# Vector Space Basis

1. Jun 5, 2015

### bugatti79

Hi Folks,

I find this link http://mathworld.wolfram.com/VectorSpaceBasis.html confusing regarding linear independence.

One of the requirement for a basis of a vector space is that the vectors in a set S are linearly independent and so this implies that the vector cannot be written in terms of the other vectors in the set S.

Yet on the first paragraph it states that the vectors form a basis if and only if every vector can be uniquely written as a linear combination of the other which to me is a contradiction!
Can someone clarify my misinterpretation
regards

2. Jun 5, 2015

### wabbit

>it states that the vectors form a basis if and only if every vector can be uniquely written as a linear combination of the other vectors in the set S.

The word "other" has nothing to do in that statement. For $x\in S$ , that unique linear combination is $x=x$.

Last edited: Jun 5, 2015
3. Jun 5, 2015

### bugatti79

ok, thanks.

On a slight tangent, why do we use this "orthonormal basis" in Quantum mechanics calculations? Ie, is the physical significance, if any, of using this?

Pardon my ignorance.

4. Jun 5, 2015

### Fredrik

Staff Emeritus
The objects on which we do measurements are represented by unit vectors. Measuring devices are represented by linear operators. A measurement changes the object so that after the measurement it's represented by an eigenvector of the linear operator that represents the measuring device. The result of the measurement is always an eigenvalue of the linear operator.

Since results are eigenvalues, a measuring device whose output (the result of the measurement) is always a real number, must be represented by a linear operator whose eigenvalues are real numbers. Such linear operators have the following properties: a) their eigenvectors span the vector space, and b) eigenvectors corresponding to different eigenvalues are orthogonal (and therefore linearly independent). So if the eigenspaces associated with the eigenvalues are all 1-dimensional, the normalized eigenvectors must form an orthonormal basis. If the eigenspaces aren't all 1-dimensional, things get more complicated, but still not very different from what I just said.

Last edited: Jun 5, 2015
5. Jun 5, 2015

### wabbit

These properties follow from the assumption that the operators are Hermitian - actually, being hermitian is equivalent to having real eigenvalues and orthogonal eigenspaces.

As to why measurement operators in QM mut be hermitian, I don't know - I think it goes together with the whole idea that QM is formulated with Hilbert spaces, with the existence of amplitudes such that the probability of an outcome is the sum of squared (moduli of) amplitudes, and with the idea of unitary evolution, but I don't know what the precise statement is.

6. Jun 5, 2015

### Fredrik

Staff Emeritus
Ah, I was remembering a theorem about equivalence of self-adjointness and having a spectrum that's a subset of $\mathbb R$, but I forgot that the theorem I had in mind is specifically for normal operators. OK, I agree that real eigenvalues alone doesn't imply self-adjointness.

I don't have all the details worked out either, but I think the strategy to answer this question should be to argue that measuring devices should be represented by projection-valued measures, and that the spectral theorem specifies a correspondence between projection-valued measures and self-adjoint operators.