### Span

##### 1. Definition and relation to existence

The columns space colA of a matrix A is all the linear combinations of the columns of A. More generally, we have

The span of vectors v1, v2, ..., vk is

span{v1, v2, ..., vk} = {c1v1 + c2v2 + ... + ckvk: all c1, c2, ..., ckR}.

Example Any vector in R3 can be expressed as follows.

x = (x1, x2, x3) = x1(1, 0, 0) + x2(0, 1, 0) + x3(0, 0, 1) = x1e1 + x2e2 + x3e3.

Thus any vector is a linear combination of the standard basis e1, e2, e3. In other words, the standard basis vectors span the euclidean space R3. By the same reason, the standard basis e1, e2, ..., en span Rn.

Example Since any polynomial a0 + a1t + a2t2 + ... + antn of degree ≤ n is a linear combination of 1, t, t2, ..., tn, Pn is spanned by 1, t, t2, ..., tn.

Example By the equality 1 = cos2t + sin2, the constant value 1 function is in the span of cos2t and sin2t. Similarly, the function cos2t = cos2t - sin2tspan{cos2t, sin2t}.

The fact that colA is a subspace can also be extended.

A span is a subspace.

Proof We only show that span{v1, v2, v3} is a subspace. The general case is similar.

For u, vspan{v1, v2, v3}, we have u = c1v1 + c2v2 + c3v3, v = d1v1 + d2v2 + d3v3, so that (see this exercise)

u + v = (c1 + d1)v1 + (c2 + d2)v2 + (c3 + d3)v3span{v1, v2, v3}.

For uspan{v1, v2, v3} and dR, we have u = c1v1 + c2v2 + c3v3, so that

du = dc1v1 + dc2v2 + dc3v3span{v1, v2, v3}.

Example (see this example) By (a + b, 3a - b, 2a + b) = a(1, 3, 2) + b(1, -1, 1), the subset

{(a + b, 3a - b, 2a + b): a, bR} ⊂ R3

is the span of (1, 3, 2) and (1, -1, 1) and is a subspace.

The subset

{a cost + b sint: a, bR} ⊂ F(R)

is a subspace because it is the span of the vectors cost and sint.

The question of whether a vector bcolA is the same as whether the system Ax = b has solutions. Correspondingly, the question of whether a vector bspan{v1, v2, ..., vk} is the same as whether the vector equation

c1v1 + c2v2 + ... + ckvk = b,

where c1, c2, ..., ck are the unknown variables, has solutions.

Example If costspan{cos2t, sin2t}, then

a cos2t + b sin2t = cost

for some numbers a and b. By taking t = 0, we get a = 1. By taking t = π, we get a = - 1. The contradiction shows that costspan{cos2t, sin2t}.

Example Consider polynomials p1 = 1 - t + t3, p2 = 3 - t + 4t2 + 3t3, p3 = 2 - t + 2t2 + 2t3, p4 = t + 4t2 - 2t3, p5 = 1 + 3t2 in P3. We would like to determine whether q = t + 3t2span{p1, p2, p3, p4, p5}. The question is the same as whether we can find c1, c2, c3, c4, c5 such that

c1(1 - t + t3) + c2(3 - t + 4t2 + 3t3) + c3(2 - t + 2t2 + 2t3) + c4(t + 4t2 - 2t3) + c5(1 + 3t2) = t + 3t2.

By comparing the coefficients of 1, t, t2, and t3, we get a system of four equations.

 x1 + 3x2 + 2x3 + x5 = 0 - x1 - x2 - x3 + x4 = 1 4x2 + 2x3 + 4x4 + 3x5 = 3 x1 + 3x2 + 2x3 - 2x4 = 0

By this example, the system has no solution. Therefore q is not in the span.

Observe that the augmented matrix for the system is

 [ 1 3 2 0 1 0 ] = [ [p1] [p2] [p3] [p4] [p5] [q] ], -1 -1 -1 1 0 1 0 4 2 4 3 3 1 3 2 -2 0 0

where we used the translation

p = a + bt + ct2 + dt3P3 ↔ [p] = (a, b, c, d) ∈ R4.

Thus problems about spans in general vector spaces can be translated into existence problems about usual systems of linear equations.

If we modify q to become r = t + 2t2, then the augmented matrix of the corresponding existence problem is changed to

 [ [p1] [p2] [p3] [p4] [p5] [r] ] = [ 1 3 2 0 1 0 ]. -1 -1 -1 1 0 1 0 4 2 4 3 2 1 3 2 -2 0 0

By an earlier example, the system has solutions, so that t + 2t2span{p1, p2, p3, p4, p5}.

In the earlier example, we also saw that the system with the augmented matrix

 [ [p1] [p2] [p4] [r] ] = [ 1 3 0 0 ] -1 -1 1 1 0 4 4 2 1 3 -2 0

has solutions. Correspondingly, we have t + 2t2span{p1, p2, p4}.