Spanning Sets

Definition. Let V be a vector space over a field F, and let S be a subset of V. The span of S is

$$\langle S \rangle = \left\{a_1 v_1 + a_2 v_2 + \cdots + a_n v_n \mid a_i \in F, v_i \in S\right\}.$$

That is, the span consists of all linear combinations of vectors in S.

S spans a subspace W of V if $W =
   \langle S\rangle$ ; that is, if every element of W is a linear combination of elements of S.

Example. Let

$$S = \left\{\left[\matrix{1 \cr 1 \cr 0 \cr}\right], \left[\matrix{0 \cr 1 \cr 1 \cr}\right]\right\} \subset \real^3.$$

(a) Prove or disprove: $(3, -1, -4)$ is in the span of S.

(b) Prove or disprove: $(5, -2, 6)$ is in the span of S.

(a) I try to find numbers a and b such that

$$a \cdot \left[\matrix{1 \cr 1 \cr 0 \cr}\right] + b \cdot \left[\matrix{0 \cr 1 \cr 1 \cr}\right] = \left[\matrix{3 \cr -1 \cr -4 \cr}\right].$$

(Remember that by default, when you convert a vector in "parenthesis form" like "$(3, -1, -4)$ " to a matrix, you convert the vector to a column vector.)

This is equivalent to the matrix equation

$$\left[\matrix{1 & 0 \cr 1 & 1 \cr 0 & 1 \cr}\right] \left[\matrix{a \cr b \cr}\right] = \left[\matrix{3 \cr -1 \cr -4 \cr}\right].$$

Row reduce the augmented matrix:

$$\left[\matrix{ 1 & 0 & 3 \cr 1 & 1 & -1 \cr 0 & 1 & -4 \cr}\right] \quad \to \quad \left[\matrix{ 1 & 0 & 3 \cr 0 & 1 & -4 \cr 0 & 0 & 0 \cr}\right]$$

The solution is $a = 3$ , $b = -4$ . That is,

$$3 \cdot \left[\matrix{1 \cr 1 \cr 0 \cr}\right] + (-4) \cdot \left[\matrix{0 \cr 1 \cr 1 \cr}\right] = \left[\matrix{3 \cr -1 \cr -4 \cr}\right].$$

The vector $(3, -1, -4)$ is in the span of S.

(b) I try to find numbers a and b such that

$$a \cdot \left[\matrix{1 \cr 1 \cr 0 \cr}\right] + b \cdot \left[\matrix{0 \cr 1 \cr 1 \cr}\right] = \left[\matrix{5 \cr -2 \cr 6 \cr}\right].$$

This is equivalent to the matrix equation

$$\left[\matrix{1 & 0 \cr 1 & 1 \cr 0 & 1 \cr}\right] \left[\matrix{a \cr b \cr}\right] = \left[\matrix{5 \cr -2 \cr 6 \cr}\right].$$

Row reduce to solve the system:

$$\left[\matrix{ 1 & 0 & 5 \cr 1 & 1 & -2 \cr 0 & 1 & 6 \cr}\right] \quad \to \quad \left[\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr}\right]$$

The last matrix says "$0 = 1$ ", a contradiction. The system is inconsistent, so there are no such numbers a and b. Therefore, $(5, -2, 6)$ is not in the span of S.


Thus, to determine whether the vector $b \in F^n$ is in the span of $v_1$ , $v_2$ , ..., $v_m$ in $F^n$ , form the augmented matrix

$$\left[\matrix{\uparrow & \uparrow & & \uparrow & \uparrow \cr v_1 & v_2 & \cdots & v_m & b \cr \downarrow & \downarrow & & \downarrow & \downarrow \cr}\right]$$

If the system has a solution, b is in the span, and coefficients of a linear combination of the v's which add up to b are given by a solution to the system. If the system has no solutions, then b is not in the span of the v's.

(In a general vector space where vectors may not be "numbers in slots", you have to got back to the definition of spanning set.)


Example. Consider the following set of vectors in $\real^3$ :

$$\left\{\left[\matrix{1 \cr 0 \cr -1 \cr}\right], \left[\matrix{2 \cr 1 \cr 3 \cr}\right], \left[\matrix{1 \cr 1 \cr -4 \cr}\right]\right\}$$

Prove that the span of the set is all of $\real^3$ .

