Prove that $\dim(T(H))\leq \dim(H)$. 
Let $V$ and $W$ be finite-dimensional vector spaces and $T$ be a linear transformation $T:V\to W$.
Let $H$ be a non-zero subspace of $V$, and let $T(H)$ be the set of images of vectors in $H$. Prove that $\dim(T(H))\leq \dim(H)$.


Proof:
Let $B=\{b_1,...,b_n\}$ be a basis for $H$.
Because $B$ is a basis and $T$ is linear, for any $x\in T(H)$, we have
$$x=T(c_1b_1+...+c_nb_n)=c_1T(b_1)+...+c_nT(b_n)$$
Therefore, $\text{span}\{T(b_1)...T(b_n)\}=T(H)$.
If $\{T(b_1)...T(b_n)\}$ is linearaly independent, then $\text{dim}(T(H))=\text{dim}(H)$ and we are done.
Otherwise, $\{T(b_1)...T(b_n)\}$ is linearaly dependent, and there exist $T(b_i)$ that can be expressed as a linear combination of the other vectors in this set.
So let $T(b_i)=d_1T(b_1)+...+d_{i-1}T(b_{i-1})+d_{i+1}T(b_{i+1})+...+d_nT(b_n)$.
Now let $y\in T(H)$,then
$$y=c_1T(b_1)+..+c_iT(b_i)+..+c_nT(b_n)=c_1T(b_1)+...+d_1T(b_1)+...+d_{i-1}T(b_{i-1})+d_{i+1}T(b_{i+1})+...+d_nT(b_n)+...+c_nT(b_n)=(c_1+d_1)T(b_1)+..+(c_{i-1}+d_{i-1})T(b_{i-1})+(c_{i+1}+d_{i+1})T(b_{i+1})+...+(c_n+d_n)T(b_n)$$
So, $\text{span}\{T(b_1)...T(b_n)\}=\text{span}\{T(b_1)...T(b_{i-1}),T(b_{i+1}),...,T(b_n)\}=T(H)$
If this new set is linearly independent then $\text{dim}(T(H))=n-1<\text{dim}(H)$.
Otherwise, we are linearly dependent and a vector can be removed without changing the span. If we get down to one remaining vector and that vector is not the $0$ vector then  $\text{dim}(T(H))=1<\text{dim}(H)$. Otherwise, it is the $0$ vector and $\text{dim}(T(H))=0<\text{dim}(H)$.
Thank you for reading the proof. Please provide feedback on the validity of my proof. Have I assumed anything I shouldn't have assumed? Any leaps in logic? I appreciate any tips suggestions and insights.
 A: This question can be found here as well.
Now about your proof, she is correct. But I probably instead of writing,

Otherwise, $\{T(b_1),...,T(b_n)\}$ is linearaly dependent, and there exist $T(b_i)$ that can be expressed as a linear combination of the other vectors in this set.

I would write,

Otherwise, $\{T(b_1),...,T(b_n)\}$ is linearaly dependent, so there is at least one vector $T(b_i)$, that can be expressed as a linear combination of the other vectors in this set.

Also note that when you write $y$, the vector $T(b_i)$ is being multiplied by $c_i$, then $y$ would be written as,
$$y=c_1T(b_1)+...+c_{i-1}T(b_{i-1})+c_i \big(\underbrace{d_1T(b_1)+...+d_{i-1}T(b_{i-1})+d_{i+1}T(b_{i+1})+...+d_nT(b_n)}_{T(b_i)}\big)+...+c_nT(b_n)$$
So in the next equality, instead of just, for example, $(c_1+d_1)$, you need to write, $(c_1+c_id_1)$, and then the proof follows as you wrote it.
A: I think you are correct. Essentially you go from the space you have control to the space you do not have control.  Now, In my humble opinion, this is too much suffering. I would argue that is easier to do the opposite i.e., Suppose $dim(H)<dim(T(H)),$ and consider a basis in $T(H)$ like $\{T(b_i)\}_{i\in [n]}.$ Then $\{b_i\}_{i\in [n]}$ has to be dependent in $H$ so WLOG $b_1=\frac{1}{c_1}\sum _{i>1}c_ib_i,$ where $c_i$ are scalars. Applying $T$ to the last equation,you get the contradiction.(Notice that essentially this is the same as your proof, but sometimes is easier to arrive to contradictions).
A: Your proof looks correct to me. Here is a much shorter proof.
Let $\{\xi_1, \dots, \xi_n\}$ be a basis of  $H$. Then $\{T\xi_1, \dots, T\xi_n\}$ spans $T(H)$. Hence, $\dim T(H) \leq  n = \dim(H)$.
Basically, my proof relies on the following fact, which you probably know:

If the vector space $V$ is linearly spanned by a set of vectors $E\subseteq V$, then
there exists a subset $F \subseteq E$ such that $F$ is a basis for
$V$.

