Re: Finding spanning sets

Hey mrmaaza123.

Hint: Try multiplying the vectors by some constant.

Re: Finding spanning sets

Quote:

Originally Posted by

**mrmaaza123** Let W = { [a+b a-b a+2b] , a,b are in R} its a 3*1 matrix, a+b , a-b and a+2b are actually the first elements of three different rows. Find any three spanning sets for W? I managed to find one,

a (1 1 1) + b(1 1 1)

Do you not mean a[1, 1, 1]+ b[1, -1, 2]

Quote:

, both of these are also 3*1 matrices where each 1 is the first element of the first three rows.

How do i find the other two ?

Sorry about my notation.

PLease help.

Re: Finding spanning sets

[1 1 1] and [1 -1 2] both are column matrices and so is [a+b a-b a+2b].

Re: Finding spanning sets

Your subspace is of the form [a+b, a- b, a+ 2b]= a[1, 1, 1]+ b[1, -1, 2] so that {[1, 1, 1], [1, -1, 2]} spans it (and in fact is a basis) Note that you could also have got that by taking a= 1, b= 0 then a= 0, b= 1. To get other sets, set a and b to other numbers: if a= b= 1, you haave [2, 0, 3] and if a= 2, b= 1, you have [3, 1, 4]. Now to show you can write [a+ b, a- b, a+ 2b]= x[2, 0, 3]+ y[3, 1, 4], you must find x and y satisfying 2x+ 3y= a+ b, y= a- b, and 3x+ 4y= a+ 2b. From the second equation, y= a- b, the first equation becomes 2x+ 3(a- b)= a+ b so that 2x= -2a+ 4b or x= a+ 2b. Putting that into the third equation, 3x+ 4y= 3(-a+ 2b)+ 4(a- b)= (-3+4)a+ (6- 4)b= a+ 2b as required so we have [a+b, a-b, a+ 2b]= (a+ 2b)[2, 0, 3]+ (a-b)[3, 1, 4].

Re: Finding spanning sets

Thank you ! I think now i can do it.