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Properties of the Kronecker product

by , PhD

The Kronecker product has several properties that are often exploited in applications.

Table of Contents

Preliminaries

In what follows, let A, $B$, $C$ and $D$ denote matrices whose dimensions can be arbitrary unless these matrices need to be multiplied or added together, in which case we require that they be conformable for addition or multiplication, as needed.

Remember that the Kronecker product is a block matrix: [eq1]where A is assumed to be $K	imes L$ and $A_{kl}$ denotes the $left( k,l
ight) $-th entry of A.

Distributive property

The distributive property holds:[eq2]

Proof

The $left( k,l
ight) $-th block of [eq3] is [eq4]Since $A_{kl}C$ is the $left( k,l
ight) $-th block of $Aotimes C$ and $B_{kl}C$ is the $left( k,l
ight) $-th block of $Botimes C$, and the above equality holds for every k and $l$, the claim is true.

It holds also for the second factor:[eq5]

Proof

The $left( k,l
ight) $-th block of [eq6] is [eq7]Since $A_{kl}B$ is the $left( k,l
ight) $-th block of $Aotimes B$ and $A_{kl}C$ is the $left( k,l
ight) $-th block of $Aotimes C$, and the above equality holds for every k and $l$, the claim is true.

Multiplication by a scalar

Let $lpha $ be a scalar. Then,[eq8]

Proof

We can see the scalar $lpha $ as a $1	imes 1$ matrix having a single entry. Then, the Kronecker product $lpha otimes B$ has a single block equal to $lpha B$.

Moreover, if $eta $ is a scalar, then[eq9]

Proof

Suppose that A is $K	imes L$. By applying the definition of Kronecker product and that of multiplication of a matrix by a scalar, we obtain [eq10]

A more general rule regarding the multiplication by scalars $lpha $ and $eta $ follows:[eq11]

Proof

Again, by applying the definition of Kronecker product and that of multiplication of a matrix by a scalar, we obtain

[eq12]

Zero matrices

Clearly, any Kronecker product that involves a zero matrix (i.e., a matrix whose entries are all zeros) gives a zero matrix as a result:[eq13]

Associativity

The associative property holds:[eq14]

Proof

Let A be $K	imes L$, $B$ be $M	imes N$ and $C$ be $O	imes P$. Let us first study the structure of [eq15]. The product $A_{kl}B_{mn}$ is the [eq16]entry of $Aotimes B$. As a consequence, the product [eq17] is the[eq18]entry of [eq19]. Let us now study the structure of [eq20]. The product $B_{mn}C_{op}$ is the [eq21]entry of $Botimes C$. Therefore, the product [eq22] is the entry of [eq20] that occupies position[eq24]Thus, the product $A_{kl}B_{mn}C_{op}$ occupies the same position in [eq25] and in [eq20] for every k, $l$, $m$, n, $o$ and p. Therefore,[eq27]

Mixed products

If A, $B$, $C$ and $D$ are such that the products $AC$ and $BD$ are well-defined, then[eq28]

Proof

Suppose A is $K	imes L$ and $C$ is $L	imes M$. [eq29]where: in step $rame{A}$ we have used the fact that the multiplication of two block matrices can be carried out as if their blocks were scalars; in step $rame{B}$ we have used the definition of matrix multiplication to deduce that[eq30]where [eq31] is the $left( k,m
ight) $-th entry of $AC$.

An often used trick is to use identity matrices (and scalar 1s) in the mixed product. For example,[eq32]

In the case in which $B$ is a column vector, the above equality becomes[eq33]

Transposition

Transposition operates as follows:[eq34]

Proof

Let A be $K	imes L$. Let us apply the rule for transposing a block matrix:[eq35]

Inverse

The rule for computing the inverse of a Kronecker product is pretty simple:[eq36]

Proof

We need to use the rule for mixed products and verify that [eq37] satisfies the definition of inverse of $Aotimes B$: [eq38]where I are identity matrices.

Block matrices

Suppose that the matrix A is partitioned into blocks as follows:[eq39]Then,[eq40]In other words, the blocks of the matrix A can be treated as if they were scalars.

Proof

It should be pretty intuitive. Suppose that A is $K	imes L$, $lpha $ is $K_{1}	imes L_{1}$, $eta $ is $K_{1}	imes L_{2}$, $gamma $ is $K_{2}	imes L_{1}$, $delta $ is $K_{2}	imes L_{2}$, $K=K_{1}+K_{2}$ and $L=L_{1}+L_{2}$. Then, we have[eq41]

Trace

If A and $B$ are square matrices, then the trace satisfies [eq42]

Proof

Remember that the trace is the sum of the diagonal entries of a matrix. As a consequence, when a matrix is partitioned, its trace can also be computed as the sum of the traces of the diagonal blocks of the matrix. Moreover, the trace is homogeneous (in the sense that it preserves multiplication by scalars). Suppose that A is $K	imes K$. Then, we have[eq43]

How to cite

Please cite as:

Taboga, Marco (2017). "Properties of the Kronecker product", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/Kronecker-product-properties.

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