Encoding Span Questions With Coefficient Vectors

From a Linear Combination to a Coefficient Vector

The span of a list of vectors consists of all the linear combinations that can be formed from that list. To decide whether a particular vector belongs to the span, we need to determine whether some choice of coefficients produces it. A coefficient vector collects those scalar choices into a single column, so the question can be written as a matrix equation.

Work over \(\mathbb{R}\). Let \(\mathbf{v}_1,\dots,\mathbf{v}_p\) be vectors in \(\mathbb{R}^m\), where \(p\) and \(m\) are positive integers. A linear combination of these vectors has the form \(x_1\mathbf{v}_1+\cdots+x_p\mathbf{v}_p\), where each \(x_j\) is a real scalar. The list of coefficients is recorded as the column vector \(\mathbf{x}=(x_1,\dots,x_p)^{\top}\). In particular, \(\mathbf{x}\) has \(p\) entries, one for each vector in the list.

Form a matrix \(\mathbf{A}\) by placing the given vectors, in their listed order, as columns. Then \(\mathbf{A}\) has \(m\) rows and \(p\) columns. The target vector \(\mathbf{b}\), if there is one, belongs to \(\mathbb{R}^m\); the coefficient vector \(\mathbf{x}\) belongs to \(\mathbb{R}^p\). These vectors play different roles, and their lengths need not be equal.

The Matrix Equation for a Linear Combination

Proof. By the definition of matrix-vector multiplication, the product of a matrix by a column vector is the sum of the matrix's columns multiplied by the corresponding entries of that vector. The columns of \(\mathbf{A}\) are \(\mathbf{v}_1,\dots,\mathbf{v}_p\), and the corresponding entries of \(\mathbf{x}\) are \(x_1,\dots,x_p\). Therefore \(\mathbf{A}\mathbf{x}=x_1\mathbf{v}_1+\cdots+x_p\mathbf{v}_p\), as required. \(\square\)

This identity explains why a single matrix records all combinations of the list: changing \(\mathbf{x}\) changes the weights, while the columns of \(\mathbf{A}\) remain the vectors being combined. For example, setting \(x_1=1\) and all other entries to \(0\) produces \(\mathbf{v}_1\); choosing different coefficients produces other combinations.

Proof. Suppose first that \(\mathbf{b}\in\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_p\}\). By the definition of span, there are real numbers \(x_1,\dots,x_p\) such that \(\mathbf{b}=x_1\mathbf{v}_1+\cdots+x_p\mathbf{v}_p\). Put those numbers into \(\mathbf{x}=(x_1,\dots,x_p)^{\top}\). The first theorem gives \(\mathbf{A}\mathbf{x}=x_1\mathbf{v}_1+\cdots+x_p\mathbf{v}_p=\mathbf{b}\), so the equation has a solution. Conversely, suppose \(\mathbf{A}\mathbf{x}=\mathbf{b}\) for some \(\mathbf{x}=(x_1,\dots,x_p)^{\top}\in\mathbb{R}^p\). By the first theorem, \(\mathbf{b}=\mathbf{A}\mathbf{x}=x_1\mathbf{v}_1+\cdots+x_p\mathbf{v}_p\). This is a linear combination of the given vectors, so \(\mathbf{b}\) belongs to their span. Both implications hold, proving the equivalence. \(\square\)

The practical translation is direct: to ask whether \(\mathbf{b}\) is in the span, put the spanning vectors into the columns of \(\mathbf{A}\) and ask whether \(\mathbf{A}\mathbf{x}=\mathbf{b}\) is solvable. The unknowns are the coefficients, not the entries of the target vector. The next examples show how this translation works in different situations.

Worked Examples: Encoding and Solving

What the Solution Set Tells You

A solution of \(\mathbf{A}\mathbf{x}=\mathbf{b}\) is a coefficient vector that expresses \(\mathbf{b}\) using the columns of \(\mathbf{A}\). If the equation is inconsistent, no such expression exists. If there is exactly one solution, there is exactly one coefficient vector for this ordered list that produces \(\mathbf{b}\). If there are multiple solutions, distinct coefficient choices produce the same target. The equivalence proved above connects each of these outcomes to span membership: membership requires at least one solution, not necessarily exactly one.

The distinction between the target and the coefficients is especially important when \(\mathbf{A}\) is not square. If there are \(p\) given vectors in \(\mathbb{R}^m\), then \(\mathbf{x}\) has \(p\) entries while \(\mathbf{b}\) has \(m\) entries. The matrix product \(\mathbf{A}\mathbf{x}\) is defined because \(\mathbf{A}\) is \(m\times p\) and \(\mathbf{x}\) is \(p\times1\); the result has \(m\) entries, as required to compare it with \(\mathbf{b}\). Checking these dimensions before calculating prevents a common setup error.

The order of the columns matters for reading the coefficients. If the list is \(\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\), then \(x_1\) multiplies the first column, \(x_2\) the second, and \(x_3\) the third. Changing the column order without changing the coefficient order changes the combination. A coefficient vector is meaningful only in relation to the chosen ordered list.

Common Mistake: Confusing the Coefficients with the Combination

A coefficient vector is not usually the vector produced by combining the spanning vectors. For example, take \(\mathbf{v}_1=(1,2)^{\top}\), \(\mathbf{v}_2=(3,1)^{\top}\), and coefficients \(\mathbf{x}=(2,-1)^{\top}\). A tempting but incorrect calculation is to report the combination as \((2,-1)^{\top}\), simply copying the coefficients. Instead, form the columns and multiply:

The correct combination is \(2\mathbf{v}_1-\mathbf{v}_2=(2,4)^{\top}-(3,1)^{\top}=(-1,3)^{\top}\). The entries of \(\mathbf{x}\) are weights; the entries of \(\mathbf{A}\mathbf{x}\) are coordinates of the resulting vector in \(\mathbb{R}^2\). Checking the product by adding the weighted columns catches this mistake.

When setting up a span question, use the same sequence each time: put the given vectors into the columns, put unknown coefficients into a column vector, and set the matrix product equal to the target. Then solve the resulting equation and check any proposed coefficient vector by multiplying it back.