Prove that $Ax=\mu x+\nu y\;$

Question

Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $(F, \langle \cdot, \cdot\rangle)$.

Let $A\in\mathcal{B}(F)$. Assume that there is unit vector $x\in F$ such that $\langle Ax, x\rangle:=\mu \in \mathbb{C}$. Why there are unit vector $y\in F$ and $\nu\geq 0$ such that $$Ax=\mu x+\nu y\;?$$

My attempt: Since $x$ is a unit vector, then $x\neq 0$. So, $$F=\text{span}(x)\oplus \text{span}(x)^\perp.$$ So, since $Tx\in F$ then $Ax=\alpha x+x_2$ with $x_2\in \text{span}(x)^\perp$. Clearly $\alpha =\mu$. Why $$x_2=\nu y \;?$$ with $\|y\|=1$ and $\nu\geq 0$.

I think one can write $x_2$ as $x_2=\frac{x_2}{\|x_2\|}\times \|x_2\|$ but perhaps $x_2$ is equal to $0$.

Answer 1

If $Ax = \mu x$, you can choose any $y$ and $\nu=0$. Otherwise, choose $$ y = \frac{Ax-\mu x}{\|Ax-\mu x\|}\quad\text{and}\quad\nu = \|Ax-\mu x\|. $$ The information that $\langle Ax,x\rangle = \mu$ is superfluous.