Multiplication Of Diagonal Matrices

Recognizing a Diagonal Matrix

A diagonal matrix has a simple pattern: entries away from the main diagonal are zero, while the entries on the main diagonal may be any real numbers. This pattern makes multiplication particularly direct. Instead of mixing entries from different rows and columns, the product of two diagonal matrices pairs entries in the same diagonal position.

Throughout, matrices have real entries. A diagonal matrix is square, so write its order as \(n\times n\). We use \(\operatorname{diag}(d_1,\dots,d_n)\) to denote the matrix whose diagonal entries are \(d_1,\dots,d_n\) and whose other entries are all zero. The values \(d_i\) are allowed to be zero.

The two descriptions are equivalent because the diagonal entries specify every entry that is not required to be zero. For \(n=1\), there are no off-diagonal positions, so every \(1\times1\) matrix is diagonal. A diagonal matrix need not have nonzero diagonal entries; for example, \(\operatorname{diag}(2,0,-1)\) is diagonal.

Scalar Multiples: Scale Each Diagonal Entry

For a real scalar \(c\), the scalar multiple \(c\mathbf{D}\) is obtained by multiplying every entry of \(\mathbf{D}\) by \(c\). Since every off-diagonal entry of \(\mathbf{D}\) is zero, scaling those entries leaves them zero. Only the diagonal entries need to be calculated.

Proof. By the definition of scalar multiplication, the entry in position \((i,j)\) of \(c\mathbf{D}\) is \(c d_{ij}\). If \(i\ne j\), then \(d_{ij}=0\), so \(c d_{ij}=0\). If \(i=j\), that entry is \(c d_{ii}=c d_i\). Thus the off-diagonal entries remain zero and the \(i\)th diagonal entry becomes \(c d_i\), as stated. \(\square\)

The scalar \(c=0\) is included in the theorem: \(0\mathbf{D}\) is the zero matrix. If \(c\ne0\), each diagonal entry is scaled by the same nonzero number; an entry that was zero remains zero. These conclusions follow directly from the entry formula and do not require any condition on the diagonal entries of \(\mathbf{D}\).

Products: Pair Entries in the Same Position

Two diagonal matrices can be multiplied when they have the same order. Indeed, if \(\mathbf{D}\) is \(n\times n\) and \(\mathbf{E}\) is \(m\times m\), their product is defined only when the number of columns of \(\mathbf{D}\) equals the number of rows of \(\mathbf{E}\), which requires \(n=m\). Let \(\mathbf{D}=\operatorname{diag}(d_1,\dots,d_n)\) and \(\mathbf{E}=\operatorname{diag}(e_1,\dots,e_n)\). The entry formula for matrix multiplication gives \((\mathbf{D}\mathbf{E})_{ij}=\sum_{k=1}^{n}d_{ik}e_{kj}\). In this sum, the diagonal pattern forces all terms to vanish unless \(i=j=k\).

Proof. Fix any row index \(i\) and column index \(j\). The entry formula gives \((\mathbf{D}\mathbf{E})_{ij}=\sum_{k=1}^{n}d_{ik}e_{kj}\). Since \(\mathbf{D}\) is diagonal, \(d_{ik}=0\) unless \(k=i\); therefore the only possible nonzero term is \(d_{ii}e_{ij}\). Since \(\mathbf{E}\) is diagonal, \(e_{ij}=0\) unless \(i=j\). Thus \((\mathbf{D}\mathbf{E})_{ij}=0\) when \(i\ne j\). When \(i=j\), the surviving term is \(d_{ii}e_{ii}=d_i e_i\). Every off-diagonal entry is zero and every diagonal entry is \(d_i e_i\), proving the formula. \(\square\)

The formula also shows that the factors commute. This is a special property of diagonal matrices: it does not mean that arbitrary matrices commute.

Proof. By the product formula, the \(i\)th diagonal entry of \(\mathbf{D}\mathbf{E}\) is \(d_i e_i\), and the \(i\)th diagonal entry of \(\mathbf{E}\mathbf{D}\) is \(e_i d_i\). Real numbers commute under multiplication, so \(d_i e_i=e_i d_i\) for each \(i\). Both products have zero off-diagonal entries. Hence their corresponding entries agree everywhere, and \(\mathbf{D}\mathbf{E}=\mathbf{E}\mathbf{D}\). \(\square\)

When Is the Product the Zero Matrix?

The product formula gives an entry-by-entry test: the product is the zero matrix exactly when every paired diagonal product is zero. This does not require either factor to be the zero matrix; different positions can have zero entries in different factors.

Proof. The product formula says \(\mathbf{D}\mathbf{E}=\operatorname{diag}(d_1e_1,\dots,d_ne_n)\). A diagonal matrix equals the zero matrix exactly when each of its diagonal entries is zero. Therefore, \(\mathbf{D}\mathbf{E}=\mathbf{O}\) exactly when \(d_i e_i=0\) for every \(i\). \(\square\)

How the Diagonal Entries Act on a Vector

The same entry pattern explains the effect of a diagonal matrix on a column vector. If \(\mathbf{x}=(x_1,\dots,x_n)^{\top}\), then the \(i\)th entry of \(\mathbf{D}\mathbf{x}\) is \(\sum_{j=1}^{n}d_{ij}x_j=d_i x_i\), because \(d_{ij}=0\) for \(j\ne i\). Thus \(\mathbf{D}\) scales each coordinate separately:

This coordinate-by-coordinate view matches the product rule: applying \(\mathbf{E}\) first scales coordinate \(i\) by \(e_i\), and then applying \(\mathbf{D}\) scales that coordinate by \(d_i\), giving \(d_i e_i\). It also highlights the limits of the rule. The product formula here applies when both factors are diagonal and of the same order; it does not say that an arbitrary matrix can be multiplied by pairing its diagonal entries.

Common Mistake: Adding Instead of Multiplying

A frequent error is to add corresponding diagonal entries when calculating a matrix product. For example, for \(\mathbf{D}=\operatorname{diag}(2,-1)\) and \(\mathbf{E}=\operatorname{diag}(3,4)\), writing \(\mathbf{D}\mathbf{E}=\operatorname{diag}(5,3)\) incorrectly adds \(2+3\) and \(-1+4\). Matrix multiplication uses products of entries, not sums of matching entries. The correct calculation is

Check: the first row times the first and second columns gives \(6\) and \(0\); the second row times those columns gives \(0\) and \(-4\). The result agrees with the diagonal product rule. Adding matching entries belongs to matrix addition, not matrix multiplication.