Proving two basic results about linear maps without requiring rank-nullity

Question

Let $U,V$ be finite dimensional vector spaces. I've proven the following statements as easy consequences of the rank-nullity theorem. But now I'm curious as to how to prove them directly without requiring this theorem. I haven't had any success, however, and would appreciate seeing how it is done.

1. There exists a surjective $T \in \mathcal{L}(U,V)$ if and only if $\dim U \geq \dim V$.

2. There exists an injective $T \in \mathcal{L}(U,V)$ if and only if $\dim V \geq \dim U$.

Note: the scalars are from $\mathbb{R}$ or $\mathbb{C}$. (I realize now that any field will work, so disregard this stipulation if it's easier.)

Answer 1

Try to use the following facts about injective and surjective linear maps.

Let $T\colon U \to V$ be a linear function and $b_1,\dots,b_n$ is a basis of the vector space $U$.

• $T$ is injective if and only if the vectors $T(b_1),\dots,T(b_n)\in V$ are linearly independent.
• $T$ is surjective if and only if the vectors $T(b_1),\dots,T(b_n)$ generate $V$, i.e., $[T(b_1),\dots,T(b_n)]=V$.

See also Prove that the linear transformation is injective iff $T(f_1),\ldots,T(f_n)$ are linearly independent for a proof of the first one.

Answer 2

1. There exists a surjective $T \in \mathcal{L}(U,V)$ if and only if $\dim U \geq \dim V$.

Let $u_1, u_2, \cdots, u_m$ be a basis of $U$ and $v_1, v_2, \cdots, v_n$ a basis of $V$.

For the "if" part:

Define $T$ as follows:

$$T(a_i) = \begin{cases}b_i & 1 \le i \le n \\ 0 & n < i \le m \end{cases}$$

The proof of its surjectivity is left as an exercise to the reader.

For the "only if" part:

(work in progress)

2. There exists an injective $T \in \mathcal{L}(U,V)$ if and only if $\dim V \geq \dim U$.

The proof is left as an exercise to the reader.