math111_logo Span

3. Properties of span

The first property concerns how span is changed when we add/delete vectors.

span{v1, v2, ..., vk} ⊂ span{v1, v2, ..., vk, vk+1}. Moreover, equality holds ⇔ vk+1span{v1, v2, ..., vk}.

The intuitive reason for the inclusion is that more directions we have, more places we can go. Moreover, the equality holds if and only if the additional direction is "wasted", or is not a "new direction" compared with the existing ones.

The second property is related to the sum of subspaces.

span{v1, v2, ..., vm, w1, w2, ..., wn} = span{v1, v2, ..., vm} + span{w1, w2, ..., wn}.

The next property is useful for computational purpose. Recall that we used row operations to simplify a matrix and then answer many questions related to the matrix. We may simplify the span by introducing similar operations for a collection of vectors.

r[vector i] + [vector j]
add a multiple of one vector to another.
[vector i] ↔ [vector j]
exchange two vectors.
d[vector i], d ≠ 0
multiply a nonzero number to a vector.

The following result allows us to use three operations to simply the vectors that generate the span.

The three operations do not change the span.

The proof of the two properties can be found here.

Example By using the three operations, we have

span{cos2t, sin2t}
= span{cos2t, 1} (1 = cos2t + sin2t)
= span{-2cos2t, 1}
= span{cos2t, 1} (cos2t = 1 - 2cos2t)
= span{cos2t, 2sin2t} (2sin2t = 1 - cos2t)
= span{cos2t, sin2t}

Example We would like to find out whether the vectors v1 = (1, -1, 0, 1), v2 = (3, -1, 4, 3), v3 = (2, -1, 2, 2), v4 = (0, 1, 4, -2), v5 = (1, 0, 3, 0) span R4 (i.e., whether the span of the five vectors is equal to R4). Note that the span is the column space of the matrix

A = [ 1 3 2 0 1 ].
-1 -1 -1 1 0
0 4 2 4 3
1 3 2 -2 0

Therefore the question is the same as whether Ax = b has solutions for any b. From the computation in an earlier example, the row echelon form (obtained by row operations on A), contains a zero row. Then by the criterion for always existence, Ax = b does not always have solutions. Therefore we conclude that span{v1, v2, v3, v4, v5} ≠ R4.

Alternatively, we may use the properties of span to find the answer. The given vectors are too complicated for us to see span{v1, v2, v3, v4, v5} clearly. Thus we apply three operations to the five vectors, which are essentially column operations on A.

By applying (-3)[col 1] + [col 2], (-2)[col 1] + [col 3], (-1)[col 1] + [col 5] to A, we get

[ 1 0 0 0 0 ].
-1 2 1 1 1
0 4 2 4 3
1 0 0 -2 -1

By (-2)[col 3] + [col 2], (-1)[col 3] + [col 4], (-1)[col 3] + [col 5], and then [col 2] ↔ [col 3], we get

[ 1 0 0 0 0 ].
-1 1 0 0 0
0 2 0 2 1
1 0 0 -2 -1

By (-2)[col 5] + [col 4] and [col 3] ↔ [col 5], we get

[ 1 0 0 0 0 ].
-1 1 0 0 0
0 2 1 0 0
1 0 -1 0 0

This is a column echelon form of A, the simplest we can get from the column operations. The computation tells us

span{v1, v2, v3, v4, v5}
= span{(1, -1, 0, 1), (0, 1, 2, 0), (0, 0, 1, -1), (0, 0, 0, 0), (0, 0, 0, 0)}
= span{(1, -1, 0, 1), (0, 1, 2, 0), (0, 0, 1, -1)}
R4

The last step is due to the fact that three vectors cannot span R4.

The example above leads to some interesting observations. First for euclidean vectors, the preservation of the span under the three operations can be rephrased in terms of column operations.

If AB by column operations, then colA = colB.

Second is the numerical consequence of spanning the whole space.

Span{v1, v2, ..., vk} = Rnkn.

The equivalent statement in terms of the column space appeared in the first part of this exercise.


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[Extra: Properties of span]