Prove that if span{u1, u2} = R2, then
span{u1 + u2, u1 − u2} = R2.
span{u1, u2} = {au1 + bu2, for any a, b ∈ R}
span{u1 + u2, u1 − u2} = { h(u1 + u2) + k(u1 − u2) for any h,k ∈ R} =
= {hu1 + hu2 + ku1 - ku2} = {(h + k)u1 + (h - k)u2 for any h,k ∈ R}
now let
x = au1 + bu2
be any element of R^2
solve
h + k = a
h - k = b
h = (a + b)/2
k = (a - b)/2
taking these values for h,k we can write x in the second form
x = h(u1 + u2) + k(u1 − u2)
indeed
x = (a + b)(u1 + u2)/2 + (a - b)(u1 - u2)/2 =
= [au1 + au2 + bu1 + bu2 + au1 - au2 - bu1 + bu2]/2 =
= [2au1 + 2bu2]/2 = au1 + bu2
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span{u1, u2} = {au1 + bu2, for any a, b ∈ R}
span{u1 + u2, u1 − u2} = { h(u1 + u2) + k(u1 − u2) for any h,k ∈ R} =
= {hu1 + hu2 + ku1 - ku2} = {(h + k)u1 + (h - k)u2 for any h,k ∈ R}
now let
x = au1 + bu2
be any element of R^2
solve
h + k = a
h - k = b
h = (a + b)/2
k = (a - b)/2
taking these values for h,k we can write x in the second form
x = h(u1 + u2) + k(u1 − u2)
indeed
x = (a + b)(u1 + u2)/2 + (a - b)(u1 - u2)/2 =
= [au1 + au2 + bu1 + bu2 + au1 - au2 - bu1 + bu2]/2 =
= [2au1 + 2bu2]/2 = au1 + bu2