Let $T:\mathbb{R}^n\to\mathbb{R}^m$ be a linear transformation. Prove that the range of T is a subspace of $\mathbb{R}^m$ with dimension at most $n$ I'm having some trouble with the second part of this question. I tried using the result that any $n+1$ vectors in $\mathbb{R}^n$ are linearly dependent, but no luck here.
The first part is easy and here is my solution.
The range of $T$ can be written as $S=\{A\mathbf{v}:\mathbf{v}\in\mathbb{R}^n\}$ where $A$ is the standard matrix of $T$. Then,
$$\mathbf{0}\in S \:\text{since} \:\mathbf{0}=A\mathbf{0}$$
$$\text{Let}\: A\mathbf{u},\,A\mathbf{v}\in S\implies A\mathbf{u}+A\mathbf{v}=A\mathbf{u+v}\in S \:\text{since} \: \mathbf{u+v}\in\mathbb{R}^n$$
$$\text{Let}\,A\mathbf{v}\in S,\: c\in\mathbb{R}\implies c(A\mathbf{v})=A(c\mathbf{v})\in S \:\text{since} \: c\mathbf{v}\in\mathbb{R}^n$$
Please provide some hints for the second part of the question regarding the upper limit of the dimension of $S$. I think the proof wouldn't be very tedious or long, but instead I need to be able to use some key concepts. (I hope that the tips/ hints doesn't involve eigenvalues and stuff because I haven't learnt it yet. Thanks a lot!)
 A: Hint: If $e_1,...,e_n$ are the vectors of a basis of $\mathbb{R^n}$ then show that the set $\{T(e_1),...,T(e_n)\}$ spans $S$.
A: Hint: How can you make a spanning set of the range using a basis of $\Bbb R^n$?
A: Hint: Range of $T$ is spanned by $\{Tv_1,Tv_2,\dots,Tv_n\}$ if  $\{v_1,v_2,\dots,v_n\}$ is a basis for $\Bbb{R}^n$.
A: The Rank-Nullity theorem states that
\begin{align*}
&\dim (\mathbb{R}^{n}) =\dim(( \ker )T) + \dim( \text{im}(T))\\
\implies& n = \dim(( \ker )T) + \dim(S)\\
\implies& \dim(S) = n - \dim(( \ker )T)
\end{align*}
And we have that $\dim(\ker(T))\geqslant 0$. This means that  $\dim(S)$ does not exceed $n$.
A: Consider the standard basis of $\mathbb{R}^{n}$ as $\{e_{1}, e_{2}, e_{3}, \dots, e_{n}\}$. Therefore as we know that the set $S$ defined by @Mathsisfun $$S = \text{im}(T) := \{ T(v)~:~v \in \mathbb{R}^{n} \}$$
At first, we need to show that $S \subseteq_{\mathcal{S}} \mathbb{R}^{m}$. To prove that
we need to show that
$\bullet~$ for $\alpha, \beta$ $\in$ $\mathbb{R}$, and $w_{1}, w_{2}$ $\in$ $S$, $ \alpha w_{1} + \beta w_{2}$ $\in$ $S$.
$\circ$ We can consider $T$ $\in$ $\mathcal{L}(\mathbb{R}^n, \mathbb{R}^m)$ sends some $v_{1} \in \mathbb{R}^{n}$ to $w_{1} \in \text{im}(T)$. Similarly $v_{2} \mapsto w_{2}$.
Therefore we have that
\begin{align*}
\alpha w_{1} + \beta w_{2} =&~ \alpha \cdot T (v_{1}) + \beta \cdot T(v_{2})\\
=&~ T(\alpha v_{1}) + T(\beta v_{2})\\
=&~ T(\alpha v_{1} + \beta v_{2})
\in~ S\quad [\text{as } \alpha v_{1} + \beta_{v_{2}} \in \mathbb{R}^{n} ] 
\end{align*}
Therefore $S$ is a subspace of $\mathbb{R}^{m}$
Now we just need to show that the set $\Omega$ defined as $$ \Omega := \{ T(e_{i})~:~i = 1, 2, 3,\dots,n \} $$
spans $S$.
Therefore for any arbitrary $v \in \mathbb{R}^n$, let $T(v)$ $\in$ $S$, we have that,
\begin{align*}
T(v) =&~ T\bigg( \sum_{i = 1}^{n} \lambda_{i} e_{i} \bigg) \quad \text{for some } \lambda_{i}\text{'s in } \mathbb{R}, \text{for } i = 1, 2 ,3,\dots,n\\
  =&~ \sum_{i = 1}^{n} \lambda_{i} T(e_{i})
\end{align*}
Therefore we have $T(v) = w \in S$ arbitrary and let $T(e_{i}) = \omega_{i}$ for all $i = 1,2, 3, \dots, n$.
Therefore we have that for any arbitrary $w \in S$, we have
$$ w = \sum_{i = 1}^{n}\lambda_{i} \omega_{i} $$
Hence the set $\{ \omega_{i}~:~ i = 1,2,3,\dots, n \} \equiv \Omega $, spans the set $S$.
$\circ$ Again, The Rank-Nullity theorem states that
\begin{align*}
&\dim (\mathbb{R}^{n}) =\dim(( \ker )T) + \dim( \text{im}(T))\\
\implies& n = \dim(( \ker )T) + \dim(S)\\
\implies& \dim(S) = n - \dim(( \ker )T)
\end{align*}
And we have that $\dim(\ker(T))\geqslant 0$. This means that  $\dim(S)$ does not exceed $n$.
These complete the claim.
A: For the first part, $\text{im}(T)=\text{span}(T(e_1),.....,T(e_n))$, where $e_i$ is the standard basis of $\mathbb{R}^n$.
Now, we will show that $\text{rank}(T)\leq m$ , where $T:\mathbb{R}^n\rightarrow \mathbb{R}^m$.
Suppose for a contradiction that $\text{rank}(T)=k>m$.
Let $\mathbb{B}=\{$ $w_1,....,w_k$ $\}$ be a basis for $\text{im}(T)$. As $\text{im}(T)\subseteq \mathbb{R}^m$, $\mathbb{B}$ is a linearly independent collection of vectors in $\mathbb{R}^m$. Clearly, $~\text{span}(\mathbb{B})=\mathbb{R}^m$ cannot hold, as this would mean that $w_1,....,w_k$ is a basis for $\mathbb{R}^m$. In which case, we may extend $\mathbb{B}$ to a basis for $\mathbb{R}^m$, however that new collection would have more than $k$ vectors, and so it must be linearly dependent. Contradiction.
