Operator with strictly positive eigenvalues

[physicus]

Homework Statement

Consider a Hilbert space with a (not necessarily orthogonal) basis \{f_i\} Show that G=\sum_i |f_i\rangle\langle f_i| has strictly positive eigenvalues.

Homework Equations

The Attempt at a Solution

I know that G=\sum_i |f_i\rangle\langle f_i| is hermitian. Therefore, the eigenvalues are real and the Hilbert space has an orthonormal basis of eigenvectors of G. However, a general hermitian matrix does not need to be positive definite. Therefore, I need another approach.

I consider an eigenvector |a\rangle = \sum_i a_i |f_i\rangle with G|a\rangle = \lambda|a\rangle
\Rightarrow \lambda\sum_j a_j |f_j\rangle = \lambda |a\rangle = G |a\rangle = \sum_{ij}a_i |f_j\rangle\langle f_j|f_i\rangle
\Rightarrow \lambda = \frac{1}{a_j}\sum_{i}a_i \langle f_j|f_i\rangle
Unfortunately, I cannot conclude \lambda>0 from that.

Can anybody help me?

 

[dextercioby]

Can you show that G and G² are related ?

 

[physicus]

Unfortunately, I don't know how to relate G² and G.

G^2=\sum_{ij}|f_i\rangle\langle f_i|f_j\rangle\langle f_j|
I would need an expression for \langle f_i|f_j\rangle But since the basis set is not orthonormal I don't see what this could be.

 

[dextercioby]

Alright. Different approach. Let psi be an eigenvector of G with eigenvalue a. Then from the spectral equation you have

a=1/normsq(psi) * <psi|G|psi> = 1/normsq(psi) * sum (i) <psi|f_i><f_i|psi> = ...

Can you now show the dots hide a number > 0 ?

 

[physicus]

\ldots = \frac{1}{||\,|\psi\rangle||^2} \sum_i |\langle \psi | f_i\rangle|^2

Since psi ist not the zero vector (as it is an eigenvector) its overlap with at least one of the basis vectors is non-zero. Therefore, the sum of squares is strictly greater than 0.

Thank you very much.