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?

  • $\begingroup$ 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

1 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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.