Let $A$ be an $m\times n$ matrix with entries in $F$ and let $T$ be the linear transformation from $F^{n\times 1}$ into $F^{m\times 1}$ defined by $T(X)=AX$. Show that if $m\lt n$ it may happen that $T$ is onto without being non-singular. Similarly, show that if $m\gt n$ we may have $T$ non-singular but not onto.
We can prove a stronger result:
Let $V$ be an $n$-dimensional vector space over field $F$ and $W$ be an $m$-dimensional vector space over field $F$. $(a)$ If $m\lt n$, then $\exists T\in L(V,W)$ such that $T$ is surjective but not injective. $(b)$ If $m\gt n$, then $\exists T\in L(V,W)$ such that $T$ is injective but not surjective.
My attempt: $(a)$ Suppose $m=\mathrm{dim}(W) \lt \mathrm{dim}(V)=n$. Since $\mathrm{dim}(V)=n\in \Bbb{N}$, we have $\exists B_V=\{v_1,…,v_n\}$ basis of $V$. Since $\mathrm{dim}(W)=m\in \Bbb{N}$, we have $\exists B_W=\{w_1,…,w_m\}$ basis of $W$. Let $\beta_i=w_i$, $\forall i\in J_m$ and $\beta_i=w_m$, $\forall i\in J_n\setminus J_m$. By theorem 1 section 3.1, $\exists !$ $T:V\to W$ such that $T$ is linear map and $T(v_i)=\beta_i=w_i$, $\forall i\in J_m$ and $T(v_i)=\beta_i=w_m$, $\forall i\in J_n\setminus J_m$. By generalize exercise 8 section 3.1, $\mathrm{span}(\beta_1,…,\beta_n)=R_T$. Thus $R_T$ $= \mathrm{span}(\beta_1,…,\beta_n)$ $= \mathrm{span}(\{\beta_1,…,\beta_m\})$ $=W$. Hence $T$ is surjective. $T(v_n)=T(v_m)$, for $v_n\neq v_m$. Hence $T$ is not injective.
$(b)$ Suppose $m=\mathrm{dim}(W) \gt \mathrm{dim}(V)=n$. Let $B_V=\{v_1,…,v_n\}$ and $B_W=\{w_1,…,w_m\}$ be basis of $V$ and $W$, respectively. Let $\beta_i=w_i$, $\forall i\in J_n$. By theorem 1 section 3.1, $\exists !$ $T:V\to W$ such that $T$ is linear map and $T(v_i)=\beta_i=w_i$, $\forall i\in J_n$. If $T(x)=T(y)$, for some $x,y\in V$. Since $\mathrm{span}(\{v_1,…,v_n\})=V$, we have $x=\sum_{i\in J_n}a_i\cdot_V v_i$ and $y=\sum_{i\in J_n}b_i\cdot_V v_i$. So $T(x)=T(y)$ $=\sum_{i\in J_n}a_i\cdot_W T(v_i)$ $=\sum_{i\in J_n}b_i\cdot_W T(v_i)$ $=\sum_{i\in J_n}a_i\cdot_W w_i$ $= \sum_{i\in J_n}b_i\cdot_W w_i$. Since $\{w_1,…,w_n\}\subseteq B_W$ is independent, $a_i-b_i=0_F$, $\forall i\in J_n$. So $a_i=b_i$, $\forall i\in J_n$. Thus $x=y$. Hence $T$ is injective. We claim $w_m\notin R_T$. Assume towards contradiction, $\exists x\in V$ such that $T(x)=w_m$. Since $\mathrm{span}(\{v_1,…,v_n\})=V$, we have $x=\sum_{i\in J_n}a_i\cdot_V v_i$. So $T(x)=T(\sum_{i\in J_n}a_i\cdot_V v_i)$ $=\sum_{i\in J_n}a_i\cdot_W T(v_i)$ $=\sum_{i\in J_n}a_i\cdot_W w_i$ $=w_m$. So $w_m\in \mathrm{span}(\{w_1,…,w_n\})$ $\subseteq \mathrm{span}(B_W\setminus \{w_m\})$. Which implies $B_W$ is dependent. Thus we reach contradiction. Hence $T$ is not surjective. Is my proof correct?
