Kronecker Product¶

Definition¶

Let $A=(a_{ij})$ be an $m\times n$ matrix and $B$ a $p\times q$ matrix. The Kronecker product $A\otimes B$ is the $mp\times nq$ block matrix

whose $(i,j)$ block is the scalar multiple $a_{ij}B$. No compatibility of sizes is required.

Interpretation¶

Unlike the ordinary matrix product, the Kronecker product is defined for matrices of any sizes, and the dimensions multiply: every entry of $A$ is replaced by a whole copy of $B$. Entry by entry, $(A\otimes B)_{(i-1)p+k,\,(j-1)q+l}=a_{ij}b_{kl}$$a_{ij}{b}_{kl}$.

The figure uses $A=\begin{pmatrix}1&2\\3&4\end{pmatrix}$ and $B=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ from the examples. In $A\otimes B$ each entry $a_{ij}$ becomes the block $a_{ij}B$; in $B\otimes A$ each entry $b_{kl}$ becomes $b_{kl}A$. Both are $4\times4$ with the same entries, but arranged differently, so the product is not commutative.

It is the natural tool for linear maps acting on matrices or on two indices at once.

The product is not commutative: $A\otimes B$ and $B\otimes A$ have the same size and the same entries, but in a different arrangement. They differ by a reordering of rows and columns (they are similar via a permutation matrix when both factors are square).

Key Results¶

• Bilinear and associative: $(A+A')\otimes B=A\otimes B+A'\otimes B$$B$, $(cA)\otimes B=c(A\otimes B)$$B)$, $(A\otimes B)\otimes C=A\otimes(B\otimes C)$$C)$.
• Mixed product: $(A\otimes B)(C\otimes D)=(AC)\otimes(BD)$ whenever $AC$ and $BD$ are defined.
• $(A\otimes B)^{\top}=A^{\top}\otimes B^{\top}$$B^{\top }$; if $A,B$ are invertible, $(A\otimes B)^{-1}=A^{-1}\otimes B^{-1}$$B^{-1}$.
• For $A$ ($n\times$$n\times n$$n$) with eigenvalues $\lambda_i$ and $B$ ($p\times$$p\times p$$p$) with eigenvalues $\mu_k$, the eigenvalues of $A\otimes B$ are the $np$ products $\lambda_i\mu_k$; if $A\boldsymbol{x}=\lambda\boldsymbol{x}$ and $B\boldsymbol{y}=\mu\boldsymbol{y}$ then $(A\otimes B)(\boldsymbol{x}\otimes\boldsymbol{y})=\lambda\mu\,(\boldsymbol{x}\otimes\boldsymbol{y})$$\mathbf{y})$.
• Hence the determinant and trace satisfy $\det(A\otimes B)=(\det A)^p(\det B)^n$ and $\operatorname{tr}(A\otimes B)=\operatorname{tr}A\cdot\operatorname{tr}B$$\mathrm{tr}B$. For any sizes, the rank satisfies $\operatorname{rank}(A\otimes B)=\operatorname{rank}A\cdot\operatorname{rank}B$$\mathrm{rank}B$.

Examples¶

A $4\times4$ product. For $A=\begin{pmatrix}1&2\\3&4\end{pmatrix}$ and $B=\begin{pmatrix}0&1\\1&0\end{pmatrix}$$\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$,

The two differ. Checks: the trace is $0=\operatorname{tr}A\cdot\operatorname{tr}B=5\cdot0$$0$, and $\det(A\otimes B)=(-2)^2(-1)^2=4$$4$.

Vectors. For $\boldsymbol{x}=(1,2)^{\top}$ and $\boldsymbol{y}=(3,4,5)^{\top}$$(3,4,5{)}^{\top }$, $\boldsymbol{x}\otimes\boldsymbol{y}=(3,4,5,6,8,10)^{\top}$$(3,4,5,6,8,10{)}^{\top }$: the block $1\cdot\boldsymbol{y}$ followed by the block $2\cdot\boldsymbol{y}$$\mathbf{y}$.

Block diagonal. $I_2\otimes B=\begin{pmatrix}B&O\\O&B\end{pmatrix}$$\left(\begin{array}{cc}B& O\\ O& B\end{array}\right)$, whereas $B\otimes I_2$ replaces each entry $b_{kl}$ by the block $b_{kl}I_2$; here that is $\begin{pmatrix}O&I_2\\I_2&O\end{pmatrix}$.

Prerequisites¶

• Matrix Product
• Matrix Scalar Multiplication
• Block Matrix

Related Concepts¶

• Eigenvalue and Eigenvector
• Determinant
• Trace
• Inverse and Singular Matrices

Sources¶

• Math V: Advanced Linear Algebra · Week 6

Aliases: tensor product of matrices, matrix direct product Keywords: A ⊗ B, otimes, mixed-product property, vec operator Symbols: lambda, mu

Last updated: 22 October 2026 19:04