# Quantum mechanics- eigenvectots of a linear transformation

1. Apr 29, 2013

### Syrus

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. The attempt at a solution

2. Apr 29, 2013

### vela

Staff Emeritus
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.

3. Apr 29, 2013

### 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'

4. Apr 30, 2013

### vela

Staff Emeritus
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.