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

In what follows, let , , and 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: where is assumed to be and denotes the -th entry of .

The distributive property holds:

Proof

The -th block of is Since is the -th block of and is the -th block of , and the above equality holds for every and , the claim is true.

It holds also for the second factor:

Proof

Let be a scalar. Then,

Proof

We can see the scalar as a matrix having a single entry. Then, the Kronecker product has a single block equal to .

Moreover, if is a scalar, then

Proof

Suppose that is . By applying the definition of Kronecker product and that of multiplication of a matrix by a scalar, we obtain

A more general rule regarding the multiplication by scalars and follows:

Proof

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

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:

The associative property holds:

Proof

Let be , be and be . Let us first study the structure of . The product is the entry of . As a consequence, the product is theentry of . Let us now study the structure of . The product is the entry of . Therefore, the product is the entry of that occupies positionThus, the product occupies the same position in and in for every , , , , and . Therefore,

If , , and are such that the products and are well-defined, then

Proof

Suppose is and is . where: in step we have used the fact that the multiplication of two block matrices can be carried out as if their blocks were scalars; in step we have used the definition of matrix multiplication to deduce thatwhere is the -th entry of .

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

In the case in which is a column vector, the above equality becomes

Transposition operates as follows:

Proof

Let be . Let us apply the rule for transposing a block matrix:

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

Proof

We need to use the rule for mixed products and verify that satisfies the definition of inverse of : where are identity matrices.

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

Proof

It should be pretty intuitive. Suppose that is , is , is , is , is , and . Then, we have

If and are square matrices, then the trace satisfies

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 is . Then, we have

Please cite as:

Taboga, Marco (2021). "Properties of the Kronecker product", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/Kronecker-product-properties.

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