an exercise about the projections.

Question

There is an exercise in operator theory that says: If P and Q are projections on H that $||P-Q||<1$ then dimension of ImP and ImQ are the same.

i cant understand what is the relation between the norm and dimension. please help me. so thanks.

Answer 1

The idea is that if the subspaces have different dimensions, then you will find a place where the distance is one. Technically, one uses the condition to prove that the two projections are unitarily equivalent. Here is my preferred argument:

Assume that $\|p-q\|<1$, with $p,q\in A$, a unital C$^*$-algebra. Let $x=pq+(1-p)(1-q)$. Then, as $2p-1$ is a unitary, $$ \|1-x\|=\|(2p-1)(p-q)\|=\|p-q\|<1. $$ So $x$ is invertible. Now let $x=uz$ be the polar decomposition, $z=(x^*x)^{1/2}\in A$. Then $u=xz^{-1}\in A$. Also, $px=pq=xq$, and $qx^*x=qpq$, so $qx^*x=x^*xq$, and then $qz=zq$. Then $$ pu=pxz^{-1}=xqz^{-1}=uzqz^{-1}=uqzz^{-1}=uq. $$ So $q=u^*pu$.

You can find this argument and others in the answers here