# Operator with strictly positive eigenvalues

1. Oct 26, 2012

### physicus

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

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.

2. Relevant equations

3. 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?

2. Oct 26, 2012

### dextercioby

Can you show that G and G2 are related ?

3. Oct 26, 2012

### 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.

4. Oct 27, 2012

### 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 ?

5. Oct 27, 2012

### 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.