Zero Vectors, Repeated Vectors, and Independence

Two Immediate Sources of Dependence

Some lists can be recognized as dependent without solving a system or comparing every vector with every other one. A zero vector contributes nothing to a linear combination, and two identical entries can cancel each other. In both cases, the defining coefficient equation gives a direct way to exhibit dependence.

Throughout, the scalars are real. A finite list \(\mathbf{v}_1,\dots,\mathbf{v}_p\) is linearly dependent if there are real numbers \(c_1,\dots,c_p\), not all zero, such that the weighted sum of the listed vectors is the zero vector. A choice of coefficients satisfying this equation with at least one nonzero coefficient is a dependence relation. If the only coefficients satisfying the equation are all zero, the list is linearly independent.

The words list and set matter here. In a list, each position has its own coefficient, even when two positions contain the same vector. Thus the entries in positions \(i\) and \(j\) can be assigned different coefficients. A set, by contrast, records whether an element occurs, not how many times it occurs; writing a set would discard the information that a vector appears twice.

A Zero Vector Supplies a Nonzero Coefficient

If \(\mathbf{v}_k=\mathbf{0}\), its coefficient can be chosen nonzero without changing the sum. Set \(c_k=1\), for example, and set every other coefficient to \(0\). Every term in the resulting combination is zero, but the coefficient list is not all zero.

Proof. Suppose \(\mathbf{v}_k=\mathbf{0}\) for some position \(k\) in the list \(\mathbf{v}_1,\dots,\mathbf{v}_p\). Choose \(c_k=1\) and \(c_i=0\) for every \(i\ne k\). Then \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=1\mathbf{v}_k=\mathbf{0}\). The coefficient \(c_k=1\) is nonzero, so this is a nontrivial dependence relation. Hence the list is dependent. \(\square\)

This argument does not require the other entries to be nonzero or distinct. If a list contains several zero vectors, choosing a coefficient of \(1\) for any one of their positions and \(0\) elsewhere still gives a dependence relation. The theorem also covers a list consisting of the zero vector alone.

Two Identical Entries Can Cancel

Now suppose two different positions contain the same vector: \(\mathbf{v}_i=\mathbf{v}_j\) with \(i\ne j\). Give the entry in position \(i\) coefficient \(1\), and the identical entry in position \(j\) coefficient \(-1\). Give all remaining entries coefficient \(0\). The two nonzero terms cancel because they are the same vector with opposite coefficients.

Proof. Suppose \(i\ne j\) and \(\mathbf{v}_i=\mathbf{v}_j\). Choose \(c_i=1\), \(c_j=-1\), and \(c_k=0\) for every other position \(k\). Since the positions are distinct, these choices are consistent, and at least one coefficient is nonzero. The weighted sum is \(1\mathbf{v}_i+(-1)\mathbf{v}_j=\mathbf{v}_i-\mathbf{v}_j=\mathbf{0}\), where the last equality follows from \(\mathbf{v}_i=\mathbf{v}_j\). Thus the coefficients give a nontrivial dependence relation, so the list is dependent. \(\square\)

How to Use These Tests Carefully

These are sufficient tests for dependence: finding a zero entry or two identical entries guarantees that the list is dependent. The tests do not say that every dependent list must have a zero vector or repeated entries. A dependence relation is defined by the full weighted-sum equation, so other coefficient patterns can also make a sum equal to zero. For this tutorial’s two cases, however, the relevant coefficients can be written down immediately.

Keep track of the positions as well as the vector values. If the same vector occurs in positions \(i\) and \(j\), then the coefficient \(1\) belongs to position \(i\) and \(-1\) belongs to position \(j\). They are two coefficient choices, even though both positions hold the same vector. If a zero vector is present, a coefficient of \(1\) at its position is enough; setting all coefficients to zero would not show dependence.

Common mistake: using only the trivial coefficient choice. For the list \(\mathbf{v}_1=\begin{bmatrix}4\\-2\end{bmatrix}\), \(\mathbf{v}_2=\begin{bmatrix}0\\0\end{bmatrix}\), the calculation \(0\mathbf{v}_1+0\mathbf{v}_2=\begin{bmatrix}0\\0\end{bmatrix}\) is correct, but it does not prove dependence: both coefficients are zero. Correct it by choosing \((c_1,c_2)=(0,1)\). Then \(0\begin{bmatrix}4\\-2\end{bmatrix}+1\begin{bmatrix}0\\0\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\), and the coefficient \(1\) makes this a nontrivial relation.

Common mistake: treating a list as a set. If a student writes the entries \(\begin{bmatrix}1\\0\end{bmatrix}\), \(\begin{bmatrix}0\\1\end{bmatrix}\), \(\begin{bmatrix}1\\0\end{bmatrix}\) as a set, the repeated entry disappears from the notation. For the original list, the correct coefficient equation includes all three positions. Taking coefficients \((1,0,-1)\) gives \(1\begin{bmatrix}1\\0\end{bmatrix}+0\begin{bmatrix}0\\1\end{bmatrix}-1\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\). The list is dependent; discarding a repeated position would hide this immediate relation.

When checking either pattern, use the same short sequence: identify the relevant position or pair of positions, assign coefficients that are not all zero, and substitute them into the defining equation. The substitution is the final check: it confirms both that the weighted sum is \(\mathbf{0}\) and that the relation is nontrivial.