Calculadora de matrices

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A (3×3)
Filas: 3
Columnas: 3
B (3×3)
Filas: 3
Columnas: 3
Supports: fractions (1/2), decimals (0.5), constants (pi, e). Empty cells are treated as 0.

Resultados del cálculo

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.


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