What the Pivot Test Does—and Does Not—Say
Row reduction gives a reliable test for whether a list of vectors is linearly independent, but its output is easy to misread. A nonzero column is not necessarily a pivot column. A free variable is not a pivot variable. And the columns displayed in a reduced matrix are generally not the original vectors. These distinctions matter because the test concerns the original list, even though its calculation uses a changed matrix.
Let \(\mathbf{v}_1,\dots,\mathbf{v}_p\) be columns in \(\mathbb{R}^m\), and form \(\mathbf{A}=\begin{bmatrix}\mathbf{v}_1&\cdots&\mathbf{v}_p\end{bmatrix}\). The list is linearly independent when the equation \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\) has only the solution in which every coefficient is zero. Writing \(\mathbf{c}=(c_1,\dots,c_p)^\top\), this is the homogeneous system \(\mathbf{A}\mathbf{c}=\mathbf{0}\).
In reduced row echelon form (RREF), each nonzero row has a leading \(1\), called a pivot, and that pivot is the only nonzero entry in its column. The pivot positions move strictly to the right as one moves down the rows; any zero rows are at the bottom. A column containing a pivot is a pivot column. A variable whose column has no pivot is a free variable.
Proof. Row operations preserve the solutions of a homogeneous system: swapping equations leaves the system unchanged, multiplying an equation by a nonzero scalar preserves its solutions, and adding a multiple of one equation to another preserves the solutions because the operation can be reversed. Therefore \(\mathbf{A}\mathbf{c}=\mathbf{0}\) and \(\operatorname{rref}(\mathbf{A})\mathbf{c}=\mathbf{0}\) have the same solutions. In the reduced system, each pivot variable is determined by the free variables in its row. If every column has a pivot, there are no free variables, and every pivot variable must equal zero. The only solution is then \(\mathbf{c}=\mathbf{0}\). If some column has no pivot, assign \(1\) to one free variable and \(0\) to every other free variable. Each pivot variable is then determined by its row, giving a solution with a coefficient equal to \(1\), so the solution is not zero. Thus the homogeneous system has only the zero solution exactly when every column has a pivot. By the definition of independence, this is exactly the stated test. \(\square\)
Read Pivot Positions, Not Just Nonzero Entries
The test asks whether each column has a pivot. It does not ask whether each column contains a nonzero entry. A nonpivot column can contain several nonzero entries; its variables are still free because no leading entry occurs in that column. This is one reason it is safer to mark the pivot positions explicitly than to judge the matrix by appearance.
Worked Example: A Nonzero Column Without a Pivot
Consider the vectors \(\mathbf{v}_1=(1,0,1)^\top\), \(\mathbf{v}_2=(0,1,1)^\top\), and \(\mathbf{v}_3=(2,-1,1)^\top\). Decide whether they are independent and find a relation if they are dependent.
Place the vectors as columns. Apply \(R_3\to R_3-R_1\), then \(R_3\to R_3-R_2\):
The pivots are in columns 1 and 2. Column 3 has nonzero entries, but it has no pivot, so it is a nonpivot column and its variable \(c_3\) is free. The reduced system gives \(c_1+2c_3=0\) and \(c_2-c_3=0\). Set \(c_3=1\); then \(c_1=-2\) and \(c_2=1\). Thus \(-2\mathbf{v}_1+\mathbf{v}_2+\mathbf{v}_3=\mathbf{0}\), so the vectors are dependent.
Check: \(-2(1,0,1)^\top+(0,1,1)^\top+(2,-1,1)^\top=(0,0,0)^\top\). Equivalently, \(\mathbf{v}_3=2\mathbf{v}_1-\mathbf{v}_2=(2,-1,1)^\top\).
A common wrong calculation is to call column 3 a pivot column because it contains \(2\) and \(-1\). Those entries are not pivots: they are not leading \(1\)s in their rows. The correction is to locate the leading entries row by row and then identify their column indices. Here the third variable is free even though its column is not zero.
Reduced Columns Are Not the Original Vectors
Row operations act on the rows of a matrix and change its columns as vectors. Consequently, a column of the RREF is usually not the corresponding vector from the original list. However, relations among columns are preserved by the same row operations. The entries in a nonpivot column of the RREF can therefore provide coefficients for a relation among the corresponding original columns.
Proof. Each row operation preserves the solutions of a homogeneous system, as shown in the proof of the pivot test. Applying the operations that transform \(\mathbf{A}\) into \(\mathbf{R}\) therefore preserves the set of coefficient columns \(\mathbf{c}\) satisfying the homogeneous equation. Hence \(\mathbf{A}\mathbf{c}=\mathbf{0}\) exactly when \(\mathbf{R}\mathbf{c}=\mathbf{0}\). Since a matrix-vector product is the linear combination of its columns with coefficients from \(\mathbf{c}\), the same coefficient relation holds for the corresponding columns of both matrices. \(\square\)
More explicitly, suppose columns \(j_1,\dots,j_r\) of \(\mathbf{R}\) are pivot columns and column \(k\) is nonpivot. In RREF, each pivot column has a \(1\) at its pivot row and zeros elsewhere. If the entries of column \(k\) in those pivot rows are \(a_1,\dots,a_r\), then \(\mathbf{R}\)'s column \(k\) is \(a_1\) times pivot column \(j_1\), plus \(\cdots\), plus \(a_r\) times pivot column \(j_r\). The same equality, with the same coefficients, holds among the corresponding original columns of \(\mathbf{A}\). The reduced columns help reveal the coefficients; they do not replace the original vectors in the final relation.
