Matrix Linear Algebra: showing one span is a subset of the other?

Question

I understand that if span A is a subset of span B, then all of the elements of A are in B. But I'm kind of confused as to how you go about showing that one span is a subset of another.

Answer 1

Long story short: to show span(A) is a subset of span(B) you just need to show that each element of A is a subset of span(B) [that is, it can be written as a linear combination of elements in B]

To see why, recall that span(A) is all of the different linear combinations using elements in A and span(B) is all of the different linear combinations of elements in B. The idea is that linear combinations of linear combinations are linear combinations (if that makes sense?). So if you could show that each of the elements of A are linear combinations of things in B then linear combinations of things in A would (in turn) be linear combinations of things in B. Maybe it would be better just to show the math...?

First, I'll do it for small A and B so that it is more obvious what is going on.

Claim: Let A = {a_1, a_2} and B = {b_1, b_2, b_3}. If A is a subset of span(B) then span(A) is a subset of span(B).

To show this, we have to show that every x in span(A) is also in span(B). So let x be in span(A). By definition,

x = c_1a_1 + c_2a_2 where c_1,c_2 are elements of the field that the span is over (probably real numbers if you are looking at normal vectors - if you don't know what a field is, just put the words "are real numbers" in wherever I write "in the field").

Now since each a_i is in span(B), it can be written as:

a_1 = d_1b_1 + d_2b_2 + d_3b_3
a_2 = e_1b_1 + e_2b_2 + e_3b_3

where d_i and e_i are all in the field. So, plugging in, we get

x = c_1 * (d_1b_1 + d_2b_2 + d_3b_3) + c_2 * (e_1b_1 + e_2b_2 + e_3b_3)

and so (rearranging)

x = (c_1d_1 + c_2e_1) * b_1 + (c_1d_2 + c_2e_2) * b_2 + (c_1d_3 + c_2e_3) * b_3

and each of the terms involving c_i, d_i and e_i are in the field, so we have expressed x as a linear combination of elements in B, so by definition x is in span(B).

Therefore, span(A) is a subset of span(B).

Now, if you wanted to extend this to general sets A and B, you would need to use summation notation (which doesn't translate well here, but here it goes)

Let x \in span(A). Then x = \sum_{a \in A} c_a * a where c_a are in the field.

Since each a \in span(B),

a = \sum_{b \in B} c_{a,b} * b where c_{a,b} are in the field.

So x = \sum_{a \in A} c_a * (\sum_{b \in B} c_{a,b} * b) and rearranging gives

x = \sum_{b \in B} [\sum{a \in A} (c_a * c_{a,b})] * b

where the coefficients of each b is in the field since each c_i and each c_{i,j} is in the field. So, by definition, x \in span(B).

Answer 2

for facilitation, call the three matrix-vectors: V1, V2 & V3 then aV1 + bV2 + cV3 = Z the place Z is the 2x2 0 matrix then evaluate entries 1a + 0b + 0c = 0 1a + 1b + 1c = 0 0a - 1b + 1c = 0 1a + 0b + 0c = 0 for this reason a = 0 b + c = 0 -b+ c = 0 for this reason a = 0, b = 0, c = 0 after fixing the linear equations.... for this reason S is a collection of linearly self reliant vectors. enable the climate of D be called [m 0] [0 m] then [m 0] = aV1 + bV2 + cV3 [0 m] this time you could confirm no depend if the equipment will yield answer in a, b and c. 1a + 0b + 0c = m 1a + 1b + 1c = 0 0a - 1b + 1c = 0 1a + 0b + 0c = m for this reason a = m b = 0 c = 0 ...... is the answer