When is an idempotent matrix the identity?

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.

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

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.
Incorrect because $$S$$ could be the zero map. Your assumption about the existence of a pre-image $$u$$ is wrong.
• Oh, I understand. So, we only have $S(u)=v$ for $v \in range(S)$. Nov 13, 2018 at 14:43