I'm reading through C* Algebras by Murphy, and the following theorem and proof is presented.
Theorem: suppose that $a,b \in A^{+}$ (are positive elements) of a C* algebra A. Then $a \leq b$ implies $a^{\frac{1}{2}} \leq b^{\frac{1}{2}}$.
Proof: we may suppose wlog that A is unital. If we show that $a^2 \leq b^2$ implies $a \leq b$ then we're done.
Let $t>0$ and Let $c,d $ be the real and imaginary hermitian parts of $(tI+b+a)(tI+b-a)$ for $I$ the unit of $A$.
It can be verified that $c \geq t^2I$ implying that $c$ is invertible and positive.
Then $I+ic^{-\frac{1}{2}}dc^{-\frac{1}{2}}$ is invertible implying that $c+id$ is invertable. Hence $(tI+b-a)$ is left invertable and hence invertible since it's hermitian.
Hence $t \notin \sigma(b-a)$, implying that $a \leq b$.
My question, why is $I+ic^{-\frac{1}{2}}dc^{-\frac{1}{2}}$ invertible? I don't see how this follows from $c$ being positive and invertible whatsoever.
If someone could explain that'd be great.
The element $d$ is selfadjoint, so its spectrum is real. Conjugating with an element and its adjoint preserves selfadjointness, so $c^{-1/2}dc^{-1/2}$ is selfadjoint and thus $\sigma(c^{-1/2}dc^{-1/2})\subset\mathbb R$. Then $$ \sigma(I+ic^{-1/2}dc^{-1/2})=1+i\sigma(c^{-1/2}dc^{-1/2}) $$ cannot contain $0$, as the real part of all its elements is $1$.