Dependence Means One Vector Comes From the Others

From a Dependence Relation to Redundancy

A dependence relation involves all the positions in a list, but it may have zero coefficients at some positions. The key is to select a position whose coefficient is nonzero: that coefficient can be divided out, leaving an expression for the vector at that position in terms of the others. Conversely, if one vector is already a linear combination of the rest, moving all terms to one side produces a dependence relation.

We work over \(\mathbb{R}\), with vectors in \(\mathbb{R}^n\) written as columns. A list \((\mathbf{v}_1,\dots,\mathbf{v}_p)\) is linearly dependent if there are scalars \(c_1,\dots,c_p\), not all zero, such that \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\). A linear combination of a list is a sum of its vectors multiplied by real scalars. In this tutorial, “the other vectors” means all the other positions in the list: repeated vectors, if present, still occupy separate positions.

For a list with one vector, dependence means \(c_1\mathbf{v}_1=\mathbf{0}\) for some \(c_1\ne0\). Dividing by \(c_1\) gives \(\mathbf{v}_1=\mathbf{0}\), which is exactly the combination of the empty list. For a list with at least two vectors, the same idea works by isolating one term from a nontrivial relation.

Proof. Suppose first that the list is dependent. Then for some scalars \(c_1,\dots,c_p\), not all zero,

At least one coefficient is nonzero; choose an index \(k\) with \(c_k\ne0\). Separate that term and divide by \(c_k\):

Thus \(\mathbf{v}_k\) is a linear combination of the remaining vectors. For the reverse direction, suppose that, for some index \(k\), \(\mathbf{v}_k=\sum_{j\ne k}d_j\mathbf{v}_j\) for real scalars \(d_j\). Move \(\mathbf{v}_k\) to the other side. The resulting equation is

This is a dependence relation because the coefficient of \(\mathbf{v}_k\) is \(-1\), which is nonzero. If \(p=1\), the sum over \(j\ne k\) is the empty sum \(\mathbf{0}\), and the same equation says \(-\mathbf{v}_1=\mathbf{0}\). Hence both directions hold, including the one-vector case. \(\square\)

Reading the Coefficients Carefully

The theorem is an existence statement: dependence guarantees that at least one vector can be written using the others. It does not assert that every vector in the list can be so written. The proof also gives the coefficients explicitly. If a relation has coefficient \(c_k\ne0\) at the chosen position, the coefficient of \(\mathbf{v}_j\) in the isolated expression is \(-c_j/c_k\), not \(c_j/c_k\). Keeping the minus sign is important.

Building a Relation From a Representation

The reverse direction is just as useful. When a problem gives an expression for one vector in terms of the others, it immediately supplies a dependence relation. The coefficient of the vector being represented becomes \(-1\) after all terms are moved to one side. That coefficient certifies that the relation is nontrivial, even if some of the other coefficients happen to be zero.

Proof. Suppose the list is dependent. By the theorem, there is an index \(k\) and scalars \(d_j\) such that \(\mathbf{v}_k=\sum_{j\ne k}d_j\mathbf{v}_j\). Every linear combination of the remaining vectors is already a linear combination of the full list, so the span after deletion is contained in the original span. In the other direction, any vector in the original span is a combination of the list. Replace each occurrence of \(\mathbf{v}_k\) in that combination by \(\sum_{j\ne k}d_j\mathbf{v}_j\); the result is a combination of the remaining vectors. Thus the original span is contained in the span after deletion, and the spans are equal. Conversely, suppose deleting \(\mathbf{v}_k\) leaves the span unchanged. Since \(\mathbf{v}_k\) belongs to the original span, it belongs to the span of the remaining vectors. By the definition of span, it is a linear combination of those vectors, so the theorem implies that the list is dependent. When \(p=1\), the remaining span is \(\{\mathbf{0}\}\), and the same reasoning says the sole vector must be zero. \(\square\)

One Redundant Vector Does Not Mean Every Vector

The index \(k\) matters. A dependence relation may have zero coefficients on some vectors, and those positions cannot be isolated from that relation. The relation certifies redundancy only at a position with a nonzero coefficient. A different relation might identify a different position, but the theorem promises at least one such position, not all of them.

Common Pitfall: Dividing by a Zero Coefficient

A dependence relation has at least one nonzero coefficient, but it may also have zero coefficients. It is an error to choose a zero coefficient and divide by it. For the last example, the relation has coefficient list \((0,-2,1)\). Trying to isolate \(\mathbf{w}_1\) by dividing by its coefficient would require division by \(0\), which is undefined. It is also wrong to conclude that \(\mathbf{w}_1\) comes from the other vectors: the coordinate calculation above rules that out.

The correction is to choose a position with a nonzero coefficient. Here the coefficient of \(\mathbf{w}_3\) is \(1\), so isolate that vector instead:

More generally, start with any nontrivial relation, locate a nonzero coefficient \(c_k\), and divide only by that coefficient. The resulting expression proves the existence required by the theorem. In the reverse direction, the coefficient \(-1\) on the represented vector ensures that a representation always produces a nontrivial relation.