Linear operator, its dual, proving surjectivity

In summary, the conversation discusses the equivalence between a continuous linear operator $T$ being surjective and its adjoint $T^*$ being injective and having a closed range. The speaker mentions a similar statement that they have proven and asks for ideas on how to prove the equivalence. Another person suggests using a classical theorem of Banach which states that a bounded linear operator between Banach spaces has closed range if and only if its adjoint also has closed range. They suggest looking in Yosida's Functional Analysis for a proof of this theorem.
  • #1
Linux
7
0
Let \(\displaystyle T: X \rightarrow Y\) be a continuous linear operator between Banach spaces.

Prove that $T$ is surjective \(\displaystyle \iff\) \(\displaystyle T^*\) is injective and \(\displaystyle im T^*\) is closed.

I've proven a "similar" statement, with \(\displaystyle imT^*\) replaced with \(\displaystyle imT\).

There I used these facts: $\overline{imT}= ^{\perp}(kerT^*)$ and $\overline{imT^*} \subset (kerT)^{\perp}$

However, I do not know how to prove the equivalence above.

Could you give me some ideas?
 
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  • #2
Linux said:
Let \(\displaystyle T: X \rightarrow Y\) be a continuous linear operator between Banach spaces.

Prove that $T$ is surjective \(\displaystyle \iff\) \(\displaystyle T^*\) is injective and \(\displaystyle im T^*\) is closed.

I've proven a "similar" statement, with \(\displaystyle imT^*\) replaced with \(\displaystyle imT\).

There I used these facts: $\overline{imT}= ^{\perp}(kerT^*)$ and $\overline{imT^*} \subset (kerT)^{\perp}$

However, I do not know how to prove the equivalence above.

Could you give me some ideas?
A classical theorem of Banach (you'll find it somewhere in his book Opérations linéaires – it's a consequence of the open mapping theorem) says that a bounded linear operator between Banach spaces has closed range if and only if its adjoint also has closed range. I think that will be the key ingredient in your problem.
 
  • #3
Thank you. Could you tell me where I can find a proof of this theorem, apart from Stefan Banach's book (which is in French)?
 
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  • #4
(From my phone by Tapatalk)
Try Yosida's Functional Analysis.
 
  • #5
Thanks. I've just found the book.
 

FAQ: Linear operator, its dual, proving surjectivity

What is a linear operator?

A linear operator is a mathematical function that maps one vector space to another, while preserving the structure of the vector space. In other words, it is a function that takes in vectors as inputs and produces vectors as outputs, and follows the rules of linearity.

What is the dual of a linear operator?

The dual of a linear operator is a function that maps from the dual space of the output vector space to the dual space of the input vector space. It can be thought of as a "reverse" linear operator, as it takes in the outputs of the original linear operator and produces inputs for the original linear operator.

How do you prove surjectivity of a linear operator?

To prove surjectivity of a linear operator, you must show that for every vector in the output vector space, there exists at least one vector in the input vector space that maps to it. This can be done by showing that the range of the linear operator is equal to the entire output vector space.

What is the importance of proving surjectivity of a linear operator?

Proving surjectivity of a linear operator is important because it ensures that every possible output in the vector space can be obtained by applying the linear operator to some input vector. It also allows for the inverse of the linear operator to exist, which is crucial in many mathematical applications.

Can a linear operator be both injective and surjective?

Yes, a linear operator can be both injective (one-to-one) and surjective (onto). In fact, a linear operator is bijective (both injective and surjective) if and only if it has an inverse. This means that every output in the vector space can be uniquely mapped to an input, and vice versa.

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