Existence of solutions and span

Section 6.1 Existence of solutions and span

We now turn to our two fundamental questions rephrased here in terms of matrix multiplication.

Existence: Is there a solution to the equation \(A\xvec=\bvec\text{?}\)
Uniqueness: If there is a solution to the equation \(A\xvec=\bvec\text{,}\) is it unique?

In this section, we focus on the existence question and see how it leads to the concept of the span of a set of vectors.

Exploration 6.1.1. The existence of solutions.

If the equation \(A\xvec = \bvec\) is inconsistent, what can we say about the pivot positions of the augmented matrix \(\left[\begin{array}{r|r} A \amp \bvec \end{array}\right]\text{?}\)
Consider the matrix \(A\)

\begin{equation*} A = \left[ \begin{array}{rrr} 1 \amp 0 \amp -2 \\ -2 \amp 2 \amp 2 \\ 1 \amp 1 \amp -3 \end{array}\right]\text{.} \end{equation*}
If \(\bvec=\threevec{2}{2}{5}\text{,}\) is the equation \(A\xvec = \bvec\) consistent? If so, find a solution.
If \(\bvec=\threevec{2}{2}{6}\text{,}\) is the equation \(A\xvec = \bvec\) consistent? If so, find a solution.
Identify the pivot positions of \(A\text{.}\)
For our two choices of the vector \(\bvec\text{,}\) one equation \(A\xvec = \bvec\) has a solution and the other does not. What feature of the pivot positions of the matrix \(A\) tells us to expect this?

Solution.

We know there must be a pivot position in the rightmost column of the augmented matrix.
We construct the augmented matrix

\begin{equation*} \left[\begin{array}{rrr|r} 1 \amp 0 \amp -2 \amp 2 \\ -2 \amp 2 \amp 2 \amp 2 \\ 1 \amp 1 \amp -3 \amp 5 \end{array}\right] \sim \left[\begin{array}{rrr|r} 1 \amp 0 \amp -2 \amp 2 \\ 0 \amp 1 \amp -1 \amp 3 \\ 0 \amp 0 \amp 0 \amp 0 \end{array}\right]\text{,} \end{equation*}

which shows that the system is consistent. The solution space is described parametrically as

\begin{equation*} \xvec=\threevec{2}{3}{0}+x_3\threevec{2}{1}{1}\text{.} \end{equation*}
Now the augmented matrix is

\begin{equation*} \left[\begin{array}{rrr|r} 1 \amp 0 \amp -2 \amp 2 \\ -2 \amp 2 \amp 2 \amp 2 \\ 1 \amp 1 \amp -3 \amp 6 \end{array}\right] \sim \left[\begin{array}{rrr|r} 1 \amp 0 \amp -2 \amp 0 \\ 0 \amp 1 \amp -1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{array}\right] \end{equation*}

showing that the equation \(A\xvec=\bvec\) is inconsistent.
There are two pivot positions in \(A\text{,}\) as shown.

\begin{equation*} \left[\begin{array}{rrr} {\mathbf 1} \amp 0 \amp -2 \\ -2 \amp {\mathbf 2} \amp 2 \\ 1 \amp 1 \amp -3 \\ \end{array}\right] \sim \left[\begin{array}{rrr} {\mathbf 1} \amp 0 \amp -2 \\ 0 \amp {\mathbf 1} \amp -1 \\ 0 \amp 0 \amp 0 \\ \end{array}\right]\text{.} \end{equation*}
Since there is a row of \(A\) that does not have a pivot position, it is possible to augment \(A\) by a vector \(\bvec\) so that we obtain a pivot position in the rightmost column of the augmented matrix. In this case, we have an inconsistent system.

In this exploration, we considered a \(3\times3\) matrix \(A\) and found that the equation \(A\xvec = \bvec\) has a solution for some vectors \(\bvec\) in \(\real^3\) and has no solution for others. We will introduce a concept called span that describes the vectors \(\bvec\) for which there is a solution.

We can write an \(m\times n\) matrix \(A\) in terms of its columns

\begin{equation*} A = \left[\begin{array}{rrrr} \vvec_1\amp\vvec_2\amp\cdots\amp\vvec_n \end{array}\right]\text{.} \end{equation*}

Proposition 6.1.1.

If \(A=\left[\begin{array}{rrrr} \vvec_1\amp\vvec_2\amp\ldots\vvec_n \end{array}\right]\) and \(\xvec=\left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \\ \end{array}\right] \text{,}\) then the following statements are equivalent.

The vector \(\xvec\) satisfies the equation \(A\xvec = \bvec \text{.}\)
The vector \(\bvec\) is a linear combination of the columns of \(A\) with weights \(x_j\text{:}\)

\begin{equation*} x_1\vvec_1 + x_2\vvec_2 + \ldots + x_n\vvec_n = \bvec\text{.} \end{equation*}
The components of \(\xvec\) form a solution to the linear system corresponding to the augmented matrix

\begin{equation*} \left[\begin{array}{rrrr|r} \vvec_1 \amp \vvec_2 \amp \cdots \amp \vvec_n \amp \bvec \end{array}\right]\text{.} \end{equation*}

Therefore the equation \(A\xvec = \bvec\) is consistent if and only if we can express \(\bvec\) as a linear combination of \(\vvec_1,\vvec_2,\ldots,\vvec_n\text{.}\)

Definition 6.1.2.

The span of a set of vectors \(\vvec_1,\vvec_2,\ldots,\vvec_n\) is the set of all linear combinations that can be formed from the vectors.

Alternatively, if \(A = \begin{bmatrix} \vvec_1 \amp \vvec_2 \amp \cdots \amp \vvec_n \end{bmatrix}\text{,}\) then the span of the vectors consists of all vectors \(\bvec\) for which the equation \(A\xvec = \bvec\) is consistent.