Extension of a basis (exchange theorem)

[latentcorpse]

hi. ok if I'm using the exchange theorem for extension to a basis. i have the standard basis of 4 dimensional real space is {e1,e2,e3,e4}. and v1=e1+e2

then i can say that the coefficient at e1 is 1 which is non zero therefore i can exchange and get {v1,e2,e3,e4} as a basis. however if v2 = v1-2e2 say then what would be the basis i could make by the exchange theorem.

i reckoned it would be "the coefficient at e2 is -2 therefore by exchange theorem {v1,-(1/2)v2,e3,e4} would be a basis."?

Unless all that matters is that the coefficient of e2 is non zero and then {v1,v2,e3,e4} would be the basis?

Help!

 

[HallsofIvy]

hi. ok if I'm using the exchange theorem for extension to a basis. i have the standard basis of 4 dimensional real space is {e1,e2,e3,e4}. and v1=e1+e2

then i can say that the coefficient at e1 is 1 which is non zero therefore i can exchange and get {v1,e2,e3,e4} as a basis. however if v2 = v1-2e2 say then what would be the basis i could make by the exchange theorem.

i reckoned it would be "the coefficient at e2 is -2 therefore by exchange theorem {v1,-(1/2)v2,e3,e4} would be a basis."?

Unless all that matters is that the coefficient of e2 is non zero and then {v1,v2,e3,e4} would be the basis?

Help!

Yes, the only important point is that the coefficient is non-zero. Since the coefficient of e2 is non-zero, you can replace e2 by v2.

Suppose v= ae1+ be2+ ce3+ de4. Since v1= e1+ e2, e1= v1- e2 so v= a(v1- e2)+ be2+ ce3+ de4= av1+ (b-a)e2+ ce3+ de4. Since v2= v1- 2e2, e2= (1/2)(v2- v1) and v= av1+ (b-a)(1/2)(v2- v1)+ ce3+ de4= (1/2)(3a+ b)v1+(1/2)bv2+ ce3+ de4. The coefficients involve "1/2" but the basis doesn't have to. In fact, {v1,-(1/2)v2,e3,e4} is a basis if and only if {v1, v2, e3, e4} is a basis.