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

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 $$0$$.

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.
• but $\mu$ is correct. – Schüler Sep 7 '19 at 15:02
• @Schüler What? What do you mean by "correct"? How can a number ($\mu$) be "correct"? – amsmath Sep 7 '19 at 15:06
• I mean the expression of $\mu$ is correct. – Schüler Sep 7 '19 at 15:24