to W, dim(V) = dim(A(V)) + dim(ker(A)). Apply this to the linear map S: U \to S(U). $\endgroup$ – vxnture Mar 10 at 17:10
  • $\begingroup$ Thanks for the reply $\endgroup$ – Yadati Kiran Mar 10 at 17:12

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