Testing Whether a Vector Lies in a Span

Membership in a Span Becomes a Linear System

A span is the set of all linear combinations of a given list of vectors. To ask whether a vector belongs to that span is therefore to ask whether some choice of coefficients produces it. The coefficients are unknowns, and matching the coordinates of the resulting vector gives a linear system. Row reduction provides a systematic way to find those coefficients—or to prove that no choice can work.

Let \(\mathbf{v}_1,\dots,\mathbf{v}_p\) and \(\mathbf{b}\) be vectors in \(\mathbb{R}^m\), where \(p\geq 1\). The vectors are columns. Put the proposed spanning vectors into the columns of a matrix \(\mathbf{A}\), and collect the unknown coefficients into a column \(\mathbf{c}\). By matrix multiplication, the entries of \(\mathbf{A}\mathbf{c}\) are exactly the coordinates of the corresponding linear combination.

This turns the question into a familiar one: is the linear system \(\mathbf{A}\mathbf{c}=\mathbf{b}\) consistent? If it is, its solutions are precisely the coefficient lists that represent \(\mathbf{b}\). If it is not, the vector is outside the span. This equivalence does not require the spanning list to be linearly independent or to have any particular number of vectors.

Proof. Suppose first that \(\mathbf{b}\) belongs to the span. By the definition of span, there are scalars \(c_1,\dots,c_p\) for which \(\mathbf{b}=c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\). The definition of matrix multiplication gives \(\mathbf{A}\mathbf{c}=c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\), so these coefficients solve \(\mathbf{A}\mathbf{c}=\mathbf{b}\). Conversely, suppose \(\mathbf{A}\mathbf{c}=\mathbf{b}\) has a solution \(\mathbf{c}=(c_1,\dots,c_p)^{\top}\). Expanding the matrix product gives \(\mathbf{b}=c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\), which is a linear combination of the listed vectors. Thus \(\mathbf{b}\) belongs to their span. \(\square\)

Row Reducing the Augmented Matrix

To solve the system, append \(\mathbf{b}\) as a final column and row reduce the augmented matrix \([\mathbf{A}\mid\mathbf{b}]\). The final column matters: it records the right-hand sides as the equations are transformed. Row operations preserve the solutions for the unknown coefficients because each operation replaces equations by equivalent equations. In particular, a zero row in the coefficient part with a nonzero entry in the final column says \(0=d\) for some \(d\ne0\), which is impossible.

Proof. Each elementary row operation is reversible: a row swap reverses itself, scaling a row by a nonzero number can be reversed by scaling by its reciprocal, and adding a multiple of one row to another can be reversed by subtracting that multiple. Applying such an operation to an augmented matrix makes the same change to both sides of the corresponding equations. Therefore, the original system and the row-reduced system have exactly the same coefficient solutions. A row with all zero coefficient entries and a nonzero final entry represents the equation \(0=d\), where \(d\ne0\); it has no solution. If there is no such row, every nonzero row has a pivot among the coefficient columns. Assign \(0\) to every free variable. Starting with the last pivot row and working upward, determine each pivot variable from its row equation; this is possible because the pivot coefficient is \(1\), and variables in later pivot positions have already been determined. The resulting values satisfy every nonzero row, and zero rows impose no condition. Hence a solution exists exactly when no contradictory row appears. \(\square\)

When the system is consistent, reduced form also describes all the coefficient choices. A variable in a coefficient column without a pivot is free: assign it a parameter, then read the pivot variables from the equations. If every coefficient column has a pivot, the equations determine all coefficients. Either way, the question of membership is answered by consistency; the number of coefficient lists is a separate issue.

Finding Coefficients for a Vector in the Span

One Vector Can Have Several Representations

A consistent system need not determine a single coefficient list. If some coefficient columns have no pivots, the associated free variables can produce a family of solutions. This still establishes membership: every one of those solutions gives a linear combination equal to the target vector.

Using Coordinates in a Polynomial Space

The method also applies when the vectors are polynomials, provided their coefficients are written in a consistent order. For polynomials of degree at most \(2\), use the coefficient column \((a_0,a_1,a_2)^{\top}\) for \(a_0+a_1x+a_2x^2\). Equating these coordinate columns is exactly the same as equating polynomial coefficients. This is a way to express the given span question as a linear system; it does not depend on the objects being written initially as coordinate tuples.

A Common Mistake: Matching Only Some Coordinates

A representation must match every coordinate, not just the coordinates that first determine the coefficients. For the nonmembership example above, one might correctly obtain \(c_1=1\) and \(c_2=1\) from the first two coordinates and then incorrectly conclude that the target is in the span. But substituting those coefficients gives \((1,1,2)^{\top}\), not \((1,1,3)^{\top}\). The final coordinate is not optional; in row reduction, the missed discrepancy appears as the contradictory row \([0\ \ 0\mid 1]\).

Another error is to put the proposed spanning vectors into rows rather than columns. The unknowns are the coefficients attached to each vector, so the product \(\mathbf{A}\mathbf{c}\) has the desired form only when the vectors are the columns of \(\mathbf{A}\). The augmented matrix must then use the target vector as its final column.

When using the method, keep the distinction between the question and its answer explicit: row reduction tests whether a representation exists; the coefficient values in a consistent system give the representation; and substituting them back checks the result. If an inconsistent row appears, it is a proof that no representation exists, not merely a failure to find one.