# Tag Info

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-...