Problems in Column Vectors, Linear Algebra

[SlowLearner007]

Dear all, the question states that:

Let x = (2, -1) and y = (2, 1) Show that (h, k) is in span {x, y} for all h and k

p.s. Numbers is red are column vectors.

What l think :

IF x+y , then span {x,y} is (4, 0) so h = 4
IF x - y, then span {x,y} is (0, -2) so k= -2

so why it says span {x,y} is consistent for all h and k ?

 

[ksdhart2]

Sorry but I find the work you've shown quite confusing. You don't seem to posted any complete thoughts on the matter. You say "If x + y" but then you don't finish that sentence. If x + y what? Similarly, if x - y what? It looks to me as if you might mean "If [span(x, y) means] x + y [then] ..." which, well, that's technically true... but the span of a set of vectors doesn't mean vector addition nor does it mean vector subtraction.

It seems like you need to start this problem back at square one, and make sure you know the definition of the terms used in the problem text. Review the definition of span in this context and try the exercise again. The span of a set of vectors is the set of all vectors that are linear combinations of the given vectors.

 

[Steven G]

Dear all, the question states that:

Let x = (2, -1) and y = (2, 1) Show that (h, k) is in span {x, y} for all h and k

p.s. Numbers is red are column vectors.

What l think :

IF x+y , then span {x,y} is (4, 0) so h = 4
IF x - y, then span {x,y} is (0, -2) so k= -2

so why it says span {x,y} is consistent for all h and k ?

You need to find real numbers a and b such that a(2,-1) + b(2,1) = (h,k). Since (h,k) is an arbitary vector in R^2, if you do find an a and b, then (2,-1) and (2,1) spans R^2

 

[HallsofIvy]

Dear all, the question states that:

Let x = (2, -1) and y = (2, 1) Show that (h, k) is in span {x, y} for all h and k

p.s. Numbers is red are column vectors.

What l think :

IF x+y , then span {x,y} is (4, 0) so h = 4
IF x - y, then span {x,y} is (0, -2) so k= -2

I would have to say it appears that you do not know what "span" means! The span of two vectors is NOT a single vector. It is the subspace or all linear combinations of the two vectors.

The span of x= (2, -1) and y= (2, 1) is all vectors of the form a(2, -1)+ b(2, 1)= (2a+ 2b, -a+ b)= (2(a+b). -a+b). A vector, (h, k), is in the span if there exist numbers a and b such that (2(a+b), -(a+b))= (h, k). That is the same as the two equations 2(a+b)= h and -a+ b= k. From the first equation a+ b= h/2. Adding -a+ b= k to that gives 2b= h/2+ k so b= h/4+ k/2 and then a= h/2- b= h/2- h/4- k/2= h/4- k/2.

A more sophisticated way of looking at this is that a "basis" for an n dimensionao vector space has three properties:
1) It spans the space.
2) It contains n vectors.
3) The vectors are independent.

Further any two of those is sufficient to prove the third!

Here we are dealing with a two dimensional space, \(\displaystyle R^2\), and the two given vectors are clearly independent so they must also span \(\displaystyle R^2\) and so any vector in \(\displaystyle R^2\) is in the span.

so why it says span {x,y} is consistent for all h and k ?