# Proof: $\ker \psi \subseteq \ker (\psi \circ \psi)$

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

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First thing you should do is write the definitions:

1. $\ker\psi=\{v\in V\mid \psi (v)=0\}$.
2. $\ker(\psi\circ\psi)=\{v\in V\mid \psi(\psi(v))=0\}$.
3. $A\subseteq B$ 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.

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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 $\ker \psi \subseteq \ker \psi^2$.

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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 $\ker\psi\subset \ker\psi\circ\psi$

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ahh i see now, that was so simple my brain needs rest – CJS May 10 '12 at 0:58