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

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?

• Yes. Take a 3x3 case and convince yourself that if the 3 bras were dependent, adding suitable first and 2nd row multiples on the third in $\cal M$ would result in a null third row, so vanishing determinant contrary to your assumption. Mar 29, 2020 at 23:07

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.)