Prove the nonzero postive operator $P$ has 1 dimensional range

Question

Let $V$ be a finite-dimensional complex inner product space, and let $P \in L(V)$ be a non-zero positive operator with the property that for all positive operators $Q, R$ such that $P = Q+R$, there is a real number $0 \le r \le 1$ such that $Q = r P$. Prove that $\dim \text{range} P = 1$.

I don't have any ideas about how to approach this problem. I tried to use a complex spectral theorem, but I failed. Then I also noticed that if $Q+R=P$ and $Q, R$ are positive, this means that $P=rP+R$ which makes $R=(1-r)P$. Thus, I think this means that $P$ can only be rewritten as $(1-r)P+rP$ and it can't be rewritten as the sum of other positive operators. This is also the given information that I'm not able to use in my proof.

Thus, any help on this? Thank!

Answer 1

Hint: I think it is easiest to prove this statement by contrapositive. If $P$ is a positive operator with range of dimension $2$ or greater, use the spectral theorem to construct positive operators $Q,R$ such that $Q + R = P$ but $Q$ and $R$ are not multiples of each other (in fact, we can select such operators $Q,R$ so that $QR = 0$).

Answer 2

If $P$ is to have rank $1$, then there must be $x,y\in V$ such that $P =\langle\cdot,x\rangle y$, with positivity of $P$ reading as $$\langle Pv,v\rangle = \langle \langle v,x\rangle y, v\rangle = \langle v,x\rangle \langle v,y\rangle \geq 0$$for all $v\in V$. Choosing $v=x$ and $v=y$ leads to $\langle x,y\rangle \geq 0$. This last condition is indeed enough to ensure that $\langle\cdot,x\rangle y$ is positive, as it spectrum is $\{\langle x,y\rangle ,0\}$ (with $0$ having multiplicity $\dim V - 1$, coming from $x^\perp$) Let's reverse engineer what $x$ and $y$ must be.

Since $P\neq 0$ by assumption, there is a non-zero $y$ in the range of $P$. Choose $x_1\in V$ with $\langle x_1,y\rangle \geq 0$ small enough so that $\langle \cdot, x_1\rangle y$ and $P - \langle\cdot,x_1\rangle y$ are both still positive. Then writing $$P = \langle \cdot, x_1\rangle y + (P-\langle\cdot,x_1\rangle y)$$yields $0\leq r \leq 1$ such that $P - \langle\cdot,x_1\rangle y = r\langle\cdot,x_1\rangle y$. Then set $x = (1+r)x_1$ and we're done.