How the Field Changes Rank and Invertibility

One Integer Matrix, Different Arithmetic

An integer matrix can be viewed over different fields by interpreting each entry in the chosen field. Over \(\mathbb{R}\), the entries retain their usual values. Over \(\mathbb{F}_p\), where \(p\) is prime, each integer is replaced by its residue class modulo \(p\). The matrix entries may look similar, but the arithmetic governing row reduction has changed: for example, \(2\) is nonzero in \(\mathbb{R}\), while its residue is zero in \(\mathbb{F}_2\). A pivot that exists over one field may disappear over another.

For a matrix over a field, its rank is the dimension of the span of its columns over that field. Row reduction computes this rank by counting pivot columns, with every row operation performed in the field being used. An integer matrix does not have a single field-independent rank: its rank is determined only after its scalar field has been specified.

For instance, modulo \(2\), the classes of \(0\) and \(2\) are both zero, while the class of \(3\) is \(1\). The field operations then use \(1+1=0\) and \(1\cdot1=1\). In particular, a nonzero integer need not give a nonzero element of \(\mathbb{F}_p\).

Invertibility and the Determinant in Each Field

For a square matrix over a field, invertibility can be tested by row reduction: the matrix is invertible exactly when there is a pivot in every column. The determinant gives the corresponding scalar test. The next theorem states both criteria for an arbitrary field \(\mathbb{F}\). Here, \(\det\mathbf{A}\ne0\) means nonzero in that field.

Proof. Each elementary row operation is multiplication on the left by an invertible elementary matrix, so it preserves whether the matrix is invertible. If the RREF is \(\mathbf{I}_n\), the row operations give \(\mathbf{E}\mathbf{A}=\mathbf{I}_n\), where \(\mathbf{E}\) is a product of invertible elementary matrices. Thus \(\mathbf{A}\) has inverse \(\mathbf{E}\). If the RREF has fewer than \(n\) pivots, the homogeneous system \(\mathbf{A}\mathbf{x}=\mathbf{0}\) has a free variable and hence a nonzero solution. An invertible matrix cannot have such a solution: multiplying by its inverse would give \(\mathbf{x}=\mathbf{0}\). This proves the equivalence of (1) and (2).

For the determinant comparison, row swaps multiply the determinant by \(-1\), scaling a row by a nonzero \(c\) multiplies it by \(c\), and adding a multiple of one row to another leaves it unchanged. These rules follow from the alternating and multilinear properties of the determinant. Each operation therefore preserves whether the determinant is zero. The RREF of a square matrix with \(n\) pivots is \(\mathbf{I}_n\), whose determinant is \(1\); if there are fewer pivots, the RREF has a zero row, so its determinant is \(0\). Consequently, \(\det\mathbf{A}\ne0\) exactly when the RREF has \(n\) pivots. This proves the equivalence with (3). \(\square\)

Now apply this criterion to an integer matrix. Its determinant is an integer, and reducing that integer modulo \(p\) gives the determinant of the reduced matrix. To see why, use the determinant formula as a signed sum of products of entries: reducing the entries before forming the sum gives the same residue as reducing the integer sum and products afterward.

Proof. By the determinant formula, every term in \(\det\mathbf{A}\) is a signed product of integer entries. Reducing all entries modulo \(p\) reduces each such product modulo \(p\), and reducing the sum gives the same result as summing the reduced terms. Therefore the determinant of the reduced matrix is the residue class of the integer \(\det\mathbf{A}\). The first theorem says that a square matrix over either field is invertible exactly when its determinant is nonzero in that field. A nonzero integer is nonzero as a real number, while its residue in \(\mathbb{F}_p\) is nonzero exactly when it is not divisible by \(p\). The integer zero has zero residue for every prime \(p\). These facts give all three claims. \(\square\)

A Determinant Can Lose Its Nonzero Value

Comparing Ranks by Row Reduction

The determinant test applies only to square matrices and answers whether the rank is full. Row reduction also detects changes in rank for square or rectangular matrices. When reducing modulo \(p\), a nonzero integer entry or pivot may become zero, and operations that were available over \(\mathbb{R}\) may no longer be valid. The examples below show both the loss of a pivot and its effect on rank.

Why the Solution Set Can Change

Rank and invertibility affect systems, but one should distinguish a matrix from a particular system using that matrix. A singular coefficient matrix does not, by itself, determine whether a system is consistent: the right-hand side matters. The first worked example had no solution modulo \(2\), but another right-hand side could yield solutions. For example, with the same reduced coefficient matrix \(\begin{bmatrix}1&1\\1&1\end{bmatrix}\) over \(\mathbb{F}_2\) and right-hand side \(\begin{bmatrix}1\\1\end{bmatrix}\), the equations agree and have solutions. The matrix has rank \(1\), so it is not invertible and cannot give a unique solution for every right-hand side.

For an invertible square matrix, every right-hand side has exactly one solution: multiplication by the inverse gives that solution, and any solution must equal it after multiplying by the inverse. If the determinant changes from nonzero to zero after reduction, that guarantee is lost. A given system may become inconsistent or may have multiple solutions; the augmented matrix must be checked to decide which.

Common Pitfall: Dividing by a Zero Residue

A frequent error is to see a nonzero integer pivot and assume it remains a valid pivot modulo every prime. For the matrix in the second worked example, a tempting but invalid step over \(\mathbb{F}_2\) is to divide the second row by \(2\), just as one could over \(\mathbb{R}\). The wrong calculation treats \(2\) as invertible and tries to turn the row \((0,2,1)\) into \((0,1,\tfrac12)\). In \(\mathbb{F}_2\), however, this row is \((0,0,1)\): the residue of \(2\) is zero, and zero has no multiplicative inverse.

The correction is to reduce every entry modulo \(p\) first, then perform row operations using only nonzero field elements as scaling factors. In \(\mathbb{F}_2\), the row \((0,0,1)\) already has its pivot in column \(3\); it cannot create a pivot in column \(2\). Likewise, a determinant that is a nonzero integer does not guarantee invertibility modulo \(p\): check whether the prime divides it.