Recovering Span Coefficients From a Reduced System

From a Target to Its Coefficients

When a target vector is in the span of given vectors, it can be expressed as a linear combination of them. The practical question is often more specific: what coefficients produce that target? There may be one coefficient vector, or there may be many. A reduced augmented system answers both questions, provided we keep track of which variables are coefficients and express every free variable in the answer.

Let \(\mathbf{v}_1,\dots,\mathbf{v}_n\) be columns in \(\mathbb{R}^m\), and let \(\mathbf{b}\in\mathbb{R}^m\) be a target. Put the given vectors into the columns of \(\mathbf{A}\), and collect their coefficients in the column vector \(\mathbf{c}\). By matrix multiplication, \(\mathbf{A}\mathbf{c}\) is exactly the linear combination of the columns with those coefficients. Thus recovering coefficients means solving \(\mathbf{A}\mathbf{c}=\mathbf{b}\).

To solve for the coefficients, row-reduce the augmented matrix \(\begin{bmatrix}\mathbf{A}&\mathbf{b}\end{bmatrix}\). The last column must stay attached throughout the reduction: it records the right-hand side of the same system. If the reduced system is consistent, its pivot equations express the pivot variables in terms of the free variables. Giving each free variable a parameter then describes all coefficient vectors, not just one.

Why Row Reduction Keeps the Coefficients

Row operations change the appearance of a system, but they do not change which coefficient vectors solve it. This is the reason a reduced augmented matrix can be used to recover coefficients for the original vectors.

Proof. A row swap exchanges two equations, so a coefficient vector satisfies the system before the swap exactly when it satisfies the system after the swap. Multiplying a row by a nonzero scalar \(k\) replaces an equation by \(k\) times that equation. Since \(k\ne0\), the new equation holds exactly when the original equation holds. Finally, adding \(k\) times one row to another replaces one equation by its sum with a multiple of another equation. Any solution of the original equations satisfies this new equation. Conversely, subtracting \(k\) times the unchanged row recovers the original equation, so any solution after the operation satisfies the original equations as well. Each elementary row operation therefore preserves the solution set; applying this fact one operation at a time proves the statement for a sequence of operations. \(\square\)

In a consistent reduced row echelon form (RREF), the coefficient columns have some pivot positions and may have nonpivot positions. A variable corresponding to a nonpivot coefficient column is free: it may be assigned any real value, and the pivot equations determine the pivot variables. The augmented column is not a coefficient variable. If it contains a pivot, the system is inconsistent, so there is no coefficient vector representing the target.

Proof. By the row-operation theorem, the RREF system has exactly the coefficient solutions of the original system. In each pivot row \(i\), the pivot entry in column \(p_i\) is \(1\), and entries in every other pivot column are \(0\). Its equation is therefore \(c_{p_i}+\sum_{j\in F}q_{ij}c_j=d_i\). Assigning \(c_j=t_j\) to each free variable and rearranging this equation gives \(c_{p_i}=d_i-\sum_{j\in F}q_{ij}t_j\). Every row below the pivot rows has zero entries in the coefficient columns. Consistency means its augmented entry is also zero, so these rows impose no further condition. Hence every choice of the free parameters gives a solution. Conversely, any solution must satisfy each pivot-row equation, so its pivot variables have exactly the displayed values for its chosen free variables. This gives all solutions, including the case \(F\) is empty, when the solution is unique. \(\square\)

Recovering a Unique Coefficient Vector

Recovering a Family of Representations

A nonpivot coefficient column does not mean that the target fails to be represented. It means that one or more coefficients can vary while the pivot equations adjust the other coefficients. To recover all representations, keep the free variables as parameters instead of selecting values for them too early.

The check also illustrates why a free variable must remain visible in the final answer. Setting \(t=0\) gives a valid representation, but it does not describe the other coefficient choices that produce the same target.

Common Pitfall: Treating One Choice as the Whole Answer

In the one-free-coefficient example, choosing \(c_3=0\) gives the correct coefficient vector \((3,4,0)^{\top}\). It would be wrong, however, to conclude that this is the only representation: the reduced equations do not require \(c_3=0\). For instance, choosing \(c_3=1\) gives \((2,3,1)^{\top}\), and \(2\mathbf{v}_1+3\mathbf{v}_2+\mathbf{v}_3=(3,4,7)^{\top}\) as well. The correction is to assign a parameter to each free variable and retain it in the pivot-variable equations.

There is a separate issue when the reduced augmented matrix is inconsistent. A row of the form \(\begin{bmatrix}0&\cdots&0&|&k\end{bmatrix}\), with \(k\ne0\), asserts \(0=k\), so no coefficient vector can represent the target. Do not read a parameterized answer from such a matrix: first check that no such row occurs. When the matrix is consistent, the pivot equations and free-variable parameters give every representation, and substitution into the original columns provides a direct check.