Reduced Row Echelon Form

From Echelon Form to Reduced Row Echelon Form

Echelon form organizes the nonzero rows of a matrix: zero rows come last, and the first nonzero entries of successive nonzero rows move to the right. This structure makes a matrix easier to read, but entries above a leading entry can still make a column harder to interpret. Reduced row echelon form adds two requirements: every leading entry is \(1\), and every other entry in its column is \(0\). Together, these conditions make each leading-entry column especially simple.

We use the convention that the leading entry of a nonzero row is its first nonzero entry, read from left to right. For example, the leading entry of \((0,0,-3,2)\) is \(-3\), in column 3. A zero row has no leading entry. As in echelon form, a matrix in reduced row echelon form has all zero rows below its nonzero rows, and the leading-entry positions move strictly to the right as we move down through the nonzero rows.

Thus, reduced row echelon form includes all the structural conditions of echelon form and imposes two further restrictions. A leading entry that is \(1\) is not enough on its own: its column must also be zero everywhere else. Conversely, a column of zeros does not cause a problem, because the definition restricts only columns that contain a leading entry.

Reading the Additional Conditions

Suppose a nonzero row has its leading entry in column \(j\). In reduced row echelon form, that entry is \(1\), and all other entries in column \(j\) are \(0\). For a matrix with \(m\) rows, the entire column is therefore the standard coordinate vector \(\mathbf{e}_i\in\mathbb{R}^m\), where \(i\) is the row containing that leading entry. For instance, if the leading entry is in row 2, its column has a \(1\) in row 2 and zeros in every other row.

This description also highlights what echelon form alone permits. A leading entry may be any nonzero number, and its column may contain nonzero entries above it. Echelon form requires the entries below a leading entry to be zero, but it does not require the entries above it to vanish. Reduced row echelon form removes both of these freedoms: it normalizes the leading entry and clears its column.

Proof. Let a leading entry occur in row \(i\), column \(j\). By the definition of reduced row echelon form, this entry equals \(1\), and every other entry in column \(j\) equals \(0\). Hence column \(j\) has exactly one nonzero entry, namely the leading entry in row \(i\). If two nonzero rows had leading entries in column \(j\), then that column would have a nonzero entry in each of those two rows. This contradicts the requirement that the leading entry be the only nonzero entry in its column. \(\square\)

Using Row Operations to Reduce a Leading-Entry Column

Elementary row operations allow us to change a matrix while preserving row equivalence. In particular, multiplying a row by a nonzero scalar can change its leading entry to \(1\), and adding a multiple of that row to another row can clear an entry in the same column. The following result explains why these operations can be used on one leading-entry column without shifting the leading entries of the other rows.

Proof. Before the operations, row \(k\) has zeros in every column before \(j\), because its first nonzero entry is in column \(j\). Multiplication of row \(k\) by the nonzero scalar \(1/a\) changes its entry in column \(j\) to \(a(1/a)=1\), leaves its earlier entries zero, and leaves its later entries in the same positions. For a row \(i<k\), the row operation subtracts a multiple of row \(k\), whose entries before column \(j\) are zero. Thus no entry of row \(i\) before column \(j\) changes, including its leading entry. For a nonzero row \(i>k\), the echelon condition places its leading entry strictly to the right of column \(j\), so its entry in column \(j\) is already zero; subtracting zero times row \(k\) leaves that row unchanged. Any zero row below row \(k\) is unchanged for the same reason. Finally, in every row \(i\ne k\), subtracting its column-\(j\) entry times row \(k\) makes the new entry there equal to \(a_{ij}-a_{ij}\cdot1=0\). Therefore column \(j\) has a single \(1\) in row \(k\), and the leading-entry positions remain unchanged. \(\square\)

This proposition justifies the two local tasks involved in making a leading-entry column satisfy the reduced conditions. It does not say that changing one column automatically makes every other column reduced; those columns must be checked as well. Each row operation used is reversible: scaling a row by \(1/a\) can be undone by scaling it by \(a\), and subtracting a multiple of one row can be undone by adding that multiple back. This is why the resulting matrix is row-equivalent to the original.

Common Mistake: A Leading One Is Not Enough

A frequent error is to stop once all leading entries equal \(1\). For example, consider

The leading entries are both \(1\), and their positions move from column 1 to column 2, so the matrix is in echelon form. A mistaken check might label it reduced because the leading entries are normalized. But column 2 contains \(3\) above its leading \(1\), so that leading entry is not the only nonzero entry in its column. The matrix is not in reduced row echelon form.

To correct the calculation, use \(R_1\to R_1-3R_2\):

Now each leading entry equals \(1\) and is the only nonzero entry in its column. The mistake was checking only the values of the leading entries, not the rest of their columns.

Use the conditions in a fixed order when checking a matrix: first verify echelon form; then inspect whether every leading entry equals \(1\); finally check that every other entry in each leading-entry column is zero. Entries in columns without leading entries are not required to vanish.