Let $\ \varphi \, : \, V\rightarrow V\ $ be a linear transformation. Prove that $\ Im(\varphi \, \circ \varphi) \subseteq Im \,\varphi\ $ Let V be a vector space and $\ \varphi \, : \, V\rightarrow V\ $ be a linear transformation. Prove that:
$$\ Im(\varphi \, \circ \varphi) \subseteq Im \,\varphi\ $$
I am struggling to see what conditions I have to verify, in order to prove it.
 A: $$x\in\text{Im}(\phi\circ\phi)\implies\;\exists\,v\in V\,\,s.t.\,\,\phi\circ\phi(v)=x$$
But $\,\phi\circ\phi(v)=\phi(\phi(v))\,$ , so in fact we got
$$x=\phi(\phi(v))\in\text{Im}(\phi)$$
A: Every element of $Im(\phi \circ \phi)$ has the form $(\phi \circ \phi) (x)$ for some $x \in V$. But $\phi : V \rightarrow V$, so $\phi (x) \in V$ as well. So setting $\phi (x) = y$ we have $(\phi \circ \phi )(x) = \phi (y) \in Im(\phi)$.
So every element of $Im(\phi \circ \phi)$ is also in $Im(\phi)$, hence $Im(\phi \circ \phi) \subseteq Im(\phi)$.
A: Recall that 
$$\mathrm{Im}(\varphi)=\{\varphi(x)\quad|\quad x\in V\}$$
Now take $y\in \mathrm{Im}(\varphi\circ \varphi)$ then there's $x\in V$ such that
$$y=\varphi\circ \varphi(x)= \varphi( \underbrace{\varphi(x)}_{z\in V})=\varphi(z)\in \mathrm{Im}(\varphi)$$
so we have
$$\ Im(\varphi \, \circ \varphi) \subseteq Im \,\varphi\ $$
A: More generally, for linear maps $U \xrightarrow{\phi} V \xrightarrow{\psi} W$, we have $\mathrm{Im}(\psi \circ \phi) \subseteq \mathrm{Im}(\psi)$.
Proof: $\mathrm{Im}(\psi \circ \phi) = (\psi \circ \phi)(U)=\psi(\phi(U)) \subseteq \psi(V)=\mathrm{Im}(\psi)$.
A: Let $x\in Im(\varphi\circ\varphi)$ and note that $x=\varphi(\varphi(y))$ for some $y\in V$ by the definition of $Im(\varphi\circ\varphi)$. But then $x=\varphi(z)$ where $z=\varphi(y)$ and so $x\in Im(\varphi)$ by the definition of $Im(\varphi)$ as $z\in V$. It follows that $Im(\varphi\circ\varphi)\subseteq Im(\varphi)$.
