What Conditions Make the Product of Two Matrices Zero?

The condition $AB=0$ does not usually force either matrix to be zero. Instead, it says that the output of $B$ always lands in directions that $A$ sends to zero.

Suppose $A$ is an $m\times n$ matrix and $B$ is an $n\times p$ matrix, with entries over a field such as $\mathbb{R}$ or $\mathbb{C}$. Their product is an $m\times p$ matrix. The key condition is

$$\operatorname{im}(B)\subseteq\ker(A).$$

Here, $\operatorname{im}(B)$ is the space spanned by the columns of $B$, and $\ker(A)$ is the set of vectors $x$ for which $Ax=0$.

The exact condition for $AB=0$

View the matrices as linear maps:

$$B: \mathbb{F}^p\to\mathbb{F}^n, \qquad A: \mathbb{F}^n\to\mathbb{F}^m.$$

The composition $AB$ first applies $B$, then applies $A$. Thus $AB=0$ precisely when $A(Bx)=0$ for every $x\in\mathbb{F}^p$. This means that each vector produced by $B$ is in the kernel of $A$.

Equivalently, check the columns individually: if $b_1,\ldots,b_p$ are the columns of $B$, then

$$AB=0 \quad\Longleftrightarrow\quad Ab_j=0\text{ for every }j.$$

This is often the quickest way to test a particular pair of matrices.

A rank condition that must hold

By rank-nullity, the dimension of the kernel of $A$ is $n-\operatorname{rank}(A)$. If $AB=0$, then the image of $B$ is contained in that kernel, so

$$\operatorname{rank}(B)\leq \dim\ker(A)=n-\operatorname{rank}(A).$$

Therefore,

$$\operatorname{rank}(A)+\operatorname{rank}(B)\leq n.$$

The number $n$ here is the shared inner dimension: the number of columns of $A$, which must equal the number of rows of $B$.

This rank inequality is necessary for a given pair to have zero product, but it is not sufficient by itself. The column space of $B$ must actually lie in the kernel of $A.

Worked example

Let

$$A=\begin{bmatrix} 1&0&0\\ 0&0&0 \end{bmatrix}, \qquad B=\begin{bmatrix} 0&0\\ 1&0\\ 0&1 \end{bmatrix}.$$

Here $A$ is $2\times3$ and $B$ is $3\times2$, so the product is defined and will be $2\times2$. The columns of $B$ are

$$\begin{bmatrix}0\\1\\0\end{bmatrix} \quad\text{and}\quad \begin{bmatrix}0\\0\\1\end{bmatrix}.$$

The kernel of $A$ consists of vectors whose first coordinate is zero. Both columns of $B$ satisfy that condition. Direct multiplication confirms it:

$$AB= \begin{bmatrix} 1&0&0\\ 0&0&0 \end{bmatrix} \begin{bmatrix} 0&0\\ 1&0\\ 0&1 \end{bmatrix} = \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix}.$$

In this example, $\operatorname{rank}(A)=1$ and $\operatorname{rank}(B)=2$, so their ranks add to $3$, the shared inner dimension.

What the condition says in special cases

• If $A$ is invertible, then $\ker(A)=\{0\}$. The condition $AB=0$ therefore forces $B=0$.
• If $B$ has full row rank $n$, then $\operatorname{im}(B)=\mathbb{F}^n$. For $AB=0$, $A$ must vanish on all of $\mathbb{F}^n$, so $A=0$.
• If both matrices are square and invertible, their product cannot be zero.
• If both matrices are nonzero, the shared dimension must be at least $2$. For inner dimension $1$, a nonzero $A$ is a nonzero column and a nonzero $B$ is a nonzero row; their product is a nonzero outer product.

In particular, for square $n\times n$ matrices, a nonzero product-zero pair requires each matrix to have rank at most $n-1$, and their ranks must add to at most $n$.

Common mistakes

Assuming one factor must be zero. This is true for scalar numbers, but not for matrices. For instance, the worked example has two nonzero factors and a zero product.

Using only the rank inequality. The inequality limits the possible ranks, but it does not establish that the vectors in the image of $B$ are in the kernel of $A$. Check the columns or compute the product.

Confusing the dimensions. The product $AB$ is defined when the number of columns of $A$ equals the number of rows of $B$. Its zero matrix has the outer dimensions, $m\times p$; it need not be square.

Reversing the order. In general, $AB=0$ does not imply $BA=0$. The two products may even have different dimensions or one may be undefined. The relevant condition for $AB=0$ is that $B$‘s image lies in $A$‘s kernel.