[edflsafoiewq]

If v is an eigenvector of T with eigenvalue λ, then (T - bI)v = λv - bv = (λ - b)v. The image is a rescaling of v (and in particular has the same eigenvalue). Therefore

    (T-λ_2) ... (T-λ_m) v_k =
    (T-λ_2) ... (T-λ_(m-1)) (λ_k - λ_m) v_k =
    (T-λ_2) ... (T-λ_(m-2)) (λ_k - λ_(m-1)) (λ_k - λ_m) v_k =
    ... =
    (λ_k-λ_2) ... (λ_k - λ_(m-1)) (λ_k - λ_m) v_k

Now if k is not 1, then the factor (λ_k - λ_k) appears in the above product, so the term drops outs. So the only term left is the v_1 one.

The eigenvalues are all distinct by hypothesis: "Non-zero eigenvectors corresponding to distinct eigenvalues...".

[T_D_K]

Great explanation, thanks.

And the "distinct eigenvalues" part is obvious in hindsight. For some reason my brain thought that we were adding them, not subtracting.