Worked Span-Membership Problems in R2

From Span Membership to a Coefficient System

In \(\mathbb{R}^2\), a span-membership question asks whether a specified target vector can be made from the given vectors using scalar coefficients. The target might be in the span with one set of coefficients, with many different sets, or not at all. Writing the vectors as columns turns these possibilities into familiar questions about solving a linear system.

Let \(\mathbf{v}_1,\dots,\mathbf{v}_p\in\mathbb{R}^2\), and let \(\mathbf{b}\in\mathbb{R}^2\) be the target. By definition, the span of the given vectors is the set of all their linear combinations. Form the matrix \(\mathbf{A}=\begin{bmatrix}\mathbf{v}_1&\cdots&\mathbf{v}_p\end{bmatrix}\), and let \(\mathbf{x}=(x_1,\dots,x_p)^{\top}\) be the coefficient vector. The matrix equation \(\mathbf{A}\mathbf{x}=\mathbf{b}\) records exactly whether the combination equals the target.

Proof. By the definition of matrix-vector multiplication, \(\mathbf{A}\mathbf{x}=x_1\mathbf{v}_1+\cdots+x_p\mathbf{v}_p\). If \(\mathbf{b}\) is in the span, then there are scalars \(x_1,\dots,x_p\) for which this sum equals \(\mathbf{b}\), so \(\mathbf{x}\) solves the system. Conversely, if \(\mathbf{A}\mathbf{x}=\mathbf{b}\), the same matrix-vector identity writes \(\mathbf{b}\) as a linear combination of the given vectors. The coefficients in the equation are precisely the coefficients in that representation. \(\square\)

Thus row reduction answers two related questions. First, does the target belong to the span? Second, if it does, what coefficient vectors represent it? The number of solutions concerns representations using this particular list of vectors; it is not a different definition of membership.

How to Read the Reduced System

To test membership, place the given vectors in the columns of \(\mathbf{A}\), append the target as the right-hand side, and row-reduce the augmented matrix \(\begin{bmatrix}\mathbf{A}&\mathbf{b}\end{bmatrix}\). Row operations preserve the system’s solutions: swapping two rows only changes equation order, multiplying an equation by a nonzero scalar gives an equivalent equation, and adding a multiple of one equation to another is reversible by subtracting that multiple. Therefore the reduced system has exactly the same coefficient solutions as the original system.

For two coefficient variables, the reduced equations have three possible outcomes. A row of the form \(0= d\), where \(d\ne0\), means that the system is inconsistent and the target is not in the span. If both variables have pivots and there is no inconsistent row, the coefficients are uniquely determined. If the system is consistent but at least one variable is free, that variable can take any real value; each choice determines the pivot variables, producing infinitely many coefficient solutions.

Proof. Row operations are reversible and so preserve the solution set. In a reduced system, an equation with all coefficient entries zero and a nonzero right-hand side cannot be satisfied, so an inconsistent row gives no solutions. If the system is consistent and every variable is a pivot variable, the pivot equations determine each variable in turn, with no free choices. They give one solution, and any solution must have those same variable values, so it is unique. If the system is consistent and there is a free variable, assign it any real value. The pivot equations then determine the pivot variables. Distinct choices of the free variable give distinct coefficient vectors, and there are infinitely many real choices. This proves all three conclusions. \(\square\)

One Target, One Representation

Here the coefficient list is unique, but that does not mean every target must have a representation. The next two examples use the same dependent pair of vectors to show how a target can have infinitely many representations or none.

A Dependent Pair: Many Representations or None

The last two examples have the same columns but different targets. The columns have a redundant coefficient direction: changing \(x_1\) by \(-2t\) while changing \(x_2\) by \(t\) leaves the combination unchanged. That freedom creates multiple representations when the system is consistent. It cannot repair a target that contradicts the equations, so the target in the inconsistent example has no representation.

Three Generators and a Free Coefficient

Common Pitfall: Confusing a Free Variable with Inconsistency

A free variable does not mean that the target is outside the span. In the three-generator example, \(x_3\) is free, but there is no contradictory row; each choice of \(x_3\) gives coefficients satisfying both equations. The correct conclusion is membership with infinitely many representations.

A concrete incorrect interpretation would be: “The reduced system has a free variable, so it has no solution.” The correction is to inspect the whole reduced system. A free variable indicates non-uniqueness only when the system is consistent. By contrast, the row \([0\quad 0\mid -1]\) in the previous example is inconsistent, regardless of any other rows, because it demands \(0=-1\).

Keep these questions in order when working a problem:

1. Is there a row with zero coefficient entries and a nonzero right-hand side? If so, the target is not in the span.
2. If there is no such row, is every coefficient variable a pivot variable? If so, the target has a unique representation using the given list.
3. If the system is consistent and at least one coefficient variable is free, what parameter describes all representations?

In every case, the coefficient matrix must use the given vectors as columns, because its matrix-vector product is the linear combination being tested. Once the coefficients are found, multiplying the original matrix by the coefficient vector is a direct check against the target.