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Let $T$ be a bounded linear operator in a Hilbert space, $T^*$ the adjoint of $T$. Then how to show that $I + T^*T$ is invertible?

Thanks.

Teodorism
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Freeman
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  • We should be able to prove that $A - I$ is a positive operator implies that $A$ is invertible. – Ben Grossmann Sep 29 '15 at 16:15
  • My thought is: Assume (I + TT) to be not invertible, then there exists some x not equals to 0 and (I + TT)x = 0. Then <(I + T*T)x, x> = 0 can lead to <x,x>+<Tx,Tx>=0, which is a contradiction. But I am not sure if it is correct. – Freeman Sep 29 '15 at 16:19
  • In infinite dimensional spaces, we can have linear operators that have a trivial kernel but fail to be invertible. – Ben Grossmann Sep 29 '15 at 16:23

1 Answers1

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Let $A=I+T^{\star}T$. Then $A=A^{\star}$ and $$ \|x\|^{2}\le \|x\|^{2}+\|Tx\|^{2}=(Ax,x) \le \|Ax\|\|x\| \\ \implies \|x\| \le \|Ax\|. $$ So $A$ is injective.

The range of $A$ is dense because $\mathcal{R}(A)^{\perp}=\mathcal{N}(A^{\star}) = \mathcal{N}(A)=\{0\}$. To see that the range is closed, suppose $\{ Ax_n \}$ converges to some $y$. Then $\{ Ax_n \}$ is a Cauchy sequence; hence $\{ x_n \}$ is a Cauchy sequence because $$ \|Ax_n - Ax_m \| \ge \|x_n-x_m\|. $$ Therefore $\{ x_n \}$ converges to some $x$ and $\{ Ax_n \}$ converges to $Ax$, which gives $Ax=y$. So the range of $A$ is closed, which proves that $A$ is surjective.

So $A$ is a linear bijection, and $\|x\|\le \|Ax\|$ implies $\|A^{-1}y\|\le \|y\|$. Hence $A$ is invertible.

Disintegrating By Parts
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