Read Pivot Information in the Right Place
Row reduction is efficient, but its conclusions depend on which feature of a matrix is being read. Pivot positions, free variables, rank, and nullity are connected; they are not interchangeable counts. Nor does row reduction preserve every feature of the original matrix: it preserves the solutions of a system and the row space, but it can change the column space. Keeping track of the matrix’s shape and the purpose of each calculation prevents several common misreadings.
Let \(\mathbf{A}\in\mathbb{R}^{m\times n}\). Its columns are vectors in \(\mathbb{R}^m\), and its rows are vectors in \(\mathbb{R}^n\). The rank of \(\mathbf{A}\), written \(\operatorname{rank}\mathbf{A}\), is the dimension of its column space \(\operatorname{Col}\mathbf{A}\). Its nullity, written \(\operatorname{nullity}\mathbf{A}\), is the dimension of its null space \(\operatorname{Nul}\mathbf{A}=\{\mathbf{x}\in\mathbb{R}^n:\mathbf{A}\mathbf{x}=\mathbf{0}\}\). A pivot column is a column containing a leading 1 in the reduced row echelon form (RREF); a free variable is the variable corresponding to a column without a pivot.
These statements concern different counts. The number of pivots is the rank; the number of free variables is the nullity; and the number of columns is their sum. The number of rows matters too: a matrix has at most \(\min(m,n)\) pivots. In particular, a zero row in the RREF is not itself a free variable. Free variables correspond to nonpivot columns.
Why Pivots Count Rank and Free Variables
First, the pivot test must be tied to the original columns. Row operations amount to multiplying on the left by an invertible matrix \(\mathbf{E}\). Thus the original and reduced matrices have the form \(\mathbf{A}\) and \(\mathbf{R}=\mathbf{E}\mathbf{A}\), where \(\mathbf{E}\) is invertible. For any coefficient vector \(\mathbf{c}\), \(\mathbf{A}\mathbf{c}=\mathbf{0}\) if and only if \(\mathbf{R}\mathbf{c}=\mathbf{0}\): one direction follows by multiplying by \(\mathbf{E}\), and the reverse by multiplying by \(\mathbf{E}^{-1}\). In other words, the same linear relations hold among the corresponding columns of both matrices.
Proof. Let \(\mathbf{R}\) be the RREF of \(\mathbf{A}\), and let the pivot-column indices be \(j_1,\dots,j_r\). In \(\mathbf{R}\), each nonpivot column is a linear combination of the pivot columns: its entries in the pivot rows give the coefficients, and its entries in all other rows are zero because those rows of \(\mathbf{R}\) are zero. Each pivot column has a 1 in its own pivot row and 0 in the other pivot rows, so the pivot columns of \(\mathbf{R}\) are linearly independent. The column relations are the same for \(\mathbf{A}\) as for \(\mathbf{R}\), since \(\mathbf{R}=\mathbf{E}\mathbf{A}\) with \(\mathbf{E}\) invertible. Therefore, the columns of \(\mathbf{A}\) at \(j_1,\dots,j_r\) are independent and span all columns of \(\mathbf{A}\). They form a basis of its column space, which consequently has dimension \(r\). \(\square\)
For the nullity count, row operations preserve the solutions of the homogeneous system, and the RREF equations make the variable count explicit. Each nonpivot variable can be assigned freely; the pivot variables are then determined by those choices. With \(n\) variables and \(r\) pivot columns, there are \(n-r\) free variables, and the solution space has dimension \(n-r\). This gives the rank-nullity relationship in this setting, including why the number of free variables is a dimension count rather than a count of zero rows.
Worked Examples: Keep the Counts and Spaces Separate
Worked Example: Rank, Nullity, and a Column-Space Basis
For the matrix below, find its rank and nullity, identify its free variables, and give a basis for its column space.
Subtract row 1 from row 3, then subtract row 2 from the new row 3:
This is the RREF. Its pivot columns are 1 and 3, so \(\operatorname{rank}\mathbf{A}=2\). There are four columns, hence \(4-2=2\) free variables: \(x_2\) and \(x_4\). The equations are \(x_1+2x_2+x_4=0\) and \(x_3+3x_4=0\). Set \(x_2=s\) and \(x_4=t\); then \(x_1=-2s-t\) and \(x_3=-3t\). Thus the solutions are \(\mathbf{x}=(-2s-t,s,-3t,t)^\top\), with two independent parameters, so \(\operatorname{nullity}\mathbf{A}=2\).
For the column-space basis, use columns 1 and 3 of the original matrix: \((1,0,1)^\top\) and \((0,1,1)^\top\). The other original columns are \((2,0,2)^\top=2(1,0,1)^\top\) and \((1,3,4)^\top=(1,0,1)^\top+3(0,1,1)^\top\), confirming that the two selected columns span all four. They are independent because \(a(1,0,1)^\top+b(0,1,1)^\top=(a,b,a+b)^\top=\mathbf{0}\) forces \(a=b=0\).
Check: For the general null-space vector, the first row gives \((-2s-t)+2s+t=0\), the second gives \(-3t+3t=0\), and the third gives \((-2s-t)+2s-3t+4t=0\). The solution satisfies \(\mathbf{A}\mathbf{x}=\mathbf{0}\).
Worked Example: Row Reduction Can Change the Column Space
Compare the row and column spaces of \(\mathbf{B}\) and its RREF. This example shows why pivot columns of the reduced matrix cannot generally be substituted for pivot columns of the original matrix when finding a basis for the original column space.
