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

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- Jan 23rd 2008, 09:31 PMlllllBasis Question
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 - Jan 23rd 2008, 11:12 PMJaneBennet
Show that {

**u**+**v**,*a***u**} and {*a***u**,*b***v**} span*V*and are sets of linearly independent vectors.

For example, for {**u**+**v**,*a**u*} …

__Span__: Let**w**=*i***u**+*j***v**be a vector in*V*. Are there scalars*k*,*l*such that**w**=*k*(**u**+**v**)+*l*(*a***u**)? Well,*k*(**u**+**v**)+*l*(*a***u**) = (*k*+*la*)**u**+*k***v**. 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*](*a***u**) is a linear combination of**u**+**v**and*a***u**.

__Linear independence__: Suppose*i*(**u**+**v**)+*j*(*a***u**) =**0**. Then (*i*+*j**a*)**u**+*i***v**=**0**. By linear independence of**u**and**v**, we have*i*= 0 and*i*+*j**a*= 0 ⇒ (as*a*≠ 0)*j*= 0. This shows that**u**+**v**and*a***u**are linearly independent.

Repeat the same procedure for {*a***u**,*b***v**}.