Worked Example: Use the Coefficients, Not the Changed Vectors
Let \(\mathbf{v}_1=(1,2)^\top\) and \(\mathbf{v}_2=(2,4)^\top\). Reduce the matrix whose columns are these vectors and interpret the result.
Apply \(R_2\to R_2-2R_1\):
There is a pivot in column 1 and no pivot in column 2, so \(c_2\) is free and the original vectors are dependent. The reduced equation is \(c_1+2c_2=0\). Taking \(c_2=1\) gives \(c_1=-2\), hence \(-2\mathbf{v}_1+\mathbf{v}_2=\mathbf{0}\).
A tempting but incorrect conclusion is that the original first vector is \((1,0)^\top\), because that is the first column displayed after reduction. It is not: the original vector is \((1,2)^\top\). The row operation changed it to \((1,0)^\top\). What remains valid is the coefficient relation.
Check: \(-2(1,2)^\top+(2,4)^\top=(-2+2,-4+4)^\top=(0,0)^\top\). The relation holds for the original vectors, not merely for the changed columns.
Count Pivot Columns, Not Pivot Rows
Another error is to check whether every row has a pivot instead of checking whether every column has one. Independence of the listed vectors depends on whether every coefficient variable is a pivot variable. A pivot in every row does not prevent extra columns from being nonpivot columns.
Worked Example: Pivots in Every Row but Dependence
Consider \(\mathbf{u}_1=(1,0)^\top\), \(\mathbf{u}_2=(0,1)^\top\), and \(\mathbf{u}_3=(1,1)^\top\). Decide whether the list is independent.
The matrix with these columns is already in RREF:
There is a pivot in each row, but only in columns 1 and 2. Column 3 is nonpivot, so \(c_3\) is free. The homogeneous equations are \(c_1+c_3=0\) and \(c_2+c_3=0\). Setting \(c_3=1\) gives \(c_1=-1\) and \(c_2=-1\), so the list is dependent.
Check: \(-\mathbf{u}_1-\mathbf{u}_2+\mathbf{u}_3=-(1,0)^\top-(0,1)^\top+(1,1)^\top=(0,0)^\top\). The wrong conclusion “a pivot in every row means independence” fails because it ignores the third coefficient variable.
A Reliable Procedure and Common Checks
Use the following sequence whenever the pivot test is applied to a list of vectors.
Construct \(\mathbf{A}=\begin{bmatrix}\mathbf{v}_1&\cdots&\mathbf{v}_p\end{bmatrix}\). The column order corresponds to the order of the vectors in the list.
In \(\operatorname{rref}(\mathbf{A})\), identify the leading \(1\) in each nonzero row. A column is a pivot column only if it contains one of these pivot positions.
If every column has a pivot, the list is independent. If at least one column has no pivot, its variable is free and the list is dependent.
Use the reduced equations to find coefficients, then substitute them into the original vectors and check that the combination is \(\mathbf{0}\).
The diagnostic distinctions can be summarized as follows.
| What you see | What it tells you | What it does not tell you |
|---|---|---|
| A column has a nonzero entry | The column is not a zero column in the displayed matrix | It need not be a pivot column |
| A variable has no pivot | It is free in the homogeneous system | It is not a pivot variable just because its column is nonzero |
| A reduced column is a combination of reduced pivot columns | The coefficients give a relation among the matching original columns | The reduced columns are not the original vectors |
| Every row has a pivot | Each nonzero row has a leading entry | Every column has a pivot, or the vectors are independent |
When a proposed dependence relation is available, direct substitution is the safest final check: multiply each original vector by its proposed coefficient, add the results coordinate by coordinate, and confirm that every coordinate is zero. When the claim is independence, verify that the pivot columns include every column index. These checks address different claims: a relation disproves independence, while a complete set of pivot columns proves it by the pivot test.
Check Your Understanding
For each question, identify pivot positions by column and use the original vectors when checking any relation.
- In \(\begin{bmatrix}1&0&3\\0&1&-2\\0&0&0\end{bmatrix}\), which variable is free, and are all three columns pivot columns?
- For the matrix in Question 1, find a nonzero coefficient vector in its null space and verify the resulting relation among its columns.
- Explain why a pivot in every row is not, by itself, a valid test for independence of the columns.
- If row reduction changes an original column from \((2,3)^\top\) to \((2,0)^\top\), which vector must appear in the final substitution check?
- A reduced matrix has a nonzero entry in every column, but one column has no pivot. What does the pivot test conclude, and why?
