I am working on an exercise for homework that is working towards proving the uniqueness and existence of a square root matrix for any strictly positive operator, without using spectral theory. As a step in my proof, I am trying to show that if we have a strictly positive matrix $A$ with $||A||<1$, then $||I-A|| < 1$. I can tell that intuitively this should be the case, since A is self-adjoint, so its operator norm is equal to its spectral radius, and since the eigenvalues of A are all between 0 and 1, so should be the eigenvalues of $I-A$. However, I want to show directly that $||I-A||<1$, i.e. to show that
$$\sup_{||x||=1} ||I-A||= \sup_{||x||=1} {(x,x)-2(Ax,x)+(Ax,Ax)} < 1.$$
Clearly the first term is 1, and we can use the strict positivity of A to conclude that the second term is negative. However it doesn't appear that there is any simple argument to be made to simplify the expression -- for example we can't just replace $(Ax,Ax)$ with $||A||$, since the supremum of the above sum in the expression is not the sum of the suprema.
Another fact I have tried to utilize is that for A self-adjoint, $$||A|| = \sup_{||x||=1} (Ax,x)$$ but I am not sure how to utilize it.
Could someone please help me with this proof?
Since $A$, and hence $I-A$, is symmetric, then $$ \|I-A\|=\max_{\|x\|=1}|\langle (I-A)x,x\rangle| =\max_{\|x\|=1}|\langle x,x\rangle-\langle Ax,x\rangle|. $$ But $\langle Ax,x\rangle< \langle x,x\rangle$, since $\|A\|<1$. Hence $$ \|I-A\| =\max_{\|x\|=1}|\langle x,x\rangle-\langle Ax,x\rangle| =\max_{\|x\|=1}\big(\langle x,x\rangle-\langle Ax,x\rangle\big) =1-\min_{\|x\|=1}\langle Ax,x\rangle<1, $$ since $A$ is positive definite.