Show that if $\rho$ is idempotent then $\rho$ acts as the identity on $\rho(V)$

Question

A linear map $V \xrightarrow{\rho} V$ is idempotent if $\rho\rho = \rho$. Show that if $\rho$ is idempotent then $\rho$ acts as the identity on $\rho(V)$. (Such linear maps are called projections: $\rho$ projects $V$ onto $\rho(V)$.)

My thoughts on this problem are that to be idempotent, $\rho$ could only be $0$ or $1$, and they would be an identity on $\rho(V)$ because $0$ is the identity for addition and $1$ is the identity for multiplication.

Answer 1

Let $x\in\rho(V)$ ($\rho(V)\neq\emptyset$ since $\rho(0)=0$). Then $x=\rho y$ for some $y\in V$. Therefore $\rho x=\rho\rho y=(\rho\rho) y=\rho y=x$.

Since $x\in\rho(V)$ was arbitrary, we conclude that $\rho=I$ when we restricted $\rho$ to $\rho(V)$.

Answer 2

$\rho(\rho(x))=\rho(x)$ for all $x$ means that, for all $s\in \rho(V)$ (a subspace of $V$), we have $\rho(s)=s$. That sounds like "acting as the identity" to me, or perhaps I would say that $\rho$'s restriction to $\rho(V)$ coincides with (or is) the identity on that space. While $0$ is the identity for addition and $1$ for multiplication in a field, the term "identity" has a further meaning which is a map that takes every element to itself.

Answer 3

It is not true that $\rho$ could only be $0$ or $1$. Let $V=\mathbb R^2$ and $\rho(x,y)=(x,0)$, then $\rho(\rho(x,y))=\rho(x,0)$. Now, suppose for any vector space $V$ that $\rho$ is idempotent. Let $\mathbf y\in\rho(V)$, what can you sat about $\rho(\mathbf y)$.