I think it's true, I just did this demo, please can you help me if I'm missing something or doing it wrong. Thanks.
Let $T\colon V \to W$ a linear transformation.
If $\dim V > \dim W$, then $T$ is not injective.
The contrapositive is: If $T$ is injective, then $\dim V \le \dim W$.
Since $T$ is injective (by hypothesis) $\ker(T) = \{0\}$, and so $\operatorname{nullity}(T) = 0$.
By the rank-nullity theorem, $$\dim V = \operatorname{nullity}(T) + \operatorname{rank}(T) = \operatorname{rank}(T) \tag{i}$$
By definition, $\operatorname{rank}(T) = \dim(\operatorname{Im}(T))$. There are two cases:
- $\dim(\operatorname{Im}(T)) = \operatorname{rank}(T) = \dim W$
- $\dim(\operatorname{Im}(T)) = \operatorname{rank}(T) < \dim W$
Then, we have by (i) that
$$\dim V = \operatorname{rank}(T) = \dim W$$
or
$$\dim V = \operatorname{rank}(T) < \dim W$$
And so $$\dim V \le \dim W$$
Sea $T\colon V \to W$ una transformación lineal.
Si $\dim V > \dim W$, entonces $T$ no es inyectiva.
La contraposición es: Si $T$ es inyectiva, entonces $\dim V \le \dim W$.
Si $T$ es inyectiva (por hipótesis) $\ker(T) = \{0\}$, así que $\operatorname{nulidad}(T) = 0$.
Por la teorema de la dimensión, $$\dim V = \operatorname{nulidad}(T) + \operatorname{rango}(T) = \operatorname{rango}(T) \tag{i}$$
Por definición, $\operatorname{rango}(T) = \dim(\operatorname{Im}(T))$. Existen dos casos:
- $\dim(\operatorname{Im}(T)) = \operatorname{rank}(T) = \dim W$
- $\dim(\operatorname{Im}(T)) = \operatorname{rank}(T) < \dim W$
Entonces, se tiene de (i) que
$$\dim V = \operatorname{rank}(T) = \dim W$$
ó
$$\dim V = \operatorname{rank}(T) < \dim W$$
Por lo tanto $$\dim V \le \dim W$$
Best Answer
This, to be sure, is one consequence of the rank nullity theorem.
However, assuming you haven't learned that yet, consider the following:
Let $\{v_1,\dots,v_n\}$ be a basis of $V$. Then $\{T(v_1),\dots,T(v_n)\}$ must span the image of $T$, which has dimension at most $\dim(W)$. So, the vectors $\{T(v_1),\dots,T(v_n)\}$ can't be linearly independent.
This allows you to deduce that $T$ is not injective. How?
In response to what you posted: your proof is correct. However, it was not necessary to break things up into two cases. It was sufficient to remark that since $Im(T) \subseteq W$, it follows that $\dim(Im(T)) \leq \dim(W)$, so that $\dim(V) = \dim(Im(t)) \leq \dim(W)$.