To shorten the proof of exercise, we can prove following:
$(a)$ If $m\lt n$ and $T\in L(V,W)$, then $T$ is not injective. $(b)$ If $m\gt n$ and $T\in L(V,W)$, then $T$ is not surjective.
My attempt: $(a)$ Suppose $m=\mathrm{dim}(W) \lt \mathrm{dim}(V)=n$. Assume towards contradiction, $T$ is injective. So $N_T=\{0_V\}$. By rank nullity theorem, $\mathrm{dim}(W)\lt \mathrm{dim}(V)$ $= \mathrm{dim}(N_T)+ \mathrm{dim}(R_T)$ $= \mathrm{dim}(R_T)$. So $\mathrm{dim}(W)\lt \mathrm{dim}(R_T)$. But by theorem 5 corollary 1 section 2.3, $\mathrm{dim}(R_T)\leq \mathrm{dim}(W)$. Thus we reach contradiction. Hence $T$ is not injective.
$(b)$ Suppose $m=\mathrm{dim}(W) \gt \mathrm{dim}(V)=n$. Assume towards contradiction, $T$ is surjective. So $R_T=W$. By rank nullity theorem, $\mathrm{dim}(W) \gt \mathrm{dim}(V)$ $= \mathrm{dim}(N_T)+ \mathrm{dim}(R_T)$ $\mathrm{dim}(N_T)+ \mathrm{dim}(W)$. So $\mathrm{dim}(W)\gt \mathrm{dim}(N_T)+ \mathrm{dim}(W)$ $\Rightarrow$ $\mathrm{dim}(N_T)\lt 0$. Thus we reach contradiction. Hence $T$ is not surjective. Is my proof correct?
The proofs seem correct, but too verbose.
To begin with, I can't understand the reference to $A$, because it's a standard exercise that every linear map $T\colon F^{n\times1}\to F^{m\times1}$ is of the form $T(X)=AX$, for a unique $m\times n$ matrix $A$.
Anyway, here's a shortened proof. It's even easier to prove the generalized form, because the first case is just a special case thereof.
Suppose $\dim V=n$ and $\dim W=m$ and take bases $\{v_1,v_2,\dots,v_n\}$ and $\{w_1,w_2,\dots,w_m$ of $V$ and $W$ respectively.
(a) If $m<n$ we can define a unique linear map $T\colon V\to W$ by declaring $$ T(v_i)=\begin{cases} w_i & 1\le i\le m \\[2ex] 0 & m<i\le n \end{cases} $$ (cite the theorem). Since the image of $T$ contains a basis of $W$, the map $T$ is surjective, but it is not injective because $T(v_n)=0$ and $v_n\ne0$.
(b) If $m>n$, we can define a unique linear map $T\colon V\to W$ by declaring $$ T(v_i)=w_i \qquad 1\le i\le n $$ The map $T$ is not surjective, because $w_m$ doesn't belong to the image of $T$. Otherwise, $$ w_m=T(\alpha_1v_1+\dots+\alpha_n v_n)=\alpha_1w_1+\dots+\alpha_nw_n $$ contradicting the linear independence of $\{w_1,\dots,w_m\}$.
The map is injective, because it transforms a basis into a linearly independent set. With more details, if $T(\alpha_1v_1+\dots+\alpha_n v_n)=0$, then $$ \alpha_1w_1+\dots+\alpha_n w_n=0 $$ and, by linear independence, $\alpha_1=\alpha_2=\dots=\alpha_n=0$.
Final note. As you see, the proof is essentially the same as yours, but shorter and more to the point.
The proof of (b) can be shortened. After defining $T\colon V\to W$ as above, we can also define a map $S\colon W\to V$ by declaring $$ S(w_i)=\begin{cases} v_i & 1\le i\le n \\[2ex] 0 & n<i\le m \end{cases} $$ and observe that $ST$ is the identity map on $V$, because $ST(v_i)=v_i$ for $1\le i\le n$. Thus $T$ is injective and it cannot be surjective because of rank-nullity.