Question

All vectors are in R^ n. Prove the following statements.

a) v·v=||v||2

b) If ||u||2 + ||v||2 = ||u + v||2, then u and v are orthogonal.

c) (Schwarz inequality) |v · w| ≤ ||v||||w||.

Answer 1

(a). Let v = (v₁,v₂,…v_n). Then v.v = v₁²+v₂²+…+v_n² = ||v||²

(b) Let u= (u₁,u₂,…u_n) and v = (v₁,v₂,…v_n). Then ||u||² = u₁²+u₂²+…+u_n² and ||v||² = v₁²+v₂²+…+v_n². Further, u+v =(u₁+v₁,u₂+v₂,…,u_n+v_n) so that ||u+v||² =(u₁+v₁ )²+(u₂+v₂ )²+…+(u_n+v_n )² = (u₁²+v₁²+2 u₁ v₁)+ (u₂²+v₂²+2u₂ v₂)+…+ (u_n²+v_n²+2u_n v_n)= (u₁²+u₂²+…+u_n²)+( v₁²+v₂²+…+v_n²)+2(u₁ v₁+ u₂ v₂+…+ u_n v_n) = ||u||²+||v||²+2(u₁ v₁+ u₂ v₂+…+ u_n v_n). Now, if ||u||²+||v||² =||u + v||², then2(u₁ v₁+ u₂ v₂+…+ u_n v_n) = 0 or, 2u.v = 0 or u.v = 0 so that u and v are orthogonal.

c).We have ||v + w||² = (v + w) · (v +wv) = v · v + w · w + 2v · w = ||v||² + ||w|² + 2v · w ≤||v||² + ||w|² + 2||v|| · ||w|| or, ||v + w||² ≤ (||v|| + ||w||)². Therefore, ||v + w|| ≤ (||v|| + ||w||).

Let v, w ∈ Rⁿ. Then for all k ∈ R, we have (v − kw, v − kw) ≥ 0. Further, (v − kw, v − kw) = (v, v) − 2 (v, w)k + (w, w)k².

Now, if we consider the quadratic function f(k) = (w,w)k² − 2 (v, w)k + (v, v), then f(k) ≥ 0.

However, a quadratic function f(k) = ak² + bk + c satisfying f(k) ≥ 0 has either two equal real roots or imaginary roots.

Hence b² − 4ac = 4 (v, w)² − 4 (v,v) (w, w) ≤ 0, so that (v, w)² ≤ (v, v) (w,w). Now, on taking the square root of both the sides, we get |(v, w)| ≤ ||v|| ||w||.