Zero Products And Failed Cancellation

When a Product Is Zero

For real numbers, a product \(ab=0\) forces at least one factor to be zero. Matrix multiplication behaves differently: a nonzero matrix can send a nonzero vector to the zero vector, because the entries in each row-column product can cancel. As a result, two nonzero matrices can have a zero product, and a nonzero matrix cannot generally be cancelled from an equation such as \(\mathbf{A}\mathbf{B}=\mathbf{A}\mathbf{C}\).

Throughout, matrices have real entries. Let \(\mathbf{A}\in\mathbb{R}^{m\times n}\) and \(\mathbf{B}\in\mathbb{R}^{n\times p}\), so their product is defined and has size \(m\times p\). The zero matrix of that size is denoted \(\mathbf{O}\). The entry formula for the product is \((\mathbf{A}\mathbf{B})_{ij}=\sum_{k=1}^{n}a_{ik}b_{kj}\). Each such entry is a sum of products; it can be zero even when many of its terms are nonzero.

A useful way to read the product is one column at a time. Write the columns of \(\mathbf{B}\) as \(\mathbf{b}_1,\dots,\mathbf{b}_p\). By the entry formula, the \(j\)th column of \(\mathbf{A}\mathbf{B}\) is \(\mathbf{A}\mathbf{b}_j\). Thus a product is zero precisely when every column of \(\mathbf{B}\) solves the homogeneous equation \(\mathbf{A}\mathbf{x}=\mathbf{0}\).

Proof. The \(j\)th column of \(\mathbf{A}\mathbf{B}\) has \(i\)th entry \(\sum_{k=1}^{n}a_{ik}b_{kj}\), which is exactly the \(i\)th entry of \(\mathbf{A}\mathbf{b}_j\). Therefore the \(j\)th column of \(\mathbf{A}\mathbf{B}\) is \(\mathbf{A}\mathbf{b}_j\). A matrix is zero exactly when all its columns are zero. Hence \(\mathbf{A}\mathbf{B}=\mathbf{O}\) exactly when \(\mathbf{A}\mathbf{b}_j=\mathbf{0}\) for every \(j\). \(\square\)

Building a Zero Product

This construction illustrates the central mechanism: the columns of \(\mathbf{B}\) can be nonzero while still being sent to zero by \(\mathbf{A}\). In the first column, the row products cancel: \(2-2=0\) and \(4-4=0\). The zero product does not arise because all the individual factors in those sums are zero.

Proof. Suppose first that a nonzero \(\mathbf{B}\) satisfies \(\mathbf{A}\mathbf{B}=\mathbf{O}\). Since \(\mathbf{B}\) is nonzero, at least one of its columns, say \(\mathbf{b}_j\), is nonzero. The preceding theorem gives \(\mathbf{A}\mathbf{b}_j=\mathbf{0}\). Thus \(\mathbf{x}=\mathbf{b}_j\) is a nonzero vector with \(\mathbf{A}\mathbf{x}=\mathbf{0}\). Conversely, suppose \(\mathbf{x}\ne\mathbf{0}\) and \(\mathbf{A}\mathbf{x}=\mathbf{0}\). Form \(\mathbf{B}\) with first column \(\mathbf{x}\) and all its other columns zero. This matrix is nonzero, and every column \(\mathbf{b}_j\) satisfies \(\mathbf{A}\mathbf{b}_j=\mathbf{0}\). The preceding theorem therefore gives \(\mathbf{A}\mathbf{B}=\mathbf{O}\). \(\square\)

Why Left Cancellation Can Fail

For real numbers, if \(a\ne0\) and \(ab=ac\), division by \(a\) gives \(b=c\). For matrices, having \(\mathbf{A}\ne\mathbf{O}\) is not enough to justify the same conclusion. Different matrices can have the same product with \(\mathbf{A}\) on the left if their columns differ by vectors that \(\mathbf{A}\) sends to zero.

The exact condition for left cancellation can be stated using the solutions of the homogeneous equation. Say that \(\mathbf{A}\) has only the trivial homogeneous solution if \(\mathbf{A}\mathbf{x}=\mathbf{0}\) implies \(\mathbf{x}=\mathbf{0}\). This condition is stronger than merely saying \(\mathbf{A}\ne\mathbf{O}\).

Proof. First suppose the only solution to \(\mathbf{A}\mathbf{x}=\mathbf{0}\) is \(\mathbf{x}=\mathbf{0}\), and suppose \(\mathbf{A}\mathbf{B}=\mathbf{A}\mathbf{C}\). For each column index \(j\), equality of the \(j\)th columns says that for every row \(i\), \(\sum_{k=1}^{n}a_{ik}b_{kj}=\sum_{k=1}^{n}a_{ik}c_{kj}\). Subtracting the right side gives \(\sum_{k=1}^{n}a_{ik}(b_{kj}-c_{kj})=0\) for every \(i\). Thus \(\mathbf{A}(\mathbf{b}_j-\mathbf{c}_j)=\mathbf{0}\). By the assumed condition, \(\mathbf{b}_j-\mathbf{c}_j=\mathbf{0}\), so \(\mathbf{b}_j=\mathbf{c}_j\). This holds for every column, and hence \(\mathbf{B}=\mathbf{C}\). Conversely, suppose there is a nonzero \(\mathbf{x}\) with \(\mathbf{A}\mathbf{x}=\mathbf{0}\). Let \(\mathbf{B}\) have first column \(\mathbf{x}\) and all remaining columns zero, and let \(\mathbf{C}=\mathbf{O}\). Then \(\mathbf{B}\ne\mathbf{C}\), while \(\mathbf{A}\mathbf{B}=\mathbf{O}=\mathbf{A}\mathbf{C}\). Therefore the stated cancellation implication cannot hold for every pair \(\mathbf{B},\mathbf{C}\). \(\square\)

The Order of the Factors Matters

A zero product in one order does not force a zero product in the other order. The dimensions of the factors must also be checked: \(\mathbf{A}\mathbf{B}\) and \(\mathbf{B}\mathbf{A}\) can both be formed when the matrices are square, but their entries can differ.

Common Mistake: Cancelling From a Zero Product

A tempting but incorrect claim is: “\(\mathbf{A}\mathbf{B}=\mathbf{O}\) and \(\mathbf{A}\ne\mathbf{O}\), so \(\mathbf{B}=\mathbf{O}\).” The first worked example directly disproves this claim. In that example, the top-left entry is computed as \(1(2)+2(-1)=2-2=0\), even though the entries \(1,2,2,-1\) involved in the calculation are not all zero. The other entries vanish as well, but the cancellation in these sums does not make \(\mathbf{B}\) the zero matrix.

The correction is to inspect the homogeneous equation \(\mathbf{A}\mathbf{x}=\mathbf{0}\). If it has a nonzero solution, that solution can be placed in a column of a nonzero matrix \(\mathbf{B}\), producing \(\mathbf{A}\mathbf{B}=\mathbf{O}\). If its only solution is zero, the left cancellation criterion proves that equal products \(\mathbf{A}\mathbf{B}=\mathbf{A}\mathbf{C}\) force \(\mathbf{B}=\mathbf{C}\). In either case, nonzeroness of \(\mathbf{A}\) alone does not settle cancellation.

When checking a proposed argument, first confirm that the matrix products have compatible dimensions. Then distinguish between the statement that a factor is nonzero and the stronger condition that its homogeneous equation has only the trivial solution. The latter condition, not the former, is what the left cancellation proof requires.