Writing Explicit Linear Dependence Relations

From a Coefficient Equation to a Relation

A dependence relation is a specific equation among a list of vectors: the vectors are multiplied by coefficients, the resulting vectors are added, and the sum is \(\mathbf{0}\). Writing down the coefficients is only part of the task. They must be attached to the correct vectors, at least one must be nonzero, and the resulting equation must actually hold. Direct substitution checks all three features.

Throughout, the vectors are in \(\mathbb{R}^n\) and are written as columns. We work with an ordered list \((\mathbf{v}_1,\ldots,\mathbf{v}_k)\), so the index on each coefficient matters: \(c_i\) is the coefficient of \(\mathbf{v}_i\). The zero vector in \(\mathbb{R}^n\) is \(\mathbf{0}\).

The nontriviality condition concerns the coefficients, not whether they look large or complicated. A relation such as \(0\mathbf{v}_1+2\mathbf{v}_2+0\mathbf{v}_3=\mathbf{0}\) is nontrivial because one coefficient is nonzero, provided the equation is true. Conversely, taking every coefficient to be zero always gives an equation equal to \(\mathbf{0}\), but it is the trivial combination and is not a dependence relation.

To translate a vector equation into standard form, move all vector terms to one side and put \(\mathbf{0}\) on the other. When a term crosses the equality sign, its coefficient changes sign. Then list the coefficients in the same order as the vectors. Finally, substitute the vectors into the equation to check it. These steps are algebraic bookkeeping; they do not require a matrix test.

Checking a Relation Coordinate by Coordinate

For vectors in \(\mathbb{R}^n\), direct substitution reduces the check to ordinary arithmetic. If \(\mathbf{v}_i=(v_{i1},\ldots,v_{in})^{\top}\), then the \(j\)th coordinate of \(c_1\mathbf{v}_1+\cdots+c_k\mathbf{v}_k\) is \(c_1v_{1j}+\cdots+c_kv_{kj}\). Each coordinate sum must equal zero.

Proof. Assume first that \(c_1\mathbf{v}_1+\cdots+c_k\mathbf{v}_k=\mathbf{0}\). Equality of vectors in \(\mathbb{R}^n\) means equality in each coordinate. The \(j\)th coordinate on the left is \(c_1v_{1j}+\cdots+c_kv_{kj}\), and the \(j\)th coordinate of \(\mathbf{0}\) is zero, so this sum is zero for every \(j=1,\ldots,n\). Conversely, assume that every one of those coordinate sums is zero. Then the vector on the left has zero in every coordinate, so it is \(\mathbf{0}\). Thus the vector equation holds. \(\square\)

This proposition gives a reliable verification routine: calculate each scalar multiple, add the resulting columns, and report the full sum. In two or three dimensions, writing the coordinate sums explicitly makes a mistaken sign or coefficient easy to spot.

Rescaling a Relation

The same dependence relation can be written in more than one way. Multiplying every coefficient by the same nonzero scalar multiplies the whole left side by that scalar. Since a scalar multiple of \(\mathbf{0}\) is still \(\mathbf{0}\), the equation remains valid; and the coefficients remain not all zero. This can make a relation easier to read, for example by clearing fractions or setting one coefficient to \(1\).

Proof. Multiply the original vector equation by \(a\). Distributivity gives \((ac_1)\mathbf{v}_1+\cdots+(ac_k)\mathbf{v}_k=a\mathbf{0}=\mathbf{0}\), so the scaled equation holds. Because the original coefficients are not all zero, there is an index \(j\) with \(c_j\ne0\). Since \(a\ne0\), \(ac_j\ne0\), so the scaled coefficients are also not all zero. Both conditions in the definition are satisfied. \(\square\)

The restriction \(a\ne0\) is essential. Multiplication by zero would turn every coefficient into zero, leaving only the trivial combination. A safe normalization therefore divides by a coefficient known to be nonzero, never by a coefficient that might vanish.

Rearranging an Equation Without Losing Its Signs

A relation may first appear as an equality between two nonzero combinations rather than as a sum equal to zero. To put it in standard form, subtract the entire right-hand side from both sides. This changes the signs of every term that came from the right. After rearranging, keep the vector order fixed and record the coefficient of each vector, including a coefficient of zero if a vector does not appear.

Common Mistake: Reversing a Sign

A frequent error is to move a term across the equality sign without changing its sign. In the preceding example, a student might incorrectly write \(2\mathbf{u}_1-\mathbf{u}_2+3\mathbf{u}_3=\mathbf{0}\). Substitution exposes the error:

So the incorrect equation is not a dependence relation for this list. The correction is to subtract the full right-hand side, yielding \(2\mathbf{u}_1-\mathbf{u}_2-3\mathbf{u}_3=\mathbf{0}\), and then check the coordinates. A coefficient tuple should never be accepted just because it looks plausible.

Another bookkeeping error is attaching coefficients to the wrong vectors. For an ordered list \((\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3)\), the tuple \((2,-1,-3)\) means \(2\mathbf{u}_1-\mathbf{u}_2-3\mathbf{u}_3\), not \(2\mathbf{u}_2-\mathbf{u}_1-3\mathbf{u}_3\). When checking a relation, write the vector equation explicitly before substituting so that each coefficient’s index stays visible.

A Reliable Record of a Dependence Relation

When presenting a relation, include enough information for another reader to check it without guessing. State the ordered list, give the equation in standard form, identify the coefficient tuple, and show the substitution. A concise write-up can follow this sequence:

An explicit relation is useful because it is a checkable certificate: it displays the exact coefficients and the exact cancellation. The definition of linear dependence asks whether a nontrivial coefficient choice produces the zero vector; a verified relation supplies such a choice directly. Keep the two checks separate: nontrivial coefficients establish that the combination is not the trivial one, while substitution establishes that the equation itself is true.