Kroneckerjev produkt (oznaka
⊗
{\displaystyle \,\otimes \,}
matrikah poljubne velikosti, in daje bločno matriko . Kroneckerjevega produkta se ne sme zamenjevati z običajnim množenjem matrik. Kroneckerjev produkt daje matriko tenzorskega produkta .
Imenuje se po nemškem matematiku in logiku Leopoldu Kroneckerju (1823–1891), čeprav ni dokazov, da ga je prvi uporabljal.
Naj bo
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{\displaystyle A\,}
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{\displaystyle m\times n\,}
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{\displaystyle B\,}
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{\displaystyle p\times q\,}
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{\displaystyle A\otimes B\,}
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{\displaystyle mp\times nq\,}
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{\displaystyle \mathbf {A} \otimes \mathbf {B} ={\begin{bmatrix}a_{11}B&\cdots &a_{1n}B\\\vdots &\ddots &\vdots \\a_{m1}B&\cdots &a_{mn}B\end{bmatrix}}.}
Bolj točno je to enako:
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{\displaystyle \mathbf {A} \otimes \mathbf {B} ={\begin{bmatrix}a_{11}b_{11}&a_{11}b_{12}&\cdots &a_{11}b_{1q}&\cdots &\cdots &a_{1n}b_{11}&a_{1n}b_{12}&\cdots &a_{1n}b_{1q}\\a_{11}b_{21}&a_{11}b_{22}&\cdots &a_{11}b_{2q}&\cdots &\cdots &a_{1n}b_{21}&a_{1n}b_{22}&\cdots &a_{1n}b_{2q}\\\vdots &\vdots &\ddots &\vdots &&&\vdots &\vdots &\ddots &\vdots \\a_{11}b_{p1}&a_{11}b_{p2}&\cdots &a_{11}b_{pq}&\cdots &\cdots &a_{1n}b_{p1}&a_{1n}b_{p2}&\cdots &a_{1n}b_{pq}\\\vdots &\vdots &&\vdots &\ddots &&\vdots &\vdots &&\vdots \\\vdots &\vdots &&\vdots &&\ddots &\vdots &\vdots &&\vdots \\a_{m1}b_{11}&a_{m1}b_{12}&\cdots &a_{m1}b_{1q}&\cdots &\cdots &a_{mn}b_{11}&a_{mn}b_{12}&\cdots &a_{mn}b_{1q}\\a_{m1}b_{21}&a_{m1}b_{22}&\cdots &a_{m1}b_{2q}&\cdots &\cdots &a_{mn}b_{21}&a_{mn}b_{22}&\cdots &a_{mn}b_{2q}\\\vdots &\vdots &\ddots &\vdots &&&\vdots &\vdots &\ddots &\vdots \\a_{m1}b_{p1}&a_{m1}b_{p2}&\cdots &a_{m1}b_{pq}&\cdots &\cdots &a_{mn}b_{p1}&a_{mn}b_{p2}&\cdots &a_{mn}b_{pq}\end{bmatrix}}}
Če sta
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{\displaystyle A\,}
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{\displaystyle B\,}
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{\displaystyle V_{1}\to W_{1}\,}
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{\displaystyle V_{2}\to W_{2}\,}
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{\displaystyle A\otimes B\,}
tenzorski produkt dveh preslikav
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{\displaystyle V_{1}\otimes W_{2}\to W_{1}\otimes W_{2}\,}
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{\displaystyle {\begin{bmatrix}1&2\\3&4\end{bmatrix}}\otimes {\begin{bmatrix}5&6\\7&8\end{bmatrix}}={\begin{bmatrix}1\cdot {\begin{bmatrix}5&6\\7&8\end{bmatrix}}&2\cdot {\begin{bmatrix}5&6\\7&8\end{bmatrix}}\\\\3\cdot {\begin{bmatrix}5&6\\7&8\end{bmatrix}}&4\cdot {\begin{bmatrix}5&6\\7&8\end{bmatrix}}\end{bmatrix}}={\begin{bmatrix}5&6&10&12\\7&8&14&16\\15&18&20&24\\21&24&28&32\end{bmatrix}}}
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{\displaystyle {\begin{bmatrix}1&3&2\\1&0&0\\1&2&2\end{bmatrix}}\otimes {\begin{bmatrix}0&5\\5&0\\1&1\end{bmatrix}}={\begin{bmatrix}1\cdot {\begin{bmatrix}0&5\\5&0\\1&1\end{bmatrix}}&3\cdot {\begin{bmatrix}0&5\\5&0\\1&1\end{bmatrix}}&2\cdot {\begin{bmatrix}0&5\\5&0\\1&1\end{bmatrix}}\\\\1\cdot {\begin{bmatrix}0&5\\5&0\\1&1\end{bmatrix}}&0\cdot {\begin{bmatrix}0&5\\5&0\\1&1\end{bmatrix}}&0\cdot {\begin{bmatrix}0&5\\5&0\\1&1\end{bmatrix}}\\\\1\cdot {\begin{bmatrix}0&5\\5&0\\1&1\end{bmatrix}}&2\cdot {\begin{bmatrix}0&5\\5&0\\1&1\end{bmatrix}}&2\cdot {\begin{bmatrix}0&5\\5&0\\1&1\end{bmatrix}}\end{bmatrix}}={\begin{bmatrix}0&5&0&15&0&10\\5&0&15&0&10&0\\1&1&3&3&2&2\\0&5&0&0&0&0\\5&0&0&0&0&0\\1&1&0&0&0&0\\0&5&0&10&0&10\\5&0&10&0&10&0\\1&1&2&2&2&2\end{bmatrix}}}
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{\displaystyle {\begin{bmatrix}1&2\\3&4\\\end{bmatrix}}\otimes {\begin{bmatrix}0&5\\6&7\\\end{bmatrix}}={\begin{bmatrix}1\cdot 0&1\cdot 5&2\cdot 0&2\cdot 5\\1\cdot 6&1\cdot 7&2\cdot 6&2\cdot 7\\3\cdot 0&3\cdot 5&4\cdot 0&4\cdot 5\\3\cdot 6&3\cdot 7&4\cdot 6&4\cdot 7\\\end{bmatrix}}={\begin{bmatrix}0&5&0&10\\6&7&12&14\\0&15&0&20\\18&21&24&28\end{bmatrix}}}
Kroneckerjev produkt je posebni primer tenzorskega produkta :
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{\displaystyle \mathbf {A} \otimes (\mathbf {B} +\mathbf {C} )=\mathbf {A} \otimes \mathbf {B} +\mathbf {A} \otimes \mathbf {C} ,}
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{\displaystyle (\mathbf {A} +\mathbf {B} )\otimes \mathbf {C} =\mathbf {A} \otimes \mathbf {C} +\mathbf {B} \otimes \mathbf {C} ,}
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{\displaystyle (k\mathbf {A} )\otimes \mathbf {B} =\mathbf {A} \otimes (k\mathbf {B} )=k(\mathbf {A} \otimes \mathbf {B} ),}
