Linear Algebra Question-Vector spaces and fields?

You have discovered that just reading math texts does not lead to math understanding. It is as if you were to watch many different people ride a unicycle. After that, do you think you could ride a unicycle. So I'm going to tell you a trick for reading math. Use a 3-5 card and cover up the proofs and the worked out examples and see if you can do them yourself. If you can't, read the proof/solution and come back to them in a few days. This takes a lot of time in the beginning, but after a while it clicks and you will find math easy. If you do this, you will find all of the above problems easy as they all just depend on the definitions. I'll start you off with problem A. Suppose v is in Span(x_1...x_n, y_1....y_n). Then v = linear combination of x1..... yn. Look at what that means. So v = linear combination of x1...xn + linear combination of y1...yn. So v is in Span(x_1..x_n) + Span(y_1....y_n). The reverse argument is similar. Good luck

You have discovered that just reading math texts does not lead to math understanding. It is as if you were to watch many different people ride a unicycle. After that, do you think you could ride a unicycle. So I'm going to tell you a trick for reading math. Use a 3-5 card and cover up the proofs and the worked out examples and see if you can do them yourself. If you can't, read the proof/solution and come back to them in a few days. This takes a lot of time in the beginning, but after a while it clicks and you will find math easy. If you do this, you will find all of the above problems easy as they all just depend on the definitions. I'll start you off with problem A. Suppose v is in Span(x_1...x_n, y_1....y_n). Then v = linear combination of x1..... yn. Look at what that means. So v = linear combination of x1...xn + linear combination of y1...yn. So v is in Span(x_1..x_n) + Span(y_1....y_n). The reverse argument is similar. Good luck

A Let x be an element of Span(x_1,...,x_n,y_1,...,y_n). Then, by definition, x=SUM_{i=1}^n x_i + SUM_{i=1}^n y_i. But Sum_{i=1}^n x_i is clearly an element of Span(x_1,...,x_n) and SUM_{i=1}^n y_i is clearly an element of Span(y_1,...,y_n), thus x is an element of Span(x_1,...,x_n) U Span(y_1,...,y_n). (Should be a union of sets, rather than a +, since we're talking about sets here). Similarly if x is an element of Span(x_1,...,x_n) U Span(y_1,...,y_n), then x is some linear combination of these vectors and is therefore an element of Span(x_1,...,x_n,y_1,...,y_n). B Let x be an element of Span(x_1,x_2,x_3) intersect Span(x_2,x_3,x_4). Then since x is an element of Span(x_1,x_2,x_3), x can be written as a linear combination of these vectors... x=ax_1+bx_2+cx_3 (a,b,c elements of F) And since x is an element of Span(x_2,x_3,x_4), it can be written as a linear combination of these vectors... x=dx_2+ex_3+fx_4 (d,e,f elements of F). So we must have... ax_1+bx_2+cx_3=dx_2+ex_3+fx_4 And therefore... ax_1+(b-d)x_2+(c-e)x_3-fx_4=0. Since the vectors are independant, each of the coefficients in the above equation must be zero. Thus, a=0, b=d, c=e and f=0, hence x is a linear combination of x_2 and x_3 and is an element of Span(x_2,x_3). Finally, if x is an element of Span(x_2,x_3), then x=ax_2+bx_3 for some a,b in F. Trivially, this is an element of Span(x_1,x_2,x_3) and Span(x_2,x_3,x_4), with the observation that, x = x+0*x_1 = x+0*x_4, hence x is an element of Span(x_1,x_2,x_3) intersect Span(x_2,x_3,x_4). C In general this piece of logic breaks down! "ax_1+(b-d)x_2+(c-e)x_3-fx_4=0. Since the vectors are independant, each of the coefficients in the above equation must be zero." Furthermore, if x_4=ax_1, for some a in F, then... Span(x_1,x_2,x_3) intersect Span(x_2, x_3, x_4) = Span(x_1,x_2,x_3) intersect Span(x_1,x_2, x_3) = Span(x_1,x_2,x_3), which does not equal Span(x_2,x_3).