Quantum mechanics- eigenvectots of a linear transformation

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Homework Help Overview

The discussion revolves around the concept of eigenvectors and their role in forming a basis for vector spaces in quantum mechanics, particularly in the context of linear transformations. The original poster questions the assertion that eigenvectors can always form a basis if they span the space, suggesting that fewer than n vectors might suffice.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the number of eigenvectors and the dimensionality of the vector space, questioning whether all eigenvectors are necessary to form a basis. The original poster raises concerns about the completeness of eigenvectors in spanning the space.

Discussion Status

The conversation is ongoing, with participants providing insights into the conditions under which eigenvectors can form a basis. There is recognition of the possibility that not all eigenvectors may be linearly independent, which could affect their ability to span the space.

Contextual Notes

There is an underlying assumption that the linear operator in question may not always have n linearly independent eigenvectors, which is central to the discussion about the nature of eigenbases.

Syrus
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Homework Statement



My quantum mechanics text (in an appendix on linear algebra) states, "f the eigenvectors span the space... we are free to use them as a basis..." and then states:

T|f1> = λ1f1
.
.
.
T|fn> = λnfn


My question is: is it not true that fewer than n vectors might constitute this "new" basis? In other words, if the eigenvectors span the space, they may not (all together, at least) form a basis.




Homework Equations





The Attempt at a Solution

 
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No, this is a basic result from linear algebra for vector spaces of finite dimension. In an n-dimensional space, every basis has to have n vectors.
 
Yes, I see. But when, then, all the talk about "An eigenbasis for a linear operator that operates on a vector space is a basis for that consists entirely of eigenvectors of (possibly with different eigenvalues). Such a basis may not exist."

See http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors under the heading 'eigenbasis'
 
It's possible that an operator will not have n linearly independent eigenvectors, so its eigenvectors can not span the vector space and therefore will not form a basis.

Your text, however, is saying if the eigenvectors span the space, they can be used as a basis. The assumption is that the linear operator has n linearly independent eigenvectors.
 

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