Mixed Independence Problems: A Row-Reduction Workshop

A Common Framework for Mixed Problems

Independence questions may be presented as coordinate lists, as columns of a matrix, or as a family that depends on a parameter. In each case, the central task is to determine whether a homogeneous system has a nonzero solution. If it does, its solution gives coefficients for a dependence relation; if it does not, the list is independent. The row-reduction work is only part of the answer: the coefficient order must match the original list, and a claimed relation should be checked in the original coordinates.

Throughout, scalars are real, vectors in \(\mathbb{R}^n\) are columns, and a list keeps track of the order of its vectors. Given \(\mathbf{v}_1,\dots,\mathbf{v}_p\in\mathbb{R}^n\), put them in that order as the columns of \(\mathbf{A}\in\mathbb{R}^{n\times p}\). A coefficient vector is \(\mathbf{c}=(c_1,\dots,c_p)^\top\).

The matrix equation \(\mathbf{A}\mathbf{c}=\mathbf{0}\) is exactly the vector equation in this definition. Thus independence asks whether the homogeneous system has only its zero solution. Row operations preserve the solution set of a homogeneous system: swapping equations, multiplying one equation by a nonzero scalar, or adding a multiple of one equation to another leaves an equivalent system, and each operation can be reversed.

Proof. By the definition of matrix multiplication, \(\mathbf{A}\mathbf{c}=c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\). Therefore \(\mathbf{A}\mathbf{c}=\mathbf{0}\) holds for exactly the coefficient vectors that give a relation among the list. The list is independent precisely when the only such coefficient vector is zero, by the definition of independence. Each elementary row operation replaces the equations by an equivalent system because the operation is reversible. Consequently row reduction does not change which coefficient vectors solve the system. \(\square\)

Reading a Reduced System Reliably

The reduced row echelon form (RREF) makes the solution choices visible. A column containing a pivot corresponds to a pivot variable; a column without one corresponds to a free variable. In a homogeneous system, there is no constant term, so when all variables are pivot variables, solving from the last pivot row upward forces every variable to be zero. A free variable, on the other hand, can be assigned a nonzero value, producing a nonzero solution.

Proof. Row operations preserve the solutions, so it is enough to solve the system whose matrix is \(\operatorname{rref}(\mathbf{A})\). If every column has a pivot, there are no free variables. Each pivot equation expresses its pivot variable using variables to its right. Starting at the last pivot and proceeding upward, each pivot variable must be zero: the last equation has no nonzero variable to its right, and each preceding equation has only variables already forced to zero. Thus the only solution is \(\mathbf{c}=\mathbf{0}\). Conversely, suppose a column has no pivot. Assign one free variable the value \(1\), and assign every other free variable the value \(0\). Solve the pivot equations from the last pivot upward to determine the pivot variables. This gives a solution of the homogeneous system, and it is nonzero because the chosen free variable is \(1\). Hence the system does not have only the zero solution. These two implications prove the statement. \(\square\)

This result gives a decision procedure, but a dependence problem asks for more than a yes-or-no answer. To find a relation, keep the variables in the same order as the original columns, choose convenient values for the free variables, solve for the pivot variables, and then substitute the resulting coefficients into the original list.

Worked Example: Four Coordinate Vectors in \(\mathbb{R}^3\)

There is more than one valid choice of free-variable values. For example, setting \(c_3=0\) and \(c_4=1\) gives \(c_1=-1\) and \(c_2=-2\), another nontrivial relation. The first choice was convenient because it avoided a coefficient on the fourth vector. A free-variable assignment is a way to produce one relation, not a claim that the relation is unique.

Worked Example: A Parameter Changes the Answer

This example illustrates why a parameter must be handled by cases. A row containing \(t-1\) cannot be treated as a pivot row when \(t=1\), because then its entries are all zero. Dividing by \(t-1\) without first stating \(t\ne1\) would discard the very case that creates dependence.

Worked Example: Pivot Test with More Coordinates Than Vectors

A Practical Order for Solving These Problems

For a mixed set of independence exercises, a consistent sequence reduces avoidable errors:

Common Pitfall: Putting the Vectors in the Rows

For an independence test, row-reduce the matrix whose columns are the vectors. A concrete error arises with \(\mathbf{e}_1=(1,0,0)^\top\) and \(\mathbf{e}_2=(0,1,0)^\top\) in \(\mathbb{R}^3\). If they are mistakenly written as rows, the matrix is \(\begin{bmatrix}1&0&0\\0&1&0\end{bmatrix}\). Seeing that its third column has no pivot and concluding “the two vectors are dependent” applies the column test to the wrong arrangement.

The correction is to place the vectors as columns: \(\mathbf{A}=\begin{bmatrix}1&0\\0&1\\0&0\end{bmatrix}\). Both columns have pivots. Equivalently, \(c_1\mathbf{e}_1+c_2\mathbf{e}_2=(c_1,c_2,0)^\top=\mathbf{0}\) forces \(c_1=c_2=0\). The vectors are independent. The example also shows why an answer should be tied back to the homogeneous system, not read from a matrix whose rows or columns do not represent the intended list.