How to prove an operator is invertible

Question

If A is a bounded, self adjoint operator on a Hilbert space, how can you see that $A^2+I$ is invertible?

Am I not sure how to even start with this, since I don't know anything about the inverse of A... (I thought about showing this by constructing the inverse, but don't see how I could do that here).

Answer 1

Suppose $(A^2+I)x=0$. Then $\langle x , (A^2+I)x \rangle= \|Ax\|^2+\|x\|^2 = 0$ and so $x=0$. Hence injective.

Now suppose $\langle y , (A^2+I)x \rangle= 0 $ for all $x$, then we see that $(A^2+I)y = 0$ and so $y=0$.

To finish, we need to show the range is closed, we have $\langle (A^2+I)x , (A^2+I)x \rangle= \langle x , (A^2+I)^2x \rangle \ge \|x\|^2$ and so $\|(A^2+I)x\| \ge \|x\|$ from which it follows that the range is closed and hence $A^2+I$ is surjective.

Answer 2

Because $A$ is selfadjoint, we have $\sigma(A)\subset \mathbb R$. Then $$ \sigma(A^2)=\{\lambda^2:\ \lambda\in\sigma(A)\}\subset[0,\infty). $$ And then $$ \sigma(A^2+I)=\{\lambda+1:\ \lambda\in\sigma(A^2)\}\subset[1,\infty), $$ so $A^2+I$ is invertible.