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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?

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  • $\begingroup$ 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. $\endgroup$ Mar 29, 2020 at 23:07

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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.

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  • $\begingroup$ Thank you very much for you help. Indeed, you are right about the misplacement of the question, my apologies. $\endgroup$
    – Valentina
    Mar 30, 2020 at 18:18

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