Homogeneous Systems as Independence Tests

From a Vector List to a Homogeneous System

Linear independence asks whether a list of vectors has any nontrivial linear combination equal to the zero vector. A homogeneous system asks whether a matrix equation with zero on the right has any solution besides the zero vector. These are two ways of asking the same question once the vectors are placed in the columns of a matrix. The coefficient vector records which linear combination is being tested.

Work over \(\mathbb{R}\). Let \(\mathbf{v}_1,\dots,\mathbf{v}_p\) be a list of vectors in \(\mathbb{R}^n\), with \(p\ge 1\). A list is ordered: its positions matter, even if two entries happen to be the same vector. Form a matrix with those vectors as columns:

A coefficient vector \(\mathbf{x}\) for this list has \(p\) entries, one for each column: \(\mathbf{x}=(c_1,\dots,c_p)^{\top}\). The product \(\mathbf{A}\mathbf{x}\) is therefore a vector in \(\mathbb{R}^n\). It combines the columns using precisely those coefficients.

The Column Combination Is the Matrix Product

The key translation follows directly from the definition of matrix multiplication. If the entries of \(\mathbf{v}_j\) are \(v_{ij}\), then the entry in row \(i\) of \(\mathbf{A}\mathbf{x}\) is \(\sum_{j=1}^{p}v_{ij}c_j\). That is also the \(i\)th entry of \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\). This holds in every row, so the two vectors are equal.

Proof. Write the entries of column \(j\) as \(v_{1j},\dots,v_{nj}\). By the definition of matrix-vector multiplication, the \(i\)th entry of \(\mathbf{A}\mathbf{x}\) is \(v_{i1}c_1+\cdots+v_{ip}c_p\). The \(i\)th entry of \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\) is the same sum. Since this is true for each \(i=1,\dots,n\), the vectors are equal. \(\square\)

For example, if there are three columns, the coefficient vector has three entries. The equation \(\mathbf{A}\mathbf{x}=\mathbf{0}\) then says that a particular combination of the three columns is zero. It does not say that the columns themselves are zero, nor does it ask whether a coefficient vector with \(n\) entries works: the number of coefficients is set by the number of columns.

The Independence Test

Proof. Suppose first that the list is linearly independent. Let \(\mathbf{x}=(c_1,\dots,c_p)^{\top}\) be any solution of \(\mathbf{A}\mathbf{x}=\mathbf{0}\). By the proposition, \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\). Independence means that the only coefficients giving this zero combination are \(c_1=\cdots=c_p=0\). Hence \(\mathbf{x}=\mathbf{0}\), and every solution is trivial.

Conversely, suppose that \(\mathbf{A}\mathbf{x}=\mathbf{0}\) has only the trivial solution. Consider any coefficients \(c_1,\dots,c_p\) such that \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\), and set \(\mathbf{x}=(c_1,\dots,c_p)^{\top}\). The proposition gives \(\mathbf{A}\mathbf{x}=\mathbf{0}\). By the assumption, \(\mathbf{x}=\mathbf{0}\), so every \(c_j=0\). This is exactly the definition of independence. Thus the list is independent if and only if the homogeneous system has only its trivial solution. \(\square\)

The two directions explain how to use the test. To establish independence, show that every solution has all coefficients zero. To establish dependence, find even one solution with at least one nonzero coefficient. The zero solution is always present, because \(\mathbf{A}\mathbf{0}=\mathbf{0}\); finding that solution alone gives no information about independence.

Worked Examples

A Common Mistake: Counting Equations Instead of Solving Them

A homogeneous system always has the trivial solution, but that does not mean it has only the trivial solution. A tempting but incorrect argument is: “The right-hand side is zero, so the only solution is the zero coefficient vector.” The first worked example disproves that claim: its right-hand side is zero, yet \((-2,1)^{\top}\) is another solution.

Another error is to conclude that a system has only the trivial solution merely because it has at least as many equations as unknowns. The number of equations alone does not settle the question. For instance, in the third example there are two equations but three unknowns, and the equations allow a nontrivial solution. In general, the test is about the actual solutions of \(\mathbf{A}\mathbf{x}=\mathbf{0}\), not just the dimensions of the displayed system.

Keep the roles of the vectors and coefficients distinct. The columns are the listed vectors in \(\mathbb{R}^n\); the unknowns form a coefficient vector in \(\mathbb{R}^p\). A nontrivial solution is a coefficient vector with at least one nonzero entry. The independence test applies to the list of columns because each such solution is exactly a nontrivial zero combination of that list.

Why This Translation Is Useful

The definition of independence quantifies over every possible choice of coefficients. The homogeneous-system formulation packages all those choices into one equation. It also makes the two possible conclusions precise: only the trivial solution means independence; any nontrivial solution supplies a dependence relation. Once a solution is found, multiplying \(\mathbf{A}\mathbf{x}\) back out is a direct check that the proposed coefficients really give zero.

For a small system, the equations may be solved by substitution, as in the examples. For larger systems, a systematic solution method is useful, but the underlying test remains the same: determine whether the homogeneous system has any solution other than zero. The matrix equation and the coefficient relation are equivalent, not competing tests.