General question about Self-adjoint operators on Hilbert Spaces so if I have a self-adjoint operator $K \in B(H,H)$. Then I found a theorem that said that 
$A$ is invertible if and only if $(A^*Ax,x) \geq c||x||$ and $(AA^*x,x) \geq c||x||$ for all $x$.
Then does that mean that if I have a self adjoint operator like $K$.
$(K^*Kx,x) = (Kx,Kx) \geq 0$, but then as long $Kx \not = 0$ for all $x$ i.e. $K = 0$, then $K$ is invertible because I have $(Kx,Kx)> 0$. But that is weird to me because that means every self-adjoing bounded linear operator is invertible.

In an infinite-dimensional vector space, there's a big difference between $(Ax,Ax)\geq c\|x\|^{2}$ for all $x\in H$ and $(Ax,Ax)>0$ for all nonzero $x\in H$. For an example, consider $\ell^{2}(\mathbb{N})$. Then if $A(e_{i})=2^{-i}e_{i}$ for all $i\geq 0,$ we can see that the inverse must be $A^{-1}(e_{i})=2^{i}e_{i}$, but this is not a bounded linear operator.
We see that $(Ax,Ax)=\sum_{i=0}^{\infty}2^{-2i}x_{i}^{2}>0$ whenever $x\neq0$, but for any $c>0,$ if $x\in\mathrm{span}(\{e_{i}\}_{i\geq I}),$ where $I$ is large enough that $2^{-2I}<c$, then $(Ax,Ax)\leq 2^{-2I}\sum_{i=0}^{\infty}x_{i}^{2}<c\|x\|^{2}$. Therefore there is no $c>0$ for which this bound holds.
EDIT: Consider the optimization problem $$\min_{x\neq0}\frac{(Ax,Ax)}{\|x\|^{2}},$$ which is equivalent to $$\min_{\|x\|=1}(Ax,Ax).$$ The unit sphere $\{x\in H:\|x\|=1\}$ is compact if and only if $H$ is finite-dimensional, so in this case, there is some $x$ which achieves this minimum $c$, which gives $(Ax,Ax)\geq c\|x\|^{2}$ for all $x\in H$. When this minimization is not over a compact set, there is no such guarantee (just consider the example I gave above).
