Independence Tests for Vectors in R^2

What the Coordinate Test Looks For

In \(\mathbb{R}^2\), each vector has two coordinates, but a list may contain one, two, or many vectors. To test the list, assign a coefficient to each vector and ask whether a nontrivial choice of coefficients can produce the zero vector. Writing the vectors as columns turns that question into a homogeneous system with two equations. Row reduction then reveals whether every coefficient must be zero or whether a free variable allows a nonzero solution.

All vectors here are columns with real coordinates. A list \(\mathbf{v}_1,\dots,\mathbf{v}_p\) is linearly independent when the equation \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\) forces every coefficient \(c_j\) to be zero. It is linearly dependent when some choice of coefficients, not all zero, gives \(\mathbf{0}\). A list is ordered: if a vector occurs twice, both appearances have their own coefficient.

Write \(\mathbf{v}_j=(x_j,y_j)\), and form the matrix whose columns are the vectors. Its homogeneous system is

Each row operation is reversible, so it preserves the solution set of this homogeneous system. In the reduced row echelon form (RREF), a variable corresponding to a column without a pivot is free: it can be assigned any real value, with the pivot variables then determined by the equations. Thus, a nonpivot column provides a way to test for a nonzero coefficient solution.

For completeness, the test follows directly from the meaning of a free variable. If every column has a pivot, there are no free variables; in a homogeneous system every pivot variable must then be zero. If some column has no pivot, choose one free variable to be \(1\) and all other free variables to be \(0\). The pivot equations determine the remaining variables, and the chosen coefficient is nonzero. This test applies to one vector just as it does to a longer list.

Testing a Pair of Vectors

For a pair \(\mathbf{v}_1=(a,c)\) and \(\mathbf{v}_2=(b,d)\), the coefficient equation is precisely the two-equation system

There is a useful interpretation of its pivot pattern. If \(\mathbf{v}_1\ne\mathbf{0}\), column 1 has a pivot. Column 2 also has a pivot exactly when \(\mathbf{v}_2\) is not a scalar multiple of \(\mathbf{v}_1\). If the first vector is zero, the pair is dependent regardless of the second vector.

Proof. First suppose the pair is independent. If \(\mathbf{v}_1=\mathbf{0}\), then \(1\mathbf{v}_1+0\mathbf{v}_2=\mathbf{0}\) is a nontrivial relation, contrary to independence. Thus \(\mathbf{v}_1\ne\mathbf{0}\). If \(\mathbf{v}_2=t\mathbf{v}_1\) for a real number \(t\), then \((-t)\mathbf{v}_1+1\mathbf{v}_2=\mathbf{0}\), also a nontrivial relation. So \(\mathbf{v}_2\) is not a scalar multiple of \(\mathbf{v}_1\).

Conversely, suppose \(\mathbf{v}_1\ne\mathbf{0}\) and \(\mathbf{v}_2\) is not a scalar multiple of \(\mathbf{v}_1\). Consider \(c_1\mathbf{v}_1+c_2\mathbf{v}_2=\mathbf{0}\). If \(c_2\ne0\), rearranging gives \(\mathbf{v}_2=-(c_1/c_2)\mathbf{v}_1\), contradicting the assumption. Therefore \(c_2=0\). The equation becomes \(c_1\mathbf{v}_1=\mathbf{0}\); since \(\mathbf{v}_1\ne\mathbf{0}\), this forces \(c_1=0\). The only relation is trivial, so the pair is independent. \(\square\)

What Changes for Longer Lists

For a list of \(p\) vectors in \(\mathbb{R}^2\), the coefficient system still has just two rows, but it has \(p\) coefficient variables. Each pivot occupies a different row, so there can be at most two pivot columns. Consequently, a list with more than two vectors must have a nonpivot column and therefore a free variable. This is the two-dimensional instance of the general principle that a list with more vectors than coordinates is dependent.

Proof. Put the vectors into the columns of a \(2\times p\) matrix and row-reduce it. There are only two rows, and each pivot must be in a different row, so there are at most two pivot columns. Since \(p>2\), at least one of the \(p\) columns has no pivot. The corresponding coefficient is free. Assign that coefficient the value \(1\), assign any other free coefficients the value \(0\), and solve for the pivot coefficients in the homogeneous RREF system. This gives a solution with a nonzero coefficient, hence a dependence relation. \(\square\)

Reading the Main Cases

The number of vectors narrows down what the pivot test can show, but it does not replace the test when there are one or two vectors. For one vector, independence means exactly that the vector is nonzero: \(c\mathbf{v}_1=\mathbf{0}\) has only \(c=0\) when \(\mathbf{v}_1\ne\mathbf{0}\), while a zero vector gives a nontrivial relation. For two vectors, independence requires a pivot in each column. For more than two vectors, independence is impossible in \(\mathbb{R}^2\).

The pivot test also explains how to obtain a relation rather than only a yes-or-no answer. Choose values for the free coefficients, then use the pivot equations to determine the remaining coefficients. A nonzero choice of a free coefficient guarantees that the resulting coefficient vector is nonzero. Finally, substitute the coefficients into the original vector combination to verify the relation. This check is useful because it confirms both the arithmetic and the order of the vectors in the list.

Common Pitfall: Two Equations Need Not Give Two Pivots

A pair gives two coordinate equations, but that does not mean the equations are independent. For \((1,2)\) and \((2,4)\), one might incorrectly reason that two equations in two unknowns must force both coefficients to zero. The actual coefficient system is

The second equation has become \(0=0\), not a new restriction. The correct reduced equation is \(c_1+2c_2=0\), so \(c_2\) is free. Taking \(c_2=1\) gives \(c_1=-2\), and the correction to the false conclusion is the explicit relation \(-2(1,2)+(2,4)=(0,0)\). Always count pivots after reduction; do not count the equations before checking whether some are redundant.

For a pair in \(\mathbb{R}^2\), the same issue can be phrased geometrically: if both vectors are nonzero but point along the same line through the origin, one is a scalar multiple of the other and the pair is dependent. In a longer list, even if no two vectors are scalar multiples, the whole list can still be dependent. The three vectors in the preceding example are pairwise nonproportional, yet their displayed relation shows dependence. The homogeneous coefficient system tests the whole list, not just its pairs.