Proof: $\ker \psi \subseteq \ker (\psi \circ \psi)$ Hi guys I'm completely clueless about this proof I came across in a uni textbook.
Let $V$ be a vector space, and $\psi:V\to V$ be a linear transformation. Prove that $\ker \psi \subseteq \ker (\psi \circ \psi)$.
Does anyone know how to construct the proof?
Cheers
 A: First thing you should do is write the definitions:


*

*$\ker\psi=\{v\in V\mid \psi (v)=0\}$.

*$\ker(\psi\circ\psi)=\{v\in V\mid \psi(\psi(v))=0\}$.

*$A\subseteq B$ if and only if for all $a\in A$, $a\in B$.


Next you should note that we always have $0\in\ker\psi$ when $\psi$ is linear. Now this is amounts to a standard element chasing proof:
Let $x\in\ker\psi$, we want to show that $x\in\ker(\psi\circ\psi)$, namely $\psi(\psi(x))=0$. However since $x\in\ker\psi$, and $\psi$ is linear we have that $$\psi(\psi(x))=\psi(0)=0$$
Therefore $x\in\ker(\psi\circ\psi)$ as wanted.
A: Let us write $\psi \circ \psi$ as $\psi^2$. Now to show that $\ker\psi \subseteq \ker \psi^2$, you need to show that given an element $x $ in the kernel of $\psi$, it is also in the kernel of $\psi^2$. In other words that if $\psi(x) = 0$ then $\psi^2(x) =0$. So suppose that $\psi(x) =0$. Then $$\psi^2(x) = \psi \Big(\psi(x) \Big).$$
However since $x$ was by assumption in the kernel of $\psi$, it must be the case that $\psi^2(x) = \psi(0)$. But then $\psi$ is a linear transformation so that $\psi(0) = 0$. Hence $\psi^2(x) =0$ implying that $x \in \ker \psi^2$. Hence $\ker \psi \subseteq \ker \psi^2$.
A: Because $\psi$ is a linear map then $\psi(0)=0$,
so if $x\in \ker\psi$ then $\psi(x)=0$ and we conclude that  $\psi(\psi(x))=\psi(0)=0$, which means that $x\in \ker\psi\circ\psi$, so $\ker\psi\subset \ker\psi\circ\psi$
