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I am solving the following question and I am not really sure about the way I approach

Question 1: Assume that $T:U\rightarrow U$ is invertible map. Prove that $(T^*)^{-1}=(T^{-1})^*$

Here is my answer:

Notice that $\langle Tv,u\rangle = \langle v, T^*u \rangle$ for all $u,v\in U$

Then $\langle T^*(T^{-1})^*v,u\rangle = \langle (T^{-1})^* v, Tu \rangle = \langle v,T^{-1}(Tu)\rangle = \langle u,v\rangle$

$T^* (T^{-1})^* = 1 $ so $(T^{-1})^* $ is the inverse of $T^*$. Hence, $(T^{-1})^*=(T^*)^{-1}$.

Question 2: Prove that for every operator $T$ the operators $T^*T$ and $T T^*$ are self-adjoint

Assumption: I have to prove $T^*T = T$ and $T T^* = T $

Suppose $T\in L (V,W)$. Fix $ w\in W$ .

$\langle Tv,u\rangle = \langle v, T^*u \rangle$ for all $u,v\in U$

Then $\langle TT^*v,u\rangle = \langle ((Tv)T^*,u \rangle = \langle Tv,Tu\rangle $ ??

$\langle T^*Tv,u\rangle = \langle (T^* (Tv),u \rangle = \langle Tv,Tu\rangle$ ???

I am not sure whether I am right or wrong for the above problem so please help me correct the above problem and understand adjoint and self-adjoint.

EpsilonDelta
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needhelp
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2 Answers2

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Another way to solve the Question 1 is:

We have $ (T^*)^{-1}T^* = 1 $ then $ ((T^*)^{-1}T^*)^* = 1^*$

Since $1^* =1 $ and $(AB) = B^*A^*$ for every operators A and B then $ T^{**}((T^*)^{-1})^* = 1 $

But $T^{**} = T$ so $((T^*)^{-1})^* = T^{-1}$ and then $(T^*)^{-1}=(T^{-1})^*$

BraQuiet
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By definition of adjoint, considering $T^*T$ as a product of operators: $$(T^*Tu,v) = (Tu,Tv) = (u,T^*Tv).$$

On other hand, if we take them as a whole: $$(T^*Tu,v) = (u,(T^*T)^*v).$$

Can you conclude now?

TZakrevskiy
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