4 ms·
I got lost in 2.2, I can't work out how applying the transformation leads to the result. Which is frustrating since it's the only non-trivial line in the proof,
by T_D_K 9y ago
I got lost in 2.2, I can't work out how applying the transformation leads to the result. Which is frustrating since it's the only non-trivial line in the proof, lol. Also, after applying the transformation, the author states that "a1(λ1 − λ2)(λ1 − λ3)...(λ1 − λm)v1 = 0" => "a1 = 0". But he never says why we know "λa != -λb for all a, b in 1..m" -- that seems non-obvious to me.
- edflsafoiewq 9y agoIf 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 9y agoGreat explanation, thanks. And the "distinct eigenvalues" part is obvious in hindsight. For some reason my brain thought that we were adding them, not subtracting.