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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-...
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