Equivalent Tests for Independent Columns

Four Ways to Express One Condition

When a vector is written as a linear combination of a fixed list, the coefficients describe how that vector is assembled from the list. If two different coefficient lists produce the same vector, then the columns contain redundancy. This tutorial connects that idea to three familiar tests: the zero-combination definition of independence, the solutions of a homogeneous system, and the pivot columns of a matrix. The connection also explains exactly when a vector in a span has unique coefficients.

Let \(\mathbf{v}_1,\dots,\mathbf{v}_p\) be a list of vectors in \(\mathbb{R}^m\), with \(p\geq 1\), and form the matrix \(\mathbf{A}=\begin{bmatrix}\mathbf{v}_1&\cdots&\mathbf{v}_p\end{bmatrix}\). For a coefficient column \(\mathbf{x}=(x_1,\dots,x_p)^{\top}\), matrix multiplication gives \(\mathbf{A}\mathbf{x}=x_1\mathbf{v}_1+\cdots+x_p\mathbf{v}_p\). Thus, each coefficient column determines a combination, and each combination of the list is obtained from some coefficient column.

The word “exactly” in unique coefficients includes two parts: a vector in the span has at least one representation, by the definition of span, and it has no second, different representation. The homogeneous system \(\mathbf{A}\mathbf{x}=\mathbf{0}\) tests the second part. If a nonzero coefficient column produces \(\mathbf{0}\), it can be added to any representation without changing the represented vector.

The Equivalence Theorem

Proof. By the definition of matrix multiplication, \(\mathbf{A}\mathbf{x}=\mathbf{0}\) says precisely that \(x_1\mathbf{v}_1+\cdots+x_p\mathbf{v}_p=\mathbf{0}\). Therefore statements 1 and 2 are equivalent: both say that the only coefficient column producing the zero vector is the zero column.

To compare statements 2 and 3, row-reduce \(\mathbf{A}\) to \(\operatorname{rref}(\mathbf{A})\). Each elementary row operation replaces a system by an equivalent system: a row swap just reorders equations, scaling a row by a nonzero number preserves its equation, and adding a multiple of one row to another preserves the solutions because the operation can be reversed. Consequently, the homogeneous systems with coefficient matrices \(\mathbf{A}\) and \(\operatorname{rref}(\mathbf{A})\) have the same solutions.

In the RREF, a variable corresponding to a column without a pivot is free. If every column is a pivot column, there are no free variables. Each variable is then a pivot variable, and the homogeneous equations force every one to be zero. So statement 3 implies statement 2. Conversely, suppose there is a column without a pivot. Assign its free variable the value \(1\), assign every other free variable the value \(0\), and solve the pivot equations for the pivot variables. The resulting coefficient column solves the homogeneous system and is nonzero, because its chosen free variable is \(1\). Statement 2 is then false. Thus statements 2 and 3 are equivalent.

It remains to compare statements 2 and 4. Suppose statement 2 holds, and let \(\mathbf{b}\) be any vector in the span. By the definition of span, there is a coefficient column \(\mathbf{x}\) such that \(\mathbf{A}\mathbf{x}=\mathbf{b}\). If another coefficient column \(\mathbf{y}\) also represents \(\mathbf{b}\), then \(\mathbf{A}\mathbf{x}=\mathbf{A}\mathbf{y}\), so \(\mathbf{A}(\mathbf{x}-\mathbf{y})=\mathbf{0}\). Statement 2 forces \(\mathbf{x}-\mathbf{y}=\mathbf{0}\), hence \(\mathbf{x}=\mathbf{y}\). Each span member therefore has unique coefficients, proving statement 4.

For the reverse implication, suppose statement 4 holds. The zero vector belongs to the span and has the representation \(\mathbf{A}\mathbf{0}=\mathbf{0}\). By uniqueness, no other coefficient column can produce the zero vector. Thus the homogeneous system has only the zero solution, which is statement 2. All four statements are equivalent. \(\square\)

This proof uses the RREF only to decide whether the homogeneous system has a nonzero solution. A pivot in every column means there are no free variables; a missing pivot means at least one free variable can be chosen nonzero. The pivot test and the coefficient-uniqueness test are not separate conditions to check after independence: they are different descriptions of the same property.

Worked Examples: Reading the Tests Together

A Useful Consequence and a Common Mistake

Proof. Since \(\mathbf{b}\in\operatorname{Col}\mathbf{A}\), it is a linear combination of the columns, so at least one solution \(\mathbf{x}\) exists. If \(\mathbf{y}\) is another solution, then \(\mathbf{A}\mathbf{x}=\mathbf{b}=\mathbf{A}\mathbf{y}\), giving \(\mathbf{A}(\mathbf{x}-\mathbf{y})=\mathbf{0}\). Independence means the only solution of this homogeneous system is \(\mathbf{0}\). Hence \(\mathbf{x}-\mathbf{y}=\mathbf{0}\), so \(\mathbf{x}=\mathbf{y}\). \(\square\)

Common mistake: “Independent columns mean \(\mathbf{A}\mathbf{x}=\mathbf{b}\) has a unique solution for every \(\mathbf{b}\).” The mistake is to infer existence from uniqueness. For the matrix in the third example, the columns are independent, but solving for \(\mathbf{b}=(1,2,3)^{\top}\) gives \(x_1=1\), \(x_2=2\), and \(0=3\). There is no solution to be unique. The correction is to separate the questions: independence ensures that there is at most one solution for any right-hand side; there is exactly one when the right-hand side also belongs to the column space.

These tests are useful in different forms. The zero-combination definition is direct when a dependence relation is easy to find. Row reduction gives a systematic pivot test. The homogeneous system identifies whether any nonzero coefficient change leaves the represented vector unchanged. Finally, for a vector already known to be in the span, independence tells us that its coefficient list is determined without ambiguity.