Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question

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.

share|cite|improve this answer

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

share|cite|improve this answer

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

share|cite|improve this answer
ahh i see now, that was so simple my brain needs rest – CJS May 10 '12 at 0:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.