Testing Spans With Parametrized Vectors

Span Membership Can Change With a Parameter

A span question asks whether a target vector can be written as a linear combination of given vectors. If the entries of those vectors depend on a real parameter \(t\), then the coefficients in that combination must satisfy a system whose entries also depend on \(t\). The system may be consistent for some parameter values and inconsistent for others. The reliable approach is to translate membership into a coefficient system, row-reduce while keeping the parameter visible, and treat separately any value that makes a potential pivot zero.

Let \(\mathbf{v}_1(t),\dots,\mathbf{v}_p(t)\) be vectors in \(\mathbb{R}^m\), and let \(\mathbf{b}\in\mathbb{R}^m\) be a target. For each fixed real value of \(t\), membership means that there are real coefficients \(c_1,\dots,c_p\) such that \(c_1\mathbf{v}_1(t)+\cdots+c_p\mathbf{v}_p(t)=\mathbf{b}\). Placing the generating vectors as columns gives the coefficient system \(\mathbf{A}(t)\mathbf{c}=\mathbf{b}\). Thus the question is whether this system is consistent at each value of \(t\). This is the column-space membership test: the target belongs to the span of the columns exactly when the coefficient system has a solution.

A Consistency Test That Keeps the Parameter Visible

Row reduction can reveal a condition on \(t\), but parameter-dependent entries require care. For each fixed \(t\), the usual row operations preserve the solutions of the augmented system. If a row becomes \([\,0\ \cdots\ 0\mid q(t)\,]\), it represents the equation \(0=q(t)\). That row makes the system inconsistent precisely when \(q(t)\ne0\). If \(q(t)=0\), it imposes no restriction. This is why a parameter value that makes a pivot zero must be checked rather than divided through.

Proof. Fix \(t\). Every row operation is reversible: swapping rows is undone by swapping them back, multiplying a row by a nonzero scalar is undone by multiplying by its reciprocal, and adding a multiple of one row to another is undone by subtracting that multiple. Therefore the row-reduced augmented matrix represents a system with exactly the same solutions as the original system. A row with zero coefficients and a nonzero augmented entry states \(0=q\) for some \(q\ne0\), which no coefficient vector can satisfy. So if such a row appears, the original system is inconsistent. Conversely, suppose no such row appears. Every row with zero coefficient entries then has augmented entry zero and gives the equation \(0=0\). In the remaining rows, assign zero to every free variable. Starting from the last nonzero row and working upward, each pivot variable can be assigned the value that makes its equation true, because its pivot coefficient is nonzero and any variables to its right have already been assigned. This produces a solution to the reduced system and hence to the original one. \(\square\)

The theorem applies separately at each fixed value of \(t\). A row operation that divides by an expression involving \(t\) is valid only when that expression is nonzero. If it may equal zero, split the work into the nonzero case and the zero case. The examples below show how this changes the conclusion about span membership.

Two Generators in \(\mathbb{R}^2\)

This example illustrates a useful general pattern: when a parameter-dependent pivot is nonzero, solve in that case; when it is zero, return to the system before division and test the resulting row. A formula obtained by dividing by \(1-t\) cannot be applied at \(t=1\).

Proof. The coefficient equations are \(c_1+c_2=r\) and \(tc_1+c_2=s\). If \(t\ne1\), subtracting the first equation from the second gives \((t-1)c_1=s-r\). Set \(c_1=(s-r)/(t-1)\) and \(c_2=r-c_1\); these values satisfy the first equation by construction, and the subtracted equation ensures they satisfy the second. Thus every target is in the span when \(t\ne1\). If \(t=1\), both equations have left side \(c_1+c_2\), so they can both hold only if \(r=s\). If \(r=s\), choosing \(c_1=r\) and \(c_2=0\) satisfies both equations. Therefore at \(t=1\) membership holds exactly when \(r=s\), proving both directions. \(\square\)

The theorem describes all targets for this particular family of generators. Away from \(t=1\), the system can produce every pair \((r,s)\). At \(t=1\), the two generators coincide, and only targets with equal coordinates can be produced.

Three Coordinates Can Impose an Extra Condition

Here the first two coordinates determine the coefficients, leaving the third coordinate as a compatibility condition. In a parameterized span problem, such a condition may select one value, several values, every value, or no value. The reduced system tells us which occurs; it is not safe to infer the answer from a visual pattern in the vectors alone.

When a Coefficient Becomes a Constraint

Notice that no division by \(t\) was needed. The row operation is valid for every fixed real \(t\), including \(t=0\), because it adds a multiple of one row to another. The final condition \(t^2=0\) then handles both the zero and nonzero cases directly.

Common Pitfall: Dividing Before Checking the Exceptional Case

In the first worked example, the reduced equation is \((1-t)c_2=3-2t\). A tempting but incorrect step is to divide by \(1-t\) and conclude that \(c_2=(3-2t)/(1-t)\) works for every \(t\). The denominator is zero at \(t=1\), so that formula does not provide a solution there. Substituting \(t=1\) into the equation before division gives \(0\cdot c_2=1\), which has no solution. The correction is to state the two cases explicitly: divide when \(t\ne1\), and test the original equation when \(t=1\).

A parameter may appear in a pivot, in a consistency row, or in both. Keep track of the equation represented by each row, and do not treat a parameter expression as nonzero until its zero values have been checked. Once the consistency cases are established, recover coefficients from the remaining equations and substitute them into the original vector combination. That final check confirms membership, not just the arithmetic of the row reduction.