Let $T:V\to W$ be a linear transformation. If $\dim V> \dim W$ then $T$ is not injective. True or false? 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 $$\dim V \le \dim W$$
 A: 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 $\dim(V) = \dim(Im(t)) \leq \dim(W)$.
A: You're right. Remember that
$$
\text{rank } T + \text{nullity } T = \text{dim } V.
$$
What is the maximum possible rank of $T$? (It may help to think of $T$ as an $m \times n$ matrix.)
A: Tu prueba es correcta. Una forma más rápida de concluir es recordando que para cualquier espacio vectorial $V$, si $U$ es un subespacio de $V$ entonces $\dim U \leq \dim V$, y, en el contexto de tu pregunta, $\operatorname{Im}(T)$ es siempre un subespacio de $W$.

Your proof is correct. A quicker way to finish the argument is by recalling that for any vector spacce $V$, if $U$ is a subspace of $V$ then $\dim U \leq \dim V$, and, in the context of your question, $\operatorname{Im}(T)$ is always a subspace of $W$.
