|Sep8-09, 01:33 AM||#1|
how can i find out that all vectors of (x,y) are a linear combination of some vectors for example (3,4) and (6,8)?
|Sep8-09, 06:45 AM||#2|
In any case, a "linear combination" of, for example, (3, 4) and (6, 8) is, by definition, a vector of the form a(3, 4)+ b(6, 8)= (3a+6b, 4a+ 8b).
Now, try to go the "other way". Suppose we want to write some vector (x, y) as a linear combination of (3, 4) and (6, 8): (x, y)= a(3, 4)+ b(6, 8)= (3a+ 6b, 4a+ 8b). Then we have the simultaneous equations 3a+ 6b= x, 4a+ 8b= y. Try to solve those for a and b in terms of x and y.
That is not a very good example because (6, 8)= 2(3,4) so any linear combination of the two is just a multiple of either. (3a+ 6b, 4a+ 8b)= (3(a+2b), 4(a+ 2b))= (a+2b)(3, 4) and (3a+ 6b, 4a+ 8b)= (6(a+ b/2), 8(a+ b/2))= (a+ b/2)(6, 8). In particular, it is NOT true that "all vectors are a linear combination of (3, 4) and (6, 8). They do NOT span R2. If you try to solve 3a+ 6b= x, 4a+ 8b= y for general x and y, it does not work. For example if I solve the first equation for a: a= (x- 6b)/3 and substitute into the second equation I get 4((x-6b)/3+ 8b= (4/3)x- 8b+ 8b= (4/3)x= y/ Both a and b have been eliminated. The equations cannot be solved for a and b.
More interesting would be two independent vectors like (3, 4) and (5, 7). Any linear combination of them would be a(3, 4)+ b(5, 7)= (3a+5b, 4a+ 7b). Now, given any (x, y), we want (x, y)= (3a+ 5b, 4a+ 7b) so 3a+ 5b= x and 4a+ 7b= y. Solve those two equations. We might say, multiply the first equation by 7 and the second equation by 5 and subtract: (21a+ 35b)- (20a+ 35b)= 7x- 5y eliminates b. a= 7x- 5y. Putting that back into the first equation, 4a+ 5b= 3(7x- 5y)+ 5b= 21x- 15y+ 5b= y. 5b= -21x+ 14y so b= -(21/5)x+ (14/5)y.
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