The rows of both matrices span \(\mathbb{R}^2\): each contains \((1,0)\) and \((0,1)\) among its rows, or has those vectors as its two nonzero rows. Their row spaces are equal. But \(\operatorname{Col}\mathbf{B}\) is spanned by \((1,0,1)^\top\) and \((0,1,1)^\top\), whereas \(\operatorname{Col}\mathbf{R}\) is spanned by \((1,0,0)^\top\) and \((0,1,0)^\top\). These are different subspaces of \(\mathbb{R}^3\): every vector in \(\operatorname{Col}\mathbf{R}\) has third entry 0, while \((1,0,1)^\top\in\operatorname{Col}\mathbf{B}\) has third entry 1.
Check: The original columns are independent: \(a(1,0,1)^\top+b(0,1,1)^\top=(a,b,a+b)^\top=\mathbf{0}\) implies \(a=b=0\). The RREF columns are also independent, but they are different vectors in a different column space. Both matrices have two pivots and rank 2; that agreement does not mean their column spaces are equal.
Worked Example: Full Rank Does Not Always Mean Zero Nullity
For this \(2\times3\) matrix, determine whether it has full rank and find its nullity.
The matrix is already in RREF, with pivots in columns 1 and 2. Thus \(\operatorname{rank}\mathbf{C}=2\). Since \(\min(2,3)=2\), this is the largest possible rank for a \(2\times3\) matrix, so \(\mathbf{C}\) has full rank. It has three variables and one free variable, \(x_3\). The equations give \(x_1=-2x_3\) and \(x_2=-3x_3\), so its nullity is \(3-2=1\).
Check: Taking \(x_3=1\) gives \(\mathbf{x}=(-2,-3,1)^\top\), and \(\mathbf{C}\mathbf{x}=(-2+2,-3+3)^\top=(0,0)^\top\). Thus the null space is not just \(\{\mathbf{0}\}\), despite full rank. The matrix has full row rank, but not full column rank: its rank equals the number of rows, not the number of columns.
What Row Operations Preserve
Each elementary row operation replaces a row by itself, a nonzero scalar multiple, or a sum involving another row. The new rows are therefore linear combinations of the old rows. The operation is reversible by another elementary row operation, so the old rows are also linear combinations of the new rows. Consequently, the two matrices have the same row space. Their null spaces also agree: an elementary row operation is multiplication by an invertible matrix \(\mathbf{E}\), and \(\mathbf{A}\mathbf{x}=\mathbf{0}\) holds exactly when \(\mathbf{E}\mathbf{A}\mathbf{x}=\mathbf{0}\).
Column space is different. Left multiplication changes the entries of the columns, and—as the example with \(\mathbf{B}\) shows—it can change the subspace they span. What remains useful is the pivot index: the original columns at the pivot indices give a basis for the original column space. Do not read the reduced columns as if they were the original ones.
Common Misreadings and Their Corrections
Wrong: “There are three rows, so the rank is 3.” In Worked Example 1, \(\mathbf{A}\) has three rows, but its RREF has only two pivots. The rank is 2, not 3. Rank counts pivots, equivalently the nonzero rows in RREF; it cannot exceed the smaller of the number of rows and columns.
Wrong: “One zero row means nullity 1.” The RREF in Worked Example 1 has one zero row but two free variables. Nullity is determined by the number of nonpivot columns: there are four columns and two pivots, so the nullity is \(4-2=2\). A zero row says one reduced equation contributes no new constraint; it does not count a free variable.
Wrong: “Full rank means every variable is a pivot variable.” For a rectangular matrix, full rank means \(\operatorname{rank}\mathbf{A}=\min(m,n)\). In Worked Example 3, rank 2 is full rank for a \(2\times3\) matrix, but the third column has no pivot and the nullity is 1. Only full column rank, meaning rank \(n\), leaves no free variables in a homogeneous system with \(n\) columns.
Wrong: “The pivot columns of the RREF give a basis for the original column space.” They give pivot-column information for the reduced matrix. As Worked Example 2 demonstrates, the reduced and original column spaces can differ. Take the pivot indices from the RREF, then select the corresponding columns of the original matrix.
| What you need | What to count or select | What not to confuse it with |
|---|---|---|
| Rank | Pivot columns, or nonzero rows in RREF | Number of rows in the original matrix |
| Nullity | Nonpivot columns: \(n-\operatorname{rank}\mathbf{A}\) | Number of zero rows |
| A basis for \(\operatorname{Col}\mathbf{A}\) | Original columns at the pivot indices | Columns of the RREF |
| Full rank | Rank equals \(\min(m,n)\) | Necessarily having no free variables |
The distinctions matter because a correct row reduction can still lead to an incorrect conclusion if the wrong feature is read from it. First note the matrix dimensions, then locate the pivots, and finally connect the count or indices to the specific space or system in question.
Check Your Understanding
For each question, identify the matrix feature that supports your conclusion.
- A \(4\times5\) matrix has three pivots. What are its rank and nullity, and how many free variables does its homogeneous system have?
- Can a \(3\times2\) matrix have full rank and a nonzero null space? Explain using the definition of full rank and the number of columns.
- What do you select from the original matrix to form a basis for its column space after row reduction?
- Give one reason a zero row in RREF does not, by itself, tell you the nullity.
- Which is preserved by elementary row operations: row space or column space? State what happens to the other in general.
