A (3×3)
Filas:
3
Columnas:
3
Enteros positivos (0 ~ 9)
Enteros (-10 ~ 10)
Decimales (-1.0 ~ 1.0)
B (3×3)
Filas:
3
Columnas:
3
Enteros positivos (0 ~ 9)
Enteros (-10 ~ 10)
Decimales (-1.0 ~ 1.0)
Supports: fractions (1/2), decimals (0.5), constants (pi, e). Empty cells are treated
as 0.
Determinante(A)
Inversa(A)
Pseudoinversa(A)
Vectores propios(A)
Traspuesta(A)
RREF(A)
SVD(A)
Valores propios(A)
Traza(A)
Rango(A)
√(A)
Forma de Jordan(A)
Diagonal(A)
Adjunta(A)
log(A)
exp(A)
Triangular superior(A)
LU(A)
sin(A)
arcsin(A)
HNF(A)
QR(A)
cos(A)
arccos(A)
LLL(A)
Cholesky(A)
tan(A)
arctan(A)
1. Definición y fórmula
The Kronecker product (also called tensor product, denoted
$\otimes$) of an $m\times n$ matrix $A=[a_{ij}]$ and a $p\times q$ matrix $B$ is the
block matrix:
$$
A \otimes B =
\begin{bmatrix}
a_{11}B & a_{12}B & \cdots & a_{1n}B \\
a_{21}B & a_{22}B & \cdots & a_{2n}B \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1}B & a_{m2}B & \cdots & a_{mn}B
\end{bmatrix},
$$
The result has size $(mp)\times(nq)$. Each entry $a_{ij}$ of $A$
multiplies the entire matrix $B$ to form the corresponding block.
2. Propiedades útiles
Mixed-product property: $(A\otimes B)(C\otimes D) =
(AC)\otimes(BD)$ when dimensions are compatible.
Transpose: $(A\otimes B)^{T} = A^{T}\otimes B^{T}$.
Associativity: $(A\otimes B)\otimes C = A\otimes(B\otimes C)$
(up to natural reshaping).
Scalar multiple: $(\alpha A)\otimes B = A\otimes(\alpha B) =
\alpha (A\otimes B)$.
Eigenvalues: If $A v = \lambda v$ and $B w = \mu w$, then
$(A\otimes B)(v\otimes w) = (\lambda\mu)(v\otimes w)$.
3. Ejemplos resueltos
Ejemplo 1: matrices pequeñas
$$
A = \begin{pmatrix}1 & 2\\ 3 & 4\end{pmatrix},\quad
B = \begin{pmatrix}0 & 5\\ 6 & 7\end{pmatrix}
$$
Compute $A\otimes B$ by replacing each $a_{ij}$ with $a_{ij}B$:
$$
A\otimes B =
\begin{bmatrix}
1\cdot B & 2\cdot B \\
3\cdot B & 4\cdot B
\end{bmatrix}
=
\begin{bmatrix}
0 & 5 & 0 & 10 \\
6 & 7 & 12 & 14 \\
0 & 15 & 0 & 20 \\
18 & 21 & 24 & 28
\end{bmatrix}.
$$
Result is $(2\cdot2)\times(2\cdot2)=4\times4$.
Ejemplo 2: con vector
$$
A = \begin{pmatrix}2 & 0\\ 1 & 3\end{pmatrix},\quad
B = \begin{pmatrix}1 & 2\end{pmatrix}
$$
$B$ is $1\times2$, $A\otimes B$ is $(2\cdot1)\times(2\cdot2)=2\times4$:
$$
A\otimes B =
\begin{bmatrix}
2B & 0B \\
1B & 3B
\end{bmatrix}
=
\begin{bmatrix}
2 & 4 & 0 & 0 \\
1 & 2 & 3 & 6
\end{bmatrix}.
$$
4. Errores comunes
Confundir la multiplicación elemento a elemento con el producto de Kronecker:
El producto de Kronecker crea grandes matrices por bloques, no una multiplicación entrada a entrada.
Esperar el mismo tamaño: $A\otimes B$ rarely has the same dimension
as $A$ or $B$ unless one is $1\times1$.
Olvidar el orden de los bloques: Blocks follow row-major order of $A$ —
each row of $A$ creates a block row of $B$-blocks.
Mismatching dimensions when using mixed-product property:
ensure matrix multiplications $AC$ and $BD$ are valid before applying the
property.
Errores de indexación: when programming, carefully map $(i,j)$ in
$A$ to blocks of size $p\times q$ in the result.
5. Practice problems
Try these Kronecker product problems. Click Mostrar respuesta to reveal the
result.
Ejercicio 1
$$
A=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix},\quad
B=\begin{pmatrix}2 & 3\\ 4 & 5\end{pmatrix}
$$
Mostrar respuesta
$$
A\otimes B = \begin{pmatrix}
2 & 3 & 0 & 0\\
4 & 5 & 0 & 0\\
0 & 0 & 2 & 3\\
0 & 0 & 4 & 5
\end{pmatrix}.
$$
Ejercicio 2
$$
A=\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix},\quad
B=\begin{pmatrix}1 & 1\\ 0 & 2\end{pmatrix}
$$
Mostrar respuesta
$$
A\otimes B =
\begin{pmatrix}
0\cdot B & 1\cdot B\\
1\cdot B & 0\cdot B
\end{pmatrix}
=
\begin{pmatrix}
0 & 0 & 1 & 1\\
0 & 0 & 0 & 2\\
1 & 1 & 0 & 0\\
0 & 2 & 0 & 0
\end{pmatrix}.
$$
Ejercicio 3
$$
A=\begin{pmatrix}1 & 2 & 3\end{pmatrix},\quad
B=\begin{pmatrix}4\\ 5\end{pmatrix}
$$
Mostrar respuesta
Here $A$ is $1\times3$, $B$ is $2\times1$, so $A\otimes B$ is
$(1\cdot2)\times(3\cdot1)=2\times3$:
$$
A\otimes B =
\begin{pmatrix}
1\cdot B & 2\cdot B & 3\cdot B
\end{pmatrix}
=
\begin{pmatrix}
4 & 8 & 12 \\
5 & 10 & 15
\end{pmatrix}.
$$
Ejercicio 4
$$
A=\begin{pmatrix}2 & -1\\ 0 & 3\end{pmatrix},\quad
B=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}
$$
Mostrar respuesta
$$
A\otimes B =
\begin{pmatrix}
2B & -1B \\
0B & 3B
\end{pmatrix}
=
\begin{pmatrix}
2 & 0 & -1 & 0 \\
0 & -2 & 0 & 1 \\
0 & 0 & 3 & 0 \\
0 & 0 & 0 & -3
\end{pmatrix}.
$$
Carefully multiply scalar blocks; keep block positions aligned with
$A$'s entries.
El producto de Kronecker también aparece junto con los valores propios , la SVD y la descomposición QR .
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