If $\varphi: V \to W$ is a $F$-linear mapping, then for every $U \subset V$ it is true that $\dim_F(\varphi(U)) \leq \dim_F(U)$

Question

Problem: Let $V,W$ be finite dimensional $F$-Vectorspaces where $F$ denotes a Field. Let $\varphi: V \to W$ be a $F$-linear mapping. Show that for every $U \subset V$ $$\dim_F(\varphi(U)) \leq \dim_F(U) \tag{*} $$

My approach: I wish that I could say that I have come up with something clever relating this easy looking exercise, but with linear algebra my efforts usually turn out in empty statements.

If $\varphi(U)\subset U$ then the statement would be trivial, but unfortunately I have $\varphi: V \to W$ so the best I can get is that $\varphi(V) \subset W$, I also tried to use some of the more famous dimension formulas such as $$ \dim_F(V) = \dim_F(\ker\varphi)+ \dim_F(\varphi(V))$$ which readily iplies that $\dim_F(\varphi(V)) \leq \dim_F(V)$ and considering that $U \subset V$ I could conclude that $$ \dim_F(\varphi(U)) \leq \dim_F(\varphi(V)) \leq \dim_F(V)$$ while at first glance this looks somewhat useful, it really seems not to do any good when it comes to verifying (*) because I have $\dim_F(U) \leq \dim_F(V)$ but not necessarily $\dim_F(\varphi(V)) \leq \dim_F(U)$

Is there a different/better way to show this statement?

Answer 1

Just apply the kernel theorem to $\varphi|_U : U \mapsto \varphi[U]$. You get that $\dim_F(U) = \dim_F(ker \varphi|_U) + \dim_F(\operatorname{im} \varphi|_U) \ge \dim_F(\operatorname{im} \varphi|_U) = \dim_F(\varphi[U])$

Answer 2

Consider the restriction $\varphi_{|U}$ of $\varphi$ on $U$ then $\varphi_{|U}$ is a linear transformation and its image is $\varphi(U)$ so by the rank-nullity theorem we have

$$\dim(U)\ge \dim(\operatorname{Im}(\varphi_{|U}))=\dim(\varphi(U))$$