Is $S_M(0,1)$ always homeomorphic to the unit sphere of $E$? Let $E$ be a infinite-dimensional complex Hilbert space, with inner product $\langle\cdot\;| \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$. Let  $M\in \mathcal{L}(E)^+$ (i.e.$M^*=M$ and $\langle Mx\;, \;x\rangle\geq0$ for all $x\in E$).
Assume that $M$ is an injective operator on $E$, so $x\longmapsto \langle Mx\;, \;x\rangle$ is a norm on $E$.

Consider
  $$S(0,1)=\{x\in E;\;\|x\|=1\}$$
  and
  $$S_M(0,1)=\{x\in E;\;\langle Mx\;, \;x\rangle=1\}$$
  Is $S_M(0,1)$ always homeomorphic to $S(0,1)$ even if $E$ is an infinite-dimensional complex Hilbert space?

Thank you.
 A: 
FINAL ANSWER [Due to Student; see comments]. The two sets are homeomorphic because the maps
  $$ h\colon x\in S\mapsto \frac{x}{\sqrt{\langle Mx, x\rangle}}\in S_M,\qquad g\colon y\in S_M\mapsto \frac{y}{\|y\|}\in S$$
  are continuous and the inverse of each other. We use the assumptions on $M$ in order to ensure that $h$ is well-defined and continuous.

Remark. If $V$ is a vector space and $\|\cdot\|_1, \|\cdot\|_2$ are two norms on it, with unit spheres $S_1, S_2$ respectively, then the maps 
$$ h_{i,j}(x_i)=\frac{x_i}{\|x_i\|_j},\qquad i\ne j,$$ 
are bijections of $S_i$ onto $S_j$ such that $h_{i,j}^{-1}=h_{j, i}$. If one of the norms is continuous with respect to the other then the maps are homeomorphisms.  

PREVIOUS VERSION OF THIS ANSWER FOLLOWS [What follows is not a satisfactory answer and it contains an error. I am leaving it here since, according to me, errors are part of the development of mathematical ideas.]

There always exists a continuous injection of $S_M$ into $S$; see 1. This injection is a homeomorphism if $0$ is not a spectral value of $M$.

1. Take the square root of $M$ and rewrite 
$$S_M=\{ \langle M^\frac12 x | M^\frac12 x\rangle=1\}, $$
so the substitution $y=M^\frac12 x$ is such that $x\in S_M$ iff $y\in S$. This substitution indicates that the map $x\in S_M\mapsto M^\frac12 x\in S$ is a continuous injection. This map is a homeomorphism if $M$ is invertible with a bounded inverse, or equivalently, if $0$ belongs to the resolvent set of $M$. Since the spectrum is the complement of the resolvent this proves the claim in the box above.

The point that follows is an attempt to a counterexample, intended to show that the condition on the spectrum of $M$ cannot be dropped. However, the example is flawed.
2. [Suggested by Daniel Wainfleet]. In the real Hilbert space $\ell^2$, consider the operator
$$M \boldsymbol x=(x_1, x_2/4, x_3/9, \ldots, x_n/n^2,\ldots), $$ 
so that
$$
S_M=\left\{ \boldsymbol x\in \ell^2\ :\ \sum_{n=1}^\infty \frac{x_n^2}{n^2}=1\right\}.$$
This set is compact [FALSE] so it cannot be homeomorphic to $S$.
