From Three Vectors to a Matrix
For three vectors in \(\mathbb{R}^3\), the independence test is a manageable row-reduction problem: put the vectors in the columns of a \(3\times3\) matrix, reduce it, and identify its pivot columns. If every column is a pivot column, the vectors are independent. If a column has no pivot, the vectors are dependent. In the dependent case, the free variables in the reduced homogeneous system tell us how to write a dependence relation.
Vectors are columns throughout. Let \(\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\in\mathbb{R}^3\), and form the matrix \(\mathbf{A}=[\mathbf{v}_1\ \mathbf{v}_2\ \mathbf{v}_3]\). A dependence relation is an equation \(c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3=\mathbf{0}\) in which the real coefficients are not all zero. Writing \(\mathbf{c}=(c_1,c_2,c_3)^\top\), this equation is exactly \(\mathbf{A}\mathbf{c}=\mathbf{0}\).
The earlier pivot test for independence says that a matrix's columns are independent exactly when every column is a pivot column in its RREF. For this topic, the practical task is to use that test on a \(3\times3\) column matrix and, if a column is free, turn the resulting equations into coefficients. Row reduction preserves the solutions of the homogeneous system, so coefficients found from the RREF give a relation among the original columns.
Proof. By the definition of matrix-vector multiplication, \(\mathbf{A}\mathbf{c}=c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3\). Thus \(\mathbf{A}\mathbf{c}=\mathbf{0}\) holds exactly when the coefficients give a relation among the columns. The zero solution gives the trivial relation, while any solution with a nonzero component gives a nontrivial dependence relation. \(\square\)
Worked Tests for Three-Vector Lists
Worked Example: Three Independent Vectors
Test \(\mathbf{v}_1=(1,0,1)^\top\), \(\mathbf{v}_2=(0,1,1)^\top\), and \(\mathbf{v}_3=(1,1,0)^\top\) for independence.
Place the vectors in columns. First use \(R_3\to R_3-R_1\), then \(R_3\to R_3-R_2\):
Scale \(R_3\) by \(-\tfrac12\), then use \(R_1\to R_1-R_3\) and \(R_2\to R_2-R_3\):
There is a pivot in each column, so the only solution of the homogeneous system is \(c_1=c_2=c_3=0\). The vectors are independent. Check: the coefficient vector is \(\mathbf{0}\), and the original matrix times that vector is \(\mathbf{A}\mathbf{0}=\mathbf{0}\), as required. The RREF shows there are no other solutions.
Worked Example: A Dependent Triple with Two Pivots
Test \(\mathbf{v}_1=(1,2,0)^\top\), \(\mathbf{v}_2=(0,1,1)^\top\), and \(\mathbf{v}_3=(1,3,1)^\top\). If dependent, find a relation.
Subtract \(2R_1\) from \(R_2\), then subtract the new \(R_2\) from \(R_3\):
Clear the entries above the second pivot using \(R_1\to R_1\) (there is already a zero above it). The displayed matrix is in RREF. The equations are \(c_1+c_3=0\) and \(c_2+c_3=0\), so \(c_1=-c_3\) and \(c_2=-c_3\). Choose the free variable \(c_3=1\). Then \((c_1,c_2,c_3)=(-1,-1,1)\), giving the relation \(-\mathbf{v}_1-\mathbf{v}_2+\mathbf{v}_3=\mathbf{0}\).
Check: \(-\mathbf{v}_1-\mathbf{v}_2+\mathbf{v}_3=-(1,2,0)^\top-(0,1,1)^\top+(1,3,1)^\top=(0,0,0)^\top\). The relation is nontrivial and verifies dependence.
Worked Example: The Free Variable Is in the Middle
Let \(\mathbf{v}_1=(1,0,0)^\top\), \(\mathbf{v}_2=(2,0,0)^\top\), and \(\mathbf{v}_3=(0,1,0)^\top\). Reduce the column matrix and find a dependence relation.
The matrix is already in RREF:
The pivot columns are 1 and 3, so \(c_2\), not \(c_3\), is free. The equations are \(c_1+2c_2=0\) and \(c_3=0\). Setting \(c_2=1\) gives \(c_1=-2\) and \(c_3=0\), hence \((-2,1,0)\) is a nonzero solution.