Let $(x, y, z) \in \real^3$ . I must find real numbers a, b, c such that

$$a \cdot\left[\matrix{1 \cr 0 \cr -1 \cr}\right] + b \cdot\left[\matrix{2 \cr 1 \cr 3 \cr}\right] + c \cdot\left[\matrix{1 \cr 1 \cr -4 \cr}\right] = \left[\matrix{x \cr y \cr z \cr}\right].$$

The matrix equation is

$$\left[\matrix{ 1 & 2 & 1 \cr 0 & 1 & 1 \cr -1 & 3 & -4 \cr}\right] \left[\matrix{a \cr b \cr c \cr}\right] = \left[\matrix{x \cr y \cr z \cr}\right].$$

Row reduce to solve:

$$\left[\matrix{ 1 & 2 & 1 & x \cr 0 & 1 & 1 & y \cr -1 & 3 & -4 & z \cr}\right] \quad \to \quad \left[\matrix{ 1 & 0 & 0 & \dfrac{1}{8} (7 x - 11 y - z) \cr \noalign{\vskip2pt} 0 & 1 & 0 & \dfrac{1}{8} (x + 3 y + z) \cr \noalign{\vskip2pt} 0 & 0 & 1 & \dfrac{1}{8} (-x + 5 y - z) \cr}\right]$$

I get the ugly solution

$$a = \dfrac{1}{8} (7 x - 11 y - z), \quad b = \dfrac{1}{8} (x + 3 y + z), \quad c = \dfrac{1}{8} (-x + 5 y - z).$$

This shows that, given any vector $(x,
   y, z)$ , I can find a linear combination of the original three vectors which equals $(x, y, z)$ .

Thus, the span of the original set of three vectors is all of $\real^3$ .


Example. Determine whether the vector is in the span of the set

$$S = \left\{(1, 2, 1), (1, 4, 2)\right\} \quad\hbox{in}\quad \integer_5^3.$$

(a) $(0, 4, 2)$ .

(b) $(1, 1, 1)$ .

(a) I want to find a and b such that

$$a \cdot \left[\matrix{1 \cr 2 \cr 1 \cr}\right] + b \cdot \left[\matrix{1 \cr 4 \cr 2 \cr}\right] = \left[\matrix{0 \cr 4 \cr 2 \cr}\right].$$

This is the system

$$\left[\matrix{1 & 1 \cr 2 & 4 \cr 1 & 2 \cr}\right] \left[\matrix{a \cr b \cr}\right] = \left[\matrix{0 \cr 4 \cr 2 \cr}\right].$$

Form the augmented matrix and row reduce:

$$\left[\matrix{ 1 & 1 & 0 \cr 2 & 4 & 4 \cr 1 & 2 & 2 \cr}\right] \quad \to \quad \left[\matrix{ 1 & 0 & 3 \cr 0 & 1 & 2 \cr 0 & 0 & 0 \cr}\right]$$

The last matrix says $a = 3$ and $b = 2$ . Therefore, $(0, 4, 2)$ is in the span of S:

$$3 \cdot \left[\matrix{1 \cr 2 \cr 1 \cr}\right] + 2 \cdot \left[\matrix{1 \cr 4 \cr 2 \cr}\right] = \left[\matrix{0 \cr 4 \cr 2 \cr}\right].\quad\halmos$$

(b) I want to find a and b such that

$$a \cdot \left[\matrix{1 \cr 2 \cr 1 \cr}\right] + b \cdot \left[\matrix{1 \cr 4 \cr 2 \cr}\right] = \left[\matrix{1 \cr 1 \cr 1 \cr}\right].$$

This is the system

$$\left[\matrix{1 & 1 \cr 2 & 4 \cr 1 & 2 \cr}\right] \left[\matrix{a \cr b \cr}\right] = \left[\matrix{1 \cr 1 \cr 1 \cr}\right].$$

Form the augmented matrix and row reduce:

$$\left[\matrix{ 1 & 1 & 1 \cr 2 & 4 & 1 \cr 1 & 2 & 1 \cr}\right] \quad \to \quad \left[\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr}\right]$$

The last row of the row reduced echelon matrix says "$0 = 1$ ". This contradiction implies that the system is has no solutions. Therefore, $(1, 1, 1)$ is not in the span of S.


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