$A$ is a bounded linear operator on a complex Hilbert space. What are the spectral properties of $I + A^*A?$

The spectral theorem for self-adjoint operators states that $$\sigma(A) \subset [r,R] \subset \mathbb{R}, \quad \text{where } r = \inf_{\|x\| = 1} (Ax,x) \text{ and } R = \sup_{\|x\| = 1} (Ax,x).$$

Clearly $I + A^*A$ is self-adjoint. Applying the spectral theorem, we know $$r = \inf_{\|x\| = 1} ((I + A^*A)x, x) = (x,x) + (Ax,Ax) = \|x\|^2 + \|Ax\|^2 \ge 0.$$ Therefore, $I+A^*A$ is a positive operator since its spectrum is nonnegative.

I'm struggling to say anything else meaningful about $\sigma(I + A^*A)$. Can we say anything else in relation to $\sigma(A)$, or perhaps provide an upper bound for values in $\sigma(I + A^*A)$? Anything we can say about the point or continuous spectrum? We know the residual spectrum is empty because $I + A^*A$ is self-adjoint.

  • $\begingroup$ I'm not sure but it seems to be the case that $$\sigma\left(I+A^*\,A\right)\subseteq \left[1+r^2,1+R^2\right]\,.$$ $\endgroup$ – Batominovski Nov 17 '17 at 8:19
  • $\begingroup$ Can you show why? Small problem is that $r^2$ is not necessarily less than $R^2$ for $r,R \in \mathbb{R}$. $\endgroup$ – stackedtritones Nov 17 '17 at 8:25
  • $\begingroup$ What can you say about $A^* A$? Then, now does the spectrum of $I+T$ relate to $T$? $\endgroup$ – Hurkyl Nov 17 '17 at 9:00
  • $\begingroup$ For an operator $A$, $\lambda \in \sigma(I + A)\ \text{iff}\ \lambda - 1 \in \sigma(A)$. I don't understand the claim about $A^*A$ having bounds $r^2$ and $R^2$ on its spectrum. $\endgroup$ – stackedtritones Nov 17 '17 at 9:09

The spectrum of a bounded selfadjoint operator $B$ is contained in the interval $[r,R]$ where $r=\inf_{\|x\|=1}\langle Bx,x\rangle$ and $R=\sup_{\|x\|=1}\langle Bx,x\rangle$. And the endpoints of the interval are always included in the spectrum. A bounded selfadjoint operator $B=I+A^*A$ has associated $r,R$ given by $$ r = 1 +\inf_{\|x\|=1}\|Ax\|^2, \;\;\; R = 1+\sup_{\|x\|=1}\|Ax\|^2=1+\|A\|^2. $$ If $A$ is not continuously invertible, then $r$ could be $1$. Otherwise, $$ \|A^{-1}x\| \le \|A^{-1}\|\|x\| \\ \|x\| \le \|A^{-1}\|\|Ax\| \\ \|A^{-1}\|^{-1}\|x\| \le \|Ax\| \\ 1+\|A^{-1}\|^{-2} \le r. $$


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