Subspaces - relation between images of linear map

Question

If $T:V\rightarrow V$ and $S:V\rightarrow V$ are linear maps and Im(S) is a subset of Im(T) and rank(T)=rank(S) then is Im(T)=Im(S) necessarily?

Answer 1

Since $Im(S)$ and $Im(T)$ are subspaces, and the rank of the operator is the dimension of this subspace, the question is the same as the following: Given subspaces $U,V$ and $\dim U = \dim V$, and $U \subset V$, is $U=V$?

In finite dimensions, yes.

Take a basis of $U$. It has exactly $\dim U$ elements, is linearly dependent over $V$ and $V$ has dimension $\dim U$ also. Hence, the same set is also a basis for $V$, which means $U=V$.

Answer 2

Yes. If $U_1$ and $U_2$ subspaces of a finite-dimensional vector space and if

$U_1 \subseteq U_2$ and $\dim U_1= \dim U_2$, then $U_1=U_2$.