Question

Let u, v belong to R^n and suppose that {2u,u-v} is a basis for subspace W. Show that {u,v} is a basis for W by showing that it satisfies the two conditions in the definition of a basis.

Any help appreciated!

Answer 1

So what are the two conditions in the definition of a basis?

Answer 2

The conditions are:
1. {v1,v2,...,vn} is linearly independent
2. {v1,v2,...,vn} span W.

Answer 3

1.) if 2u and u-v form a basis for subspace of W, then 2u and u-v are linearly independent, that means also that 2u and u-v are not scalar multiples of each other, or that
\[2u \neq c_1(u-v)\]
\[u-v \neq c_22u\]
where c1 and c2 are any constants. now from this inequalities we get:
\[2u \neq c_1u-c_2v\]
\[2u-c_1u \neq -c_2v\]
\[(2-c1)u=-c_2v\]
\[u \neq (-c_2/2-c_1)v\]
(-c_2/2-c_1) is also just any constant so let's just name that constant c
\[u \neq cv\]
using the same method to the 2nd inequality at the top we get:
\[v \neq (1-c_22)u\]
now (1-c_22) is just also any constant c' so that can be as:
\[v \neq c'u\]
and so we have proven that u and v cannot be written as a scalar multiple of each other, hence they are linearly independent.
2.) so now we have to prove that for any vector w in W, it can be expressed as a linear combination of v and u.
let's begin with the known facts, we know that 2u and u-v span W, so any vector w in W can be written as:
w=c3(2u)+c4(u-v)
where c3 and c4 are any constants, now if we rearange this:
w=2c3u+c4u-c4v
w=(2c3+c4)u-c4v
now 2c3+c4 and -c4 are any constant also, and so we can write that as,
w=cu+c'v
where c and c' are any constant. But wait a minute, isn't this just a linear combination of u and v? and the w is just any vector in W! so we have shown here that any vector w in W could be written as a linear combination of u and v, hence u and v span W, therefore u and v span W.

So since u and v are linearly independent and that it spans W, then u and v form a basis for W.

Answer 4

Suppose that u and v are linearly dependent, then v = a u,
u-v= u - a u = (1-a)u
then 2 u and u-v will be linearly dependent a contradiction.
Since u and v are in W, because
u = 1/2 (2u)
v= 1/2 (2u) -(u-v)
Hence u and v form a basis for W