Quantum mechanics- eigenvectots of a linear transformation

[Syrus]

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|f_n> = λ_nf_n


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

 

[vela]

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.

 

[Syrus]

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'

 

[vela]

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.

 

[paranormal]

Yes, it is possible for fewer than n eigenvectors to constitute a basis for the space. This is because eigenvectors can be linearly independent, meaning they can form a basis for the space without needing all n vectors. This is known as a "minimal eigenbasis" and is often used in quantum mechanics calculations for simplicity and ease of use. However, it is important to note that the number of eigenvectors needed for a minimal eigenbasis will vary depending on the specific transformation and the dimension of the space.