I'm having trouble solving this statement:
Let u & v be distinct vectors of a vector space V. Show that if {u,v} is a basis for V and a & b are nonzero scalars, then both {u+v,au} and {au,bv} are also basis for V.
Thanks in advance
Show that {u+v,au} and {au,bv} span V and are sets of linearly independent vectors.
For example, for {u+v,au} …
Span: Let w = iu+jv be a vector in V. Are there scalars k, l such that w = k(u+v)+l(au)? Well, k(u+v)+l(au) = (k+la)u+kv. It turns out that we can take k = j and k+la = i, i.e. l = (i−j)/a (since a ≠ 0). Hence w = j(u+v)+[(i−j)/a](au) is a linear combination of u+v and au.
Linear independence: Suppose i(u+v)+j(au) = 0. Then (i+ja)u+iv = 0. By linear independence of u and v, we have i = 0 and i+ja = 0 ⇒ (as a ≠ 0) j = 0. This shows that u+v and au are linearly independent.
Repeat the same procedure for {au,bv}.