Dependence Relations and Their Coefficients

What a Dependence Relation Records

A statement that a list of vectors is dependent says that some nontrivial combination of them gives the zero vector. A dependence relation makes that statement explicit: it records the coefficients, so we can check the equation and see which vectors participate. Its coefficients also tell us how to rearrange the equation to express one vector as a combination of the others.

Throughout, let \(V\) be a vector space over a field \(\mathbb{F}\), and consider an ordered list \(\mathbf{v}_1,\dots,\mathbf{v}_p\) in \(V\). The list is ordered because each coefficient belongs to a particular position. A coefficient list is the \(p\)-tuple \((c_1,\dots,c_p)\) of scalars in \(\mathbb{F}\). The list of vectors is linearly independent when the only coefficient list satisfying the equation below is the all-zero list; it is dependent when at least one other coefficient list satisfies it.

The distinction between a relation and a nontrivial relation matters: choosing every coefficient to be zero always gives an equation, but that equation reveals no dependence. Conversely, a single checked nontrivial relation is enough to prove dependence. You do not need to show that it is the only relation or that every coefficient is nonzero.

Using a Relation to Isolate a Vector

Proof. Suppose the displayed relation holds and \(c_k\ne0\). Move all terms except the \(k\)-th to the other side, then divide by \(c_k\), which is permitted because it is nonzero: \(\mathbf{v}_k=-\sum_{i\ne k}(c_i/c_k)\mathbf{v}_i\). This expresses \(\mathbf{v}_k\) using the other vectors. Conversely, suppose \(\mathbf{v}_k=\sum_{i\ne k}d_i\mathbf{v}_i\). Subtract \(\mathbf{v}_k\) from both sides to obtain \(\sum_{i\ne k}d_i\mathbf{v}_i-\mathbf{v}_k=\mathbf{0}\). This is a dependence relation whose coefficient on \(\mathbf{v}_k\) is \(-1\), so it is nontrivial. If \(p=1\), the same argument says that \(\mathbf{v}_1=\mathbf{0}\) gives the relation \(-\mathbf{v}_1=\mathbf{0}\), consistent with the empty-combination convention. \(\square\)

This theorem explains why the nonzero coefficients matter. To isolate a vector, choose an index \(k\) for which \(c_k\ne0\). A coefficient equal to zero cannot be divided by, and its vector does not appear in the expression obtained by rearranging that relation. A relation may therefore show that one vector is redundant without showing that every vector in the list is redundant.

Reading Coefficients in a Different Vector Space

The same method works in abstract vector spaces: check equality in the space being used, then use a nonzero coefficient to solve for its vector. For polynomials, equality means equality of every coefficient of each power of the variable. The vectors need not be columns in \(\mathbb{R}^n\).

Coefficient Lists Can Be Combined

For a fixed list of vectors, some coefficient lists give relations and others do not. The relation lists have a useful closure property: adding two of them, or multiplying one by a scalar, produces another relation list. The result follows directly from the vector-space operations.

Proof. The all-zero coefficient list belongs to the set because \(\sum_{i=1}^{p}0\mathbf{v}_i=\mathbf{0}\). Suppose \((c_1,\dots,c_p)\) and \((d_1,\dots,d_p)\) belong to it. Then \(\sum_{i=1}^{p}(c_i+d_i)\mathbf{v}_i=\sum_{i=1}^{p}c_i\mathbf{v}_i+\sum_{i=1}^{p}d_i\mathbf{v}_i=\mathbf{0}+\mathbf{0}=\mathbf{0}\), so their sum belongs to the set. For any scalar \(a\in\mathbb{F}\), if \((c_1,\dots,c_p)\) belongs to the set, then \(\sum_{i=1}^{p}(ac_i)\mathbf{v}_i=a\sum_{i=1}^{p}c_i\mathbf{v}_i=a\mathbf{0}=\mathbf{0}\). Thus the set contains zero and is closed under addition and scalar multiplication, which are the subspace conditions. \(\square\)

This closure property can produce more than one relation from a known one: every nonzero scalar multiple of a nontrivial relation remains nontrivial. For example, multiplying \((2,1,-1)\) by \(3\) gives \((6,3,-3)\), another witness for the vectors in the first example. Adding relations is also permitted, but the sum may be the zero coefficient list. Therefore addition alone does not guarantee a new nontrivial witness.

A Zero Coefficient Does Not Involve That Vector

Common Coefficient Errors

A frequent mistake is to find a plausible expression for one vector but move it to the wrong side when writing the relation. For the vectors in the first worked example, a mistaken calculation might claim \(\mathbf{v}_1+\mathbf{v}_2-\mathbf{v}_3=\mathbf{0}\). Substitution exposes the error:

The correct expression is \(\mathbf{v}_3=2\mathbf{v}_1+\mathbf{v}_2\), so moving all terms to the left gives \(2\mathbf{v}_1+\mathbf{v}_2-\mathbf{v}_3=\mathbf{0}\). Substitution gives \(\begin{bmatrix}2\\4\end{bmatrix}+\begin{bmatrix}3\\1\end{bmatrix}-\begin{bmatrix}5\\5\end{bmatrix}=\mathbf{0}\), verifying the corrected coefficients.

Another error is to treat a zero coefficient as though it helps isolate that vector. The isolation step requires division by a nonzero coefficient; if the coefficient is zero, no such division is valid. Always identify a nonzero coefficient before choosing which vector to express, and substitute the proposed relation back into the original vector equation. In \(\mathbb{R}^n\), this means checking every coordinate; for polynomials, it means checking every power that appears.