📂Matrix Algebra

Kronecker Product of Matrices

Definition¹

The Kronecker product of two matrices $A = [a_{ij}] \in M_{m \times n}$, $B \in M_{p \times q}$ is defined as follows.

$$ A \otimes B := \begin{bmatrix} a_{11} B & \cdots & a_{1n} B \\ \vdots & \ddots & \vdots \\ a_{m1} B & \cdots & a_{mn} B \end{bmatrix} \in M_{mp \times nq} $$

Explanation

The matrix representation of the tensor product of two linear transformations is defined by the Kronecker product of each matrix representation of the linear transformations.

Unlike multiplication, it is well-defined regardless of the size of the matrices.

Properties

For a matrix $A, B, C, D$ and a scalar $k$, the following hold:

• Scalar multiplication: $(kA) \otimes B = k (A \otimes B) = A \otimes (kB) \\[0.5em]$
• Matrix multiplication: $(A \otimes B) (C \otimes D) = (AC) \otimes (BD) \\[0.5em]$
• Associative law: $(A \otimes B) \otimes C = A \otimes (B \otimes C) \\[0.5em]$
• Distributive law: $A \otimes ( B + C ) = A \otimes B + A \otimes C \\[0.5em]$
• Transpose: $(A \otimes B)^{T} = A^{T} \otimes B^{T} \\[0.5em]$
• Trace: $\tr(A \otimes B) = \tr(A) \tr(B) \\[0.5em]$
• Inverse: $(A \otimes B)^{-1} = A^{-1} \otimes B^{-1}$
  • if $A \otimes B$ is invertible, then $\iff$ $A$ and $B$ are invertible, then $\\[0.5em]$
• Determinant: $\det (A \otimes B) = \det(A)^{n} \det(B)^{p},\quad A\in M_{n \times n}, B \in B_{p \times p} \\[0.5em]$

Like matrix multiplication, the commutative law does not hold. Simply look at the multiplication of $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and the identity matrix $I_{2}$,

$$ A \otimes I_{2} = \begin{bmatrix} a & 0 & b & 0 \\ 0 & a & 0 & b \\ c & 0 & d & 0 \\ 0 & c & 0 & d \end{bmatrix} \ne \begin{bmatrix} a & b & 0 & 0 \\ c & d & 0 & 0 \\ 0 & 0 & a & b \\ 0 & 0 & c & d \end{bmatrix} = I_{2} \otimes A $$