# Spectrum for a bounded linear operator and its adjoint on a Banach space are same.

I have to show that spectrum for a bounded linear operator and its adjoint on a Banach space are the same. Spectrum is defined as $$\sigma(T)=\{\lambda\in \mathbb{K}\ :\ T-\lambda I \ \text{is invertible}.\}$$

I have to show $$\sigma(T)=\sigma(T^*)$$. Let $$\lambda \notin \sigma(T)$$; then $$(T-\lambda I )$$ is invertible and bounded. This implies $$(T-\lambda I)^*$$ is also invertible, since $$(T^*-\lambda I)^{-1}=[(T-\lambda I)^*]^{-1}\implies T^*-\lambda I \ \text{is invertible}.$$ So $$\lambda\notin \sigma(T^*).$$

I am unable to prove the other part. Can anyone help me please?

Thanks.

$T-\lambda I$ is invertible if and only if $(T-\lambda I)^*=T^*-\lambda I$ is invertible: Since for every linear operator $A$ invertibility of $A$ and of $A^*$ are equivalent, which follows by taking the adjoints of, e.g., $AA^{-1}=I$ and $A^{-1}A=I$.

• I've never heard about this theorem. Anyway, what version are you using? In the wikipedia article I think there are more assumptions than that on the OP. – Filburt Nov 28 '17 at 19:13
• This has nothing to do with closed range theorem. Modified answer – daw Nov 28 '17 at 20:35
• Could you help me to prove that, if the adjoint $T*$ of a operator is invertible, then $T$ is surjective? – Filburt Nov 30 '17 at 15:25

We want to prove that if $$X$$ is a Banach space and $$T^*\in B(X^*)$$ is invertible, then $$T$$ is invertible in $$B(X)$$.

We go through a few steps.

• Note that $$\operatorname{ran} T$$ is closed. indeed, let $$W$$ be the inverse of $$T^*$$, and let $$\{Tx_n\}$$ be a Cauchy sequence. Then \begin{align} \|x_n-x_m\| &=\sup\{|f(x_n-x_m)|:\ f\in X^*,\ \|f\|=1\}\\ \ \\ &=\sup\{|T^*Wf\,(x_n-x_m)|:\ f\in X^*,\ \|f\|=1\}\\ \ \\ &=\sup\{|(Wf)\,(Tx_n-Tx_m)|:\ f\in X^*,\ \|f\|=1\}\\ \ \\ &\leq\|Tx_n-Tx_m\|\,\sup\{\|Wf\|:\ f\in X^*,\ \|f\|=1\}\\ \ \\ &=\|W\|\,\|Tx_n-Tx_m\|. \end{align} So $$\{x_n\}$$ is Cauchy; there exists $$x\in X$$ with $$x=\lim x_n$$. As $$T$$ is bounded, $$Tx=\lim Tx_n$$, and $$\operatorname{ran} T$$ is closed.

• $$T$$ is injective. Indeed, if $$Tx=0$$, then for any $$f\in X^*$$ we have $$f=T^*g$$ (since $$T^*$$ is surjective). Then $$f(x)=T^*g(x)=g(Tx)=g(0)=0.$$ Thus $$f(x)=0$$ for all $$f\in X^*$$, and so $$x=0$$.

• $$T$$ is surjective. Indeed, if $$y\in X\setminus \operatorname{ran} T$$, using Hahn-Banach (and the fact that $$\operatorname{ran} T$$ is closed) there exists $$g\in X^*$$ with $$g(y)=1$$, $$g(Tx)=0$$ for all $$x$$. But then $$0=g(Tx)=T^*g(x)$$ for all $$x$$, and so $$T^*g=0$$. As $$T^*$$ is injective, $$g=0$$; this is a contradiction. So $$X=$$\operatorname{ran} T$$, and$$T\$ is surjetive.

• Finally, since $$T$$ is bijective and bounded, by the Inverse Mapping Theorem it is invertible.

• Very nice proof: maybe it is just the first inequality that is a bit unclear, how can we put the norm of W outside, if it is evaluating at one point (x_n - x_m) (instead of the norm of functions themselves). – Philimathmuse Mar 31 at 20:10
• You are definitely right: that equation is wrong. I'll see if it can be saved. – Martin Argerami Mar 31 at 20:24
• I see, maybe it is easier to first compose T* with W, instead of W with T*, and then view Wf as g, whose norm is bdd above by that of W. All the estimates of your original argument still hold. – Philimathmuse Mar 31 at 20:25
• Yes, you are definitely right. I'll edit that into the answer. Thanks for noticing. – Martin Argerami Mar 31 at 20:31