Check: \(-2\mathbf{v}_1+\mathbf{v}_2+0\mathbf{v}_3=-2(1,0,0)^\top+(2,0,0)^\top+(0,0,0)^\top=(0,0,0)^\top\). The second column is twice the first, matching the computed relation.
Worked Example: A Triple with Only One Pivot
Test \(\mathbf{u}_1=(1,2,-1)^\top\), \(\mathbf{u}_2=(2,4,-2)^\top\), and \(\mathbf{u}_3=(-1,-2,1)^\top\), and find a dependence relation.
Apply \(R_2\to R_2-2R_1\) and \(R_3\to R_3+R_1\):
Only column 1 is a pivot column. The equation is \(c_1+2c_2-c_3=0\), so there are two free variables. Choose \(c_2=1\) and \(c_3=0\); then \(c_1=-2\), giving \((-2,1,0)\).
Check: \(-2\mathbf{u}_1+\mathbf{u}_2=-2(1,2,-1)^\top+(2,4,-2)^\top=(-2+2,-4+4,2-2)^\top=(0,0,0)^\top\). This confirms a nontrivial relation.
What the Number of Pivots Tells You
For a \(3\times3\) matrix, there can be at most three pivot columns. Three pivots mean independence; fewer than three mean dependence. The number of free variables is the number of nonpivot columns: one free variable when there are two pivots, two when there is one pivot, and three when there are no pivots. When there are no pivots, the matrix is the zero matrix, and any nonzero coefficient vector gives a dependence relation.
The two-pivot case has a useful feature: every dependence relation is a scalar multiple of one relation. This can help check that a relation extracted by row reduction has not missed another independent choice.
Proof. Exactly two pivot columns leave exactly one nonpivot column, whose coefficient is the one free variable; call it \(t\). Each pivot variable is determined by the reduced equations. Since the equations are homogeneous, setting \(t=0\) makes every pivot variable zero. Setting \(t=1\) produces a solution \(\mathbf{c}_0\) whose free coordinate is 1, so \(\mathbf{c}_0\ne\mathbf{0}\). For any real \(t\), the vector \(t\mathbf{c}_0\) has free coordinate \(t\) and satisfies the same homogeneous equations. The pivot variables are uniquely determined by that free coordinate, so every solution with free coordinate \(t\) must equal \(t\mathbf{c}_0\). Therefore all solutions, including the dependence relations, are scalar multiples of \(\mathbf{c}_0\). \(\square\)
Common Pitfall: Pairwise Nonproportional Is Not Enough
A tempting but incorrect shortcut is to check that no two vectors are scalar multiples and then conclude that three vectors in \(\mathbb{R}^3\) are independent. For example, in the second worked example, \((1,2,0)^\top\), \((0,1,1)^\top\), and \((1,3,1)^\top\) are pairwise nonproportional: their zero coordinates and coordinate ratios rule out any scalar multiple between each pair. But the conclusion of independence is wrong, because \(\mathbf{v}_3=\mathbf{v}_1+\mathbf{v}_2\).
The correction is to test the whole list using its column matrix. Pairwise comparisons do not test whether one vector is a combination of both of the others. Row reduction does: a nonpivot column gives a free variable, and a nonzero choice of that variable produces an explicit relation to verify against the original vectors.
Check Your Understanding
For each triple, form the column matrix. When a triple is dependent, find a nonzero coefficient vector and check the resulting relation in the original coordinates.
- Test \((1,0,0)^\top\), \((0,1,0)^\top\), and \((0,0,1)^\top\) for independence by row reduction.
- Find a dependence relation for \((1,1,0)^\top\), \((0,1,1)^\top\), and \((1,2,1)^\top\).
- For the matrix \(\begin{bmatrix}1&0&2\\0&1&-1\\0&0&0\end{bmatrix}\), identify the free variable and give one nonzero solution of the homogeneous system.
- Explain why a triple with only one pivot column has at least two free variables.
- Give a concrete reason why checking only whether pairs of vectors are proportional cannot establish independence of a triple.
