Use mathematical induction to prove that if L is a linear transformation from V to W, then L (α1v1 + α2v2 +· · ·+αnvn)= α1L (v1) + α2L (v2)+· · ·+αnL (vn) g

The exercise will show that [tex]L^{n}[/tex] is a linear operator on V and that [tex]L^{n+1}[/tex] is also a linear operator on V. This, follows that:

[tex]L^{n+1} (av) = L(L^{m}(v_{1}+v_{2})\\ = L(L^{m} (v_{1} + L^{m}v_{2})\\ = L(L^{m}v_{1} + L(L^{m}v_{2})\\ = L^{m+1}(v_{1}) + L^{m+1}(v_{2})[/tex]

eve24great2001 eve24great2001 · Answer 2 · 2020-03-23T23:07:49+01:00

Answer/Step-by-step explanation:

For the mathematical induction,

We show that the equation

L (α1v1 + α2v2 +· · ·+αnvn)= α1L (v1) + α2L (v2)+· · ·+αnL (vn) is true for

L = 1,

Assume it is true for L = n and show that it is true for L = n + 1.

If L = 1, the equation become

(α1v1 + α2v2 +· · ·+αnvn)= α1(v1) + α2 (v2)+· · ·+αn (vn). Therefore, the Right Hand side(RHS) = Left Hand side(LHS)

When L = n, we assume the following is true

(α1nv1 + α2nv2 +· · ·+αnvn)= α1n(v1) + α2n (v2)+· · ·+αn (vn)

Then, when L = n + 1,

n +1 (α1v1 + α2v2 +· · ·+αvn)= α1(n +1) (v1) + α2(n +1) (v2)+· · ·+αn(n + 1)(vn).

Open the bracket,

n(α1v1 + α2v2 +· · ·+αvn) + α1v1 + α2v2 +· · ·+αnvn = α1n (v1) + α2v2 +· · ·+αvn ) + α1(v1) + α2v2+· · ·+αn(vn)

Since we assume the the equation is true for L = n, and eliminating some terms, then

L (α1v1 + α2v2 +· · ·+αnvn)= α1L (v1) + α2L (v2)+· · ·+αnL (vn)