Distributive Laws For Matrix Products

What Distribution Means for Matrix Products

For real numbers, multiplication distributes over addition. Matrix multiplication has a corresponding rule, but the sizes of the matrices must allow every expression to be formed. There are two versions: addition can occur in the right factor, or in the left factor. The formulas look similar, but their dimensions and the entries being compared are different. We will prove each version by expanding the entry formula for a product.

For \(\mathbf{A}\in\mathbb{R}^{m\times n}\) and \(\mathbf{B}\in\mathbb{R}^{n\times p}\), the product \(\mathbf{A}\mathbf{B}\) is an \(m\times p\) matrix. Its entry in row \(i\), column \(j\) is \(\sum_{k=1}^{n}a_{ik}b_{kj}\). Matrix addition is defined for matrices of the same size, entry by entry. Thus any distributive law must respect both the product size requirements and the requirement that the matrices being added have the same size.

Proof by Comparing Entries

The proof relies on equality of matrices being equality of all corresponding entries. We first check that the expressions in each law are defined and have matching sizes. In the first law, \(\mathbf{B}+\mathbf{C}\) is \(n\times p\), so \(\mathbf{A}(\mathbf{B}+\mathbf{C})\) is \(m\times p\). Both \(\mathbf{A}\mathbf{B}\) and \(\mathbf{A}\mathbf{C}\) are also \(m\times p\), and they can be added. For the second law, \(\mathbf{A}+\mathbf{D}\) is \(m\times n\), so \((\mathbf{A}+\mathbf{D})\mathbf{C}\) is \(m\times p\); each product on the right is \(m\times p\) as well.

Proof. For the first law, fix any row index \(i\) with \(1\le i\le m\) and column index \(j\) with \(1\le j\le p\). By the product-entry formula and the definition of matrix addition, the \((i,j)\) entry on the left is

Distributing multiplication over addition for real numbers, then separating the finite sums, gives

The last expression is the \((i,j)\) entry of \(\mathbf{A}\mathbf{B}+\mathbf{A}\mathbf{C}\). This equality holds for every permitted \(i\) and \(j\), so the matrices are equal.

For the second law, fix the same kinds of indices \(i\) and \(j\). This time, the entries being added are those of \(\mathbf{A}\) and \(\mathbf{D}\). The product-entry formula gives

Again using real-number distributivity and separating the sums, we obtain

This is the \((i,j)\) entry of \(\mathbf{A}\mathbf{C}+\mathbf{D}\mathbf{C}\), and it holds for every row and column index. Therefore the matrices are equal. \(\square\)

These proofs show why the laws hold: the same summation index runs through the matching row and column entries, and scalar multiplication distributes inside each term. The order of the factors is preserved throughout. The first proof has \(a_{ik}\) multiplying each of \(b_{kj}\) and \(c_{kj}\); the second has \(c_{kj}\) multiplying each of \(a_{ik}\) and \(d_{ik}\). Neither proof switches the order of two matrix factors.

Computing Both Sides

The Laws Also Hold for Rectangular Matrices

The dimension conditions do not require square matrices. In particular, a matrix with \(m\) rows and \(n\) columns can multiply a matrix with \(n\) rows and \(p\) columns, producing an \(m\times p\) result. Each of the two distributive laws compares matrices of precisely that output size. The indices in the proof also explain why: \(i\) chooses one of the \(m\) rows, \(j\) one of the \(p\) columns, and \(k\) ranges through the \(n\) matched entries used in the product.

Common Mistake: Changing the Factor That Is Shared

Distribution allows a factor to be shared across the two terms, but the shared factor must stay in its original position. A tempting incorrect statement is \(\mathbf{A}(\mathbf{B}+\mathbf{C})=\mathbf{A}\mathbf{B}+\mathbf{B}\mathbf{C}\). The second term on the right has changed the left factor from \(\mathbf{A}\) to \(\mathbf{B}\); the proved law instead gives \(\mathbf{A}\mathbf{B}+\mathbf{A}\mathbf{C}\).

Here is a concrete calculation showing the error. Let \(\mathbf{A}=\begin{bmatrix}1&2\\0&1\end{bmatrix}\), \(\mathbf{B}=\begin{bmatrix}1&0\\1&1\end{bmatrix}\), and \(\mathbf{C}=\begin{bmatrix}0&1\\2&0\end{bmatrix}\). The left side of the incorrect claim is

But the proposed right side is \(\mathbf{A}\mathbf{B}+\mathbf{B}\mathbf{C}\). Direct multiplication gives \(\mathbf{A}\mathbf{B}=\begin{bmatrix}3&2\\1&1\end{bmatrix}\) and \(\mathbf{B}\mathbf{C}=\begin{bmatrix}0&1\\2&1\end{bmatrix}\), so

The correction is to keep \(\mathbf{A}\) as the common left factor in both terms: \(\mathbf{A}(\mathbf{B}+\mathbf{C})=\mathbf{A}\mathbf{B}+\mathbf{A}\mathbf{C}\). A reliable check is to write down the factor that appears in both summands before multiplying: it must be the same matrix on the same side.

Why the Entry Proof Is Useful

The laws make expressions with sums easier to compute and simplify. If two matrices are added before multiplying, distribution lets us compute two products and then add them; conversely, if the same factor occurs in two products, it can be factored out. The entry proof also supplies a dependable way to check proposed matrix identities: verify that the expressions have compatible sizes, then compare their entries using the product formula.

Do not infer more than the laws state. They justify distributing a single product over a sum in either factor. They do not say that matrix multiplication is commutative, and they do not permit rearranging the order of factors. The formulas preserve order: \(\mathbf{A}\mathbf{B}\) remains \(\mathbf{A}\mathbf{B}\), not \(\mathbf{B}\mathbf{A}\). Always check dimensions before applying a law; an expression that is not defined cannot be made valid by a symbolic rearrangement.