Linear independence of a set of states, and the non-vanishing determinant of the matrix comprised of their inner products

Question

So I have a very basic question in linear algebra, but I'll phrase it in terms of QM.

Suppose we are given a set of $N$ states $\{ | \psi_i \rangle\}$. Construct the $N \times N$ matrix

$$\mathcal{M}_{ij} \equiv \langle \psi_i | \psi_j \rangle,$$

and suppose $\det \mathcal{M} \neq 0$. This implies that the set of $N$ column vectors $\{ | \phi_i \rangle\}$, given by

$$ | \phi_i \rangle = \begin{pmatrix} \langle \psi_1 | \psi_i \rangle \\ \langle \psi_2 | \psi_i \rangle \\ \vdots \\ \langle \psi_N | \psi_i \rangle \end{pmatrix} $$

are linearly independent. Does this also imply linear independence of the set $\{ | \psi_i \rangle\}$? If yes, why?

Answer 1

Apply the argument you applied to the N $\{ |\phi_i\rangle \}$ s this time to the matrix $\cal N$ of the $\{ |\psi_i\rangle \}$ s.

For linear dependence, $\cal N$ must have a null eigenvalue, and so should $\cal {N^\dagger N}=\cal{M} $, contrary to assumption.

($\cal N$ is T×N, for $N\leq T\leq\infty$, and your φ construction ensured you worked in the N-subspace of the T-Hilbert space.)

The question should be in the math SE.