Proving Spanning with Pivot and Coordinate Arguments

What a Spanning Proof Must Show

To prove that a list spans a space, the central task is to show how to produce an arbitrary vector in that space from the list. When the space is \(\mathbb{R}^m\), a matrix whose columns are the listed vectors turns that task into a linear system with an arbitrary right-hand side. Row reduction then gives a practical test: the columns span \(\mathbb{R}^m\) exactly when the reduced matrix has a pivot in every row. If the vectors lie in a subspace with a known basis, their coordinates turn the same question into a smaller system.

Throughout this tutorial the scalar field is \(\mathbb{R}\). A linear combination of a list \(\mathbf{v}_1,\dots,\mathbf{v}_p\) is a vector \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\), where each \(c_j\in\mathbb{R}\). The span of the list is the set of all such linear combinations.

Both requirements matter. The list must lie in the space it is said to span, and it must be sufficient to produce every vector in that space. A proof should therefore identify an arbitrary target \(\mathbf{w}\in W\) and find coefficients that give \(\mathbf{w}\) as a combination. When useful, those coefficients can depend on the coordinates of \(\mathbf{w}\).

The Pivot-Row Test for Spanning \(\mathbb{R}^m\)

Let \(\mathbf{A}=\begin{bmatrix}\mathbf{v}_1&\cdots&\mathbf{v}_p\end{bmatrix}\in\mathbb{R}^{m\times p}\), so its columns are the vectors in the list. For \(\mathbf{c}=(c_1,\dots,c_p)^{\top}\), the product \(\mathbf{A}\mathbf{c}\) equals \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\). Thus, the columns span \(\mathbb{R}^m\) precisely when the system \(\mathbf{A}\mathbf{c}=\mathbf{b}\) has a solution for every \(\mathbf{b}\in\mathbb{R}^m\).

Proof. Let \(\mathbf{R}\) be the RREF of \(\mathbf{A}\). Row reduction consists of invertible elementary row operations, so there is an invertible matrix \(\mathbf{E}\) such that \(\mathbf{R}=\mathbf{E}\mathbf{A}\). Applying those same operations to a right-hand side preserves whether the system is solvable: \(\mathbf{A}\mathbf{c}=\mathbf{b}\) is equivalent to \(\mathbf{R}\mathbf{c}=\mathbf{E}\mathbf{b}\). If \(\mathbf{R}\) has a pivot in every row, then each row contains a leading variable. Starting from the bottom pivot row and solving upward gives values for the pivot variables for every right-hand side \(\mathbf{E}\mathbf{b}\); any variables without pivots may be set to zero. Hence every \(\mathbf{b}\) has a solution, so the columns span \(\mathbb{R}^m\). Conversely, suppose row \(i\) of \(\mathbf{R}\) has no pivot. Because \(\mathbf{R}\) is in RREF, that row is all zeros. Let \(\mathbf{e}_i\in\mathbb{R}^m\) be the vector with entry \(1\) in position \(i\) and zeros elsewhere, and choose \(\mathbf{b}=\mathbf{E}^{-1}\mathbf{e}_i\). Then \(\mathbf{E}\mathbf{b}=\mathbf{e}_i\), so row \(i\) of \(\mathbf{R}\mathbf{c}=\mathbf{E}\mathbf{b}\) reads \(0=1\). The system is inconsistent for this \(\mathbf{b}\), so the columns do not span \(\mathbb{R}^m\). \(\square\)

The test concerns rows because the target vector has \(m\) coordinates, one equation for each row. It does not require a pivot in every column. A column without a pivot can be redundant while the columns still span the whole target space.

Proving Spanning with an Arbitrary Right-Hand Side

This calculation is a proof because \(x,y,z\) were arbitrary, not because one particular vector was successfully represented. The coefficients explicitly describe a way to construct every target from the given list.

Using Coordinates in a Known Subspace

Sometimes the target space is a subspace \(W\subseteq\mathbb{R}^n\), rather than the whole coordinate space. Suppose \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_d)\) is a basis of \(W\): it is a linearly independent list that spans \(W\). Each vector \(\mathbf{w}\in W\) has a unique coordinate vector \([\mathbf{w}]_{\mathcal{B}}\in\mathbb{R}^d\), recording its coefficients in this basis.

Proof. First, coordinates in \(\mathcal{B}\) are unique: if \(\sum_{i=1}^d a_i\mathbf{b}_i=\sum_{i=1}^d a'_i\mathbf{b}_i\), subtracting gives \(\sum_{i=1}^d(a_i-a'_i)\mathbf{b}_i=\mathbf{0}\). Independence of the basis forces \(a_i-a'_i=0\) for every \(i\). Now, for any scalars \(c_1,\dots,c_p\), the coordinate vector of \(\sum_{j=1}^p c_j\mathbf{v}_j\) is \(\mathbf{C}\mathbf{c}\), by collecting the coefficients of each \(\mathbf{b}_i\). If the list spans \(W\), then each \(\mathbf{w}\in W\) is such a combination, so every coordinate vector \([\mathbf{w}]_{\mathcal{B}}\in\mathbb{R}^d\) is of the form \(\mathbf{C}\mathbf{c}\). Conversely, suppose the columns of \(\mathbf{C}\) span \(\mathbb{R}^d\). For any \(\mathbf{w}\in W\), choose \(\mathbf{c}\) such that \(\mathbf{C}\mathbf{c}=[\mathbf{w}]_{\mathcal{B}}\). The corresponding combination \(\sum_{j=1}^p c_j\mathbf{v}_j\) has the same coordinates as \(\mathbf{w}\), so uniqueness of coordinates implies it equals \(\mathbf{w}\). Since every \(\mathbf{v}_j\) belongs to \(W\), these combinations are in \(W\), and the list spans \(W\). The pivot-row test applied to \(\mathbf{C}\) gives the final equivalence. \(\square\)

The membership condition \(\mathbf{v}_j\in W\) is essential: it ensures that the list actually consists of vectors in the space being spanned. Coordinates relative to \(\mathcal{B}\) are available only for vectors in \(W\).

When the Pivot Test Shows Spanning Fails

Common Pitfalls and a Reliable Proof Pattern

Wrong pivot condition: A common error is to conclude that a list cannot span because not every column has a pivot. For the matrix \(\begin{bmatrix}1&0&1\\0&1&1\end{bmatrix}\), the third column is not a pivot column, but both rows have pivots. As shown above, the columns do span \(\mathbb{R}^2\): \((x,y)^{\top}=x(1,0)^{\top}+y(0,1)^{\top}\). For spanning the full target \(\mathbb{R}^m\), check every row, not every column.

Wrong membership argument: It is not enough to observe that every vector in \(W\) is a combination of a list; the list must also consist of vectors in \(W\). For example, the zero vector is \(0(1,0)^{\top}\), but the list \(((1,0)^{\top})\) does not span \(W=\{\mathbf{0}\}\), because \((1,0)^{\top}\notin W\) and the list’s span is the entire horizontal axis. The definition requires both containment and production of every target.

For a complete spanning proof, a dependable sequence is:

The pivot test is a compact way to prove that the coefficient system is solvable for every target. When a row lacks a pivot, row reduction can also reveal a specific obstruction: an inconsistent right-hand side or a coordinate relation that every combination must satisfy. In a subspace, coordinates relative to a known basis provide the same kind of proof in the dimension of that subspace rather than in the ambient coordinate space.