2.2 Vector Equations and Spans¶

The span of a set of vectors¶

If the vector \(\vec b\) is in the span of the vectors \(\{\vec v_1, \vec v_2, \dots \vec v_n\}\), then there is some linear combination that makes the vector equation true.

\[c_1 \vec v_1 + c_2 \vec v_2 + \dots + c_n \vec v_n = \vec b\]

As shown in Margalit and Robanoff, we can find the weights \(c_1, c_2, \dots , c_n\) by row-reducing an augmented matrix. Let’s rewrite the equation above slightly by calling the weights \(x_1, x_2, \dots , x_n\).

\[x_1 \vec v_1 + x_2 \vec v_2 + \dots + x_n \vec v_n = \vec b\]

If the columns of the matrix \(A\) are comprised of the vectors \(\vec v_1, \vec v_2, \dots ,\vec v_n\), we would write

\[A = [\vec v_1, \vec v_2, \dots , \vec v_3]\]

and then the above vector equations are equivalent to the matrix equations

\[A\vec x = \vec b\]

We already know from Chapter 1 that the solution vector \(\vec x\) can be found by row-reducing the augmented matrix \([A | \vec b]\).

Example 1¶

Determine if the vector \(\vec b\) in the span of the vectors

\[\begin{split}\vec v_1 = \left[\begin{array}{r}5\\3\\-2\\\end{array}\right], \vec v_2 = \left[\begin{array}{r}0\\2\\5\\\end{array}\right], \vec v_3 = \left[\begin{array}{r}3\\4\\3\\\end{array}\right]\end{split}\]

where

\[\begin{split}\vec b = \left[\begin{array}{r}4\\0\\2\\\end{array}\right]\end{split}\]

v1 = [5 ; 3 ; -2]
v2 = [ 0; 2 ; 5 ]
v3 = [3 ; 4 ; 3]
b = [ 4 ; 0 ; 2]

We create the matrix \(A\) using the vectors as its columns.

A = [v1, v2, v3]

A =

     5     0     3
     3     2     4
    -2     5     3

We want to augment \(A\) with the vector \(\vec b\) and then row-reduce.

[A,b]

ans =

     5     0     3     4
     3     2     4     0
    -2     5     3     2

rref([A,b])

ans =

    1.0000         0         0    5.2308
         0    1.0000         0    6.9231
         0         0    1.0000   -7.3846

The components of the vector \(\vec x\) are the weights of the linear combination. Let’s use the rats function to better understand what rational numbers we’re talking about.

rats(rref([A,b]))

ans =

  3x56 char array

    '       1             0             0           68/13    '
    '       0             1             0           90/13    '
    '       0             0             1          -96/13    '

Mathematicians often write vectors like this as scalar multiples because they look much better.

\[\begin{split}\vec x = \frac{1}{13}\left[\begin{array}{rrr}68 \\ 90 \\ -96\\\end{array}\right]\hspace{1cm} \text{instead of}\hspace{1cm}\vec x = \left[\begin{array}{rrr}\frac{68}{13} \\ \frac{90}{13} \\ \frac{-96}{13}\\\end{array}\right]\end{split}\]

Example 2¶

Determine if the vector \(\vec b\) in the span of the vectors \(\vec v_1, \vec v_2\) where

\[\begin{split}\vec v_1 = \left[\begin{array}{r}0\\4\\0\\4\\\end{array}\right], \vec v_2 = \left[\begin{array}{r}-1\\5\\0\\-1\\\end{array}\right] \hspace{1cm} \text{and} \hspace{1cm} \vec b = \left[\begin{array}{r}5\\-17\\0\\3\\\end{array}\right]\end{split}\]

We create the vectors first, then row-reduce the augemented matrix.

v1 = [0;4;0;4]
v2 = [-1;5;0;-1]
b = [5 ; -17 ; 0 ; 3]

M = [v1,v2]

rref([M, b])

We see the third row is impossible since \((0)x_1 + (0)x_2 \neq 1\). Since the matrix equations are inconsistent, there is no linear cobination of the vectors \(\vec v_1,\vec v_2\) that is equal to \(\vec b\) and, thus,

\[\vec b \notin \text{Span}\{\vec v_1, \vec v_2\}\]

Linear Algebra with MATLAB

2.2 Vector Equations and Spans¶

The span of a set of vectors¶

Example 1¶

Example 2¶