Let $A$ be a positive definite matrix. I am trying to prove that there are $n$ linearly independent vectors $v_1, v_2, \dots, v_n$ such that $A_{ij} = v_i^{\top}v_j$.
From the Spectral Theorem, we know that $A$ can be factored as $A = V\Lambda V^T = V^T \Lambda V$, where $V$ is a diagonal matrix whose columns are the unit eigenvectors of $A$ and $\Lambda$ is a diagonal matrix containing the eigenvalues of $A$ on its diagonal. This gives the desired result immediately, with the exception that $\Lambda$ must the identity matrix, so $A$ must have eigenvalue $1$ with multiplicity $n$, which of course is not true in general.
I'm looking for a way to extend this idea to cover the general case, but cannot figure out a way to break up $\Lambda$ to give the desired result.
Does anyone have a suggestion on how to proceed?
Hint: if $A$ can be orthogonally diagonalized as $A = U\Lambda U^T$, then set $V = U \sqrt{\Lambda}$. Now, what could I mean by $\sqrt{\Lambda}$?
Alternatively, if you know about positive semidefinite square roots, it certainly suffices to set $V = \sqrt A = U\sqrt{\Lambda}U^T$.