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I was doing my linear algebra homework and I came across a result which doesn't make much sense to me (this isn't really related to my homework, I'm just curious).

It would appear that given $V$ a finite-dimensional vector space and some $S \in \mathcal{L}(V)$ such that $S^2=S$, we have that for $\forall v \in V$, $ S(v) = v$. (Note: We have $S^2 = S \circ S$)

This appears to be true given the following simple proof:

Assume $u \in V$ such that $S(u) = v, v \in V$

Then, we have $S(S(u)) = S(u) = v$

And therefore, $S(v) = v$ for any $v \in V$

Finally, given that $P$ will give us back as an output any vector we give it as an input, do we not have that $P$ must necessarily be the identity $I_V$ in V? Now, my question is whether or not this result that I've found is accurate, as, if it is, my homework question becomes trivial, and, additionally, it wouldn't make sense to notate $S$ in this way when we could simply write $I_V$?

So, my overall question is simply whether or not this result is accurate and if not what is the major error? Thanks for any help anyone can give.

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  • $\begingroup$ What if $S=0$?? $\endgroup$
    – amd
    Nov 13, 2018 at 22:34

2 Answers 2

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You made a mistake in the proof: You assume there exists an element $u\in V$ so that $Su=v$ for any $v\in V$. This is wrong. What You actually proved is that $S|_{im(S)}$ i.e. the restriction of $S$ to its image is $id_{im(S)}$. As a counterexample You can take projections on subspaces.

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  • $\begingroup$ Alternatively, the OP proved that a surjective idempotent matrix is the identity, which is true. $\endgroup$ Nov 13, 2018 at 23:07
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Incorrect because $S$ could be the zero map. Your assumption about the existence of a pre-image $u$ is wrong.

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  • $\begingroup$ Oh, I understand. So, we only have $S(u)=v$ for $v \in range(S)$. $\endgroup$
    – beeselmane
    Nov 13, 2018 at 14:43

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