Membership in a Span Versus Spanning the Whole Space

One Target or Every Target?

Given a collection of vectors, two questions can sound similar: does their span contain a particular target, and do they span the entire ambient space? The first asks about one vector. The second asks whether every vector in that space can be formed from the collection. A successful answer to the first question does not, by itself, settle the second.

Let \(\mathbf{A}\in\mathbb{R}^{m\times n}\) have columns \(\mathbf{a}_1,\dots,\mathbf{a}_n\). The column space is the set of all linear combinations of those columns. A system is consistent if it has at least one solution. These definitions turn the two questions into tests about systems of equations.

For a single target \(\mathbf{b}\in\mathbb{R}^m\), membership means that some choice of coefficients produces exactly \(\mathbf{b}\). Collecting those coefficients into \(\mathbf{x}\in\mathbb{R}^n\) gives the equation \(\mathbf{A}\mathbf{x}=\mathbf{b}\). The right-hand side is important: it is part of the particular question being tested.

Proof. By definition, \(\mathbf{b}\in\operatorname{Col}\mathbf{A}\) means there are real numbers \(x_1,\dots,x_n\) such that \(\mathbf{b}=x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n\). The definition of matrix multiplication gives \(\mathbf{A}\mathbf{x}=x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n\), so such coefficients exist exactly when \(\mathbf{A}\mathbf{x}=\mathbf{b}\) has a solution. For the row-reduction test, row operations preserve the solution set of the augmented system, as established in the earlier material on augmented systems. A row \([0\ \cdots\ 0\mid c]\) with \(c\ne0\) represents the impossible equation \(0=c\), so it rules out a solution. If there is no such row, every nonzero row gives an equation with a leading variable. Assign zero to each free variable, then solve the pivot equations from the last nonzero row upward. Each pivot variable is determined by an equation whose right-hand side is a real number, so this produces a solution. Thus the reduced augmented matrix has a solution exactly when it has no contradictory row. \(\square\)

Testing One Target by Row Reduction

A useful discipline is to keep the target attached to the coefficient matrix throughout row reduction. The question is not whether a reduced matrix has a column equal to the target. It is whether the augmented system, with the target on its right-hand side, is consistent.

When Do the Columns Span the Whole Space?

To span \(\mathbb{R}^m\), the columns must be able to produce every right-hand side in \(\mathbb{R}^m\). That is a universal requirement, not a test of one selected target. There is a finite set of targets that captures this requirement: the standard basis vectors \(\mathbf{e}_1,\dots,\mathbf{e}_m\), where \(\mathbf{e}_i\) has a \(1\) in position \(i\) and zeros elsewhere.

Proof. Suppose first that the columns span \(\mathbb{R}^m\). Each standard basis vector \(\mathbf{e}_i\) is in \(\mathbb{R}^m\), so it is a linear combination of the columns. The single-target membership test then says \(\mathbf{A}\mathbf{x}=\mathbf{e}_i\) is consistent for every \(i\). Conversely, suppose each of these systems is consistent. For every \(i\), choose a vector \(\mathbf{x}_i\) such that \(\mathbf{A}\mathbf{x}_i=\mathbf{e}_i\). Any \(\mathbf{b}\in\mathbb{R}^m\) has the standard-coordinate expression \(\mathbf{b}=b_1\mathbf{e}_1+\cdots+b_m\mathbf{e}_m\). Therefore, by linearity of matrix multiplication, \(\mathbf{A}(b_1\mathbf{x}_1+\cdots+b_m\mathbf{x}_m)=b_1\mathbf{e}_1+\cdots+b_m\mathbf{e}_m=\mathbf{b}\). Thus every \(\mathbf{b}\in\mathbb{R}^m\) belongs to \(\operatorname{Col}\mathbf{A}\), which means the columns span \(\mathbb{R}^m\). \(\square\)

This is the consistency form of the every-right-hand-side criterion: spanning the whole codomain means that \(\mathbf{A}\mathbf{x}=\mathbf{b}\) is consistent for every \(\mathbf{b}\in\mathbb{R}^m\). The standard-basis test replaces infinitely many possible targets with \(m\) particular systems. The criterion concerns consistency, not uniqueness: it does not require each target to have only one coefficient vector.

Several Targets and Separate Coefficients

Sometimes the task is to test several targets against the same column space. Put the targets into a matrix \(\mathbf{B}=[\mathbf{b}_1\ \cdots\ \mathbf{b}_k]\). Then \(\mathbf{A}\mathbf{X}=\mathbf{B}\) asks for one coefficient vector for each target: the \(j\)-th column of \(\mathbf{X}\) supplies the coefficients used to form \(\mathbf{b}_j\). The coefficients for one target need not equal the coefficients for another.

Common Pitfall: Row-Reducing Only the Columns

Row operations preserve solutions only when they are applied to the entire augmented matrix. They do not preserve the original column space as a subset of the original coordinate space. For example, let \(\mathbf{A}=\begin{bmatrix}1\\0\end{bmatrix}\). Its column space consists of vectors \((t,0)\), so \(\mathbf{b}=(0,1)\) is not in that space.

A wrong calculation swaps the two rows of \(\mathbf{A}\) to obtain \(\begin{bmatrix}0\\1\end{bmatrix}\), then concludes that \(\mathbf{b}\) is a column of the new matrix and so belongs to the original column space. This conclusion confuses the transformed columns with the original ones. The correct test row-reduces the augmented matrix:

The first row of the final augmented matrix represents \(0=1\), so the system is inconsistent. The target is not in the original column space. The correct rule is to place the target in the augmented column before row reduction, then use consistency to answer the membership question.

The distinction is useful whenever a calculation gives a particular solution. A solution for one target certifies membership of that target only. To conclude that the columns span the whole space, the test must cover every right-hand side, or equivalently the standard basis targets. Keeping those quantifiers separate prevents a single successful system from being mistaken for a general spanning result.