0

Proof that:
Linear operator ($P:X\to X$) is projective ($P^2=P$) IFF
$\exists$ direct sum decomposition of X ($X=V\oplus U$), such that $\forall u\in U:Pu=u$ and $\forall v\in V:Pv=0$.

My proof:

$(\Leftarrow)$

Let $x\in X$. As of direct sum definition there exist only one decomposition $x=v+u$.
Then $P(P(x))=P(P(u+v))=P(P(u)+P(v))=P(u)=P(u)+P(v)=P(u+v)=P(x)$

$(\Rightarrow)$

Let $V$ be the nullspace of $P$, and $U=X\backslash V\cup \{0\}$.
For non-trivial case $U\supset\{0\}:$
Suppose $\exists u \in U : P(u)=v \in V, v\neq 0$, but $P(P(u))=P(v)=0\neq P(u)$
This contradicts the projective property of $P$. Thus $range(P)\subseteq U$.
Suppose $u \in U$ and $P(u)=u'$, then $P(P(u))=P(u')$. By projective property $P(u')=u'=P(u)$. Thus $P(u)-P(u')=P(u-u')=0$. Because of $u-u'\notin V $ the only way to satisfy it is $u-u'=0$. So $P(u)=u$ and $range(P)=U$

Please check correctness of my proof or suggest another way. Thanks.

user102417
  • 111

1 Answers1

0

You don't need a basis for $(\Rightarrow)$ (the proof for $(\Leftarrow)$ is perfect). Suppose $X = U \oplus V$ satisfies the conditions. Let $x \in X$ and $V$ be the null space of $P$ and $U$ be the image of $P$.

  • By definition of the null space, if $v \in V, Pv = 0$.
  • If $u \in U$, then $u = Px$ for some $x$, so $Pu = PPx = Px = u$.
  • Let $x \in X$, then $P(x - Px) = Px - PPx = Px - Px = 0$, and so $x = Px + (x - Px)$ belongs to the sum $U + V$. Therefore $X = U + V$.
  • Let $x \in U \cap V$. Then $x = Pz$ for some $z$, and $Px =0$. But $0 = Px = PPz = Pz = x$, therefore $U \cap V = 0$.

This finally proves that $X = U \oplus V$ with the required conditions.

Najib Idrissi
  • 54,185