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:
Since $H\neq\{0\}$, then some subset $B$ of $H$ is a basis for $H$ (by the spanning set theorem). Let $B=\{b_1,\dots,b_p\}$ be a basis for $H$, so $\dim(H)=p$.
Then for all $x\in$Span$(H)$, there exists scalars $c_1,\dots,c_p$ such that $x=c_1b_1+\cdots+c_pb_p$.
Since $T$ is linear,
$$x=c_1b_1+\cdots+c_pb_p\implies T(x)=c_1T(b_1)+\cdots+c_pT(b_p)$$
and the vectors $T(b_1),\dots,T(b_p)$ are linearly independent. The vectors $T(b_1),\dots,T(b_p)$ also span $T(H)$, since for all $T(x)\in$Span$(T(H))$, we can find an $x\in$Span$(H)$ such that $T(x)\in$Span$(T(H))$.
So $\{T(b_1),\dots,T(b_p)\}$ is a basis for $T(H)$. Thus, $\dim(T(H))\leq\dim(H)$.
Some questions:
I'm convinced with: "The vectors $T(b_1),\dots,T(b_p)$ also span $T(H)$, since for all $T(x)\in$Span$(T(H))$, we can find an $x\in$Span$(H)$ such that $T(x)\in$Span$(T(H))$" but I am not sure if this is true.
Also, it seems to be that $\dim(T(H))=\dim(H)$, is there a case where $\dim(T(H))<\dim(H)$?
Best Answer
Note that: $$T(B)=\{T(b_1),\dots,T(b_p)\}$$ is spanning set of $T(H)$ but it is not necessarily linearly independent.