Trouble w/ Lemma: Nullity U Ignored in Finite Vector Space

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In summary, the text discusses the equivalence of one-to-one and onto in linear transformations between equidimensional vector spaces. However, in a later lemma, the text seems to forget this result and focuses on other properties of linear operators. The lemma states that if two linear operators have finite-dimensional null spaces and one is onto, then the null space of their composition is also finite-dimensional. The proof mentions that one of the operators is also one-to-one, but it does not consider the possibility that the vector space may not be finite-dimensional. This is where the lemma may be misleading, as it assumes the vector space is finite-dimensional without explicitly stating it. In summary, the lemma serves as an auxiliary proposition to the main discussion on linear transformations
  • #1
genxhis
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The text I am using has proved the following thereom near the beginning of the chapter: If two vector spaces V, W are equidimensional (finite) and T is a linear transformation from V to W, then one-to-one and onto are equivalent. It has also used the result liberally in latter sections.

Trouble oocurs when it comes to a lemma near the end of the chapter. The text suddenly seems to "forget" the preceding thereom. The lemma is: "Let V be a vector space, and suppose that T and U are linear operators [transformations onto same vector space] on V such that U is onto and the null spaces of T and U are finite-dimensional. Then the null space of TU is finite-dimensional, and nullity TU = nullity T + nullity U." This is followed by a lengthy proof. But, according the the previous result, U is also one-to-one. This in turn readily means the nullity U is always zero. I don't understand why this is wholly ignored.
 
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  • #2
oh.. nm. The lemma never states V is finite dimensional.
 
  • #3
What does Lemma mean?
 
  • #4
JasonRox said:
What does Lemma mean?
Auxiliary proposition.
 

FAQ: Trouble w/ Lemma: Nullity U Ignored in Finite Vector Space

What is the lemma about nullity U in a finite vector space?

The lemma states that in a finite vector space, the nullity of a linear transformation U is equal to the dimension of the null space of U.

Why is the nullity of U important in a finite vector space?

The nullity of U is important because it provides information about the structure of U and the dimension of the null space, which is essential for solving systems of linear equations and understanding the behavior of linear transformations.

How is the nullity of U related to the rank of U?

The nullity of U and the rank of U are related by the rank-nullity theorem. This theorem states that the sum of the nullity and rank of U is equal to the dimension of the vector space on which U operates.

Can the nullity of U be greater than the dimension of the vector space?

No, the nullity of U cannot be greater than the dimension of the vector space. This is because the null space of U is a subspace of the vector space, so its dimension cannot exceed that of the vector space.

How is the nullity of U calculated?

The nullity of U can be calculated by finding the dimension of the null space of U. This can be done by solving the homogeneous system of linear equations represented by U=0 or by finding the dimension of the kernel of U.

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