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{\displaystyle (\mathbf {A} \otimes \mathbf {B} )\otimes \mathbf {C} =\mathbf {A} \otimes (\mathbf {B} \otimes \mathbf {C} ),}
kjer je
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{\displaystyle A\,}
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{\displaystyle B\,}
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{\displaystyle C\,}
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{\displaystyle k\,}
skalar Kroneckerjev produkt ni komutativen . To pomeni da sta matriki
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{\displaystyle A\otimes B\,}
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{\displaystyle B\otimes A\,}
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{\displaystyle A\otimes B\neq B\otimes A}
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{\displaystyle P\,}
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{\displaystyle Q\,}
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{\displaystyle \mathbf {A} \otimes \mathbf {B} =\mathbf {P} \,(\mathbf {B} \otimes \mathbf {A} )\,\mathbf {Q} \,}
Č e pa sta matriki
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{\displaystyle A\,}
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{\displaystyle B\,}
kvadratni , potem sta
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{\displaystyle A\otimes B\,}
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{\displaystyle B\otimes A\,}
podobni , kar pomeni, da je
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{\displaystyle P=Q^{T}\,}
Če so matrike
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{\displaystyle A\,}
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{\displaystyle B\,}
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{\displaystyle C\,}
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{\displaystyle D\,}
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{\displaystyle AC\,}
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{\displaystyle BD\,}
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{\displaystyle (\mathbf {A} \otimes \mathbf {B} )(\mathbf {C} \otimes \mathbf {D} )=\mathbf {AC} \otimes \mathbf {BD} }
Transponiranje Kroneckerjevega produkta da:
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{\displaystyle (A\otimes B)^{T}=A^{T}\otimes B^{T}}
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{\displaystyle {\overline {A\otimes B}}={\overline {A}}\otimes {\overline {B}}}
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{\displaystyle (A\otimes B)^{*}=A^{*}\otimes B^{*}}
sled je za kvadratne matrike enaka:
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{\displaystyle \mathrm {sl} (A\otimes B)=\mathrm {sl} (A)\cdot \mathrm {sl} (B)}
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{\displaystyle \mathrm {rank} (A\otimes B)=\mathrm {rank} (A)\cdot \mathrm {rank} (B)}
če ima matrika
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{\displaystyle A\,}
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{\displaystyle n\times n\,}
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{\displaystyle B\,}
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{\displaystyle m\times m\,}
determinanto velja:
det
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{\displaystyle \det(A\otimes B)={\det }^{m}(A)\,{\det }^{n}(B)}
če so
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{\displaystyle (\lambda _{i})_{i=1..n}\,}
lastne vrednosti matrike
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{\displaystyle A\,}
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{\displaystyle (\mu _{j})_{j=1..m}\,}
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{\displaystyle B\,}
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{\displaystyle (\lambda _{i}\,\mu _{j})_{i=1..n \atop j=1..m}}
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{\displaystyle A\otimes B}
kadar sta matriki
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{\displaystyle A\,}
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{\displaystyle B\,}
obrnljivi velja tudi:
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{\displaystyle (A\otimes B)^{-1}=A^{-1}\otimes B^{-1}}
kadar imajo matrike
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{\displaystyle A,B,C\,}
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{\displaystyle D\,}
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{\displaystyle A:m\times n\,}
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{\displaystyle B:p\times q\,}
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{\displaystyle C:n\times r\,}
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{\displaystyle D:q\times s\,}
in sta matriki
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{\displaystyle AC\,}
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{\displaystyle BD\,}
[ 1]
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{\displaystyle AC\otimes BD=>(A\otimes B)(C\otimes D)}
