L'équation 7.10 est : 

` ([bbA] otimes [bbB])([bba] otimes [bb"b"]) = [bbA] [bba] otimes [bbB] [bb"b"]`

avec :  

` [bbA] = [(A_(11),A_(12)),(A_(21),A_(22))]  [bbB] = [(B_(11),B_(12)),(B_(21),B_(22))]  [bba] = ((a_1),(a_2))  [bb"b"] = ((b_1),(b_2))  `

Partie a) Son écriture en forme en composantes s'écrit alors :

` ([(A_(11),A_(12)),(A_(21),A_(22))]otimes[(B_(11),B_(12)),(B_(21),B_(22))]) ( ((a_1),(a_2))otimes((b_1),(b_2)) ) = ( [(A_(11),A_(12)),(A_(21),A_(22))]((a_1),(a_2)) otimes [(B_(11),B_(12)),(B_(21),B_(22))]((b_1),(b_2)) )`

Partie b) Effectuons les deux multiplications ` bbA bba ` et ` bbB bb"b" ` du membre de droite :

` [bbA](bba) = [(A_(11),A_(12)),(A_(21),A_(22))]((a_1),(a_2)) = [(A_(11)a_1 + A_(12)a_2),(A_(21)a_1 + A_(22)a_2)] `

` [bbB](bb"b") = [(B_(11),B_(12)),(B_(21),B_(22))]((b_1),(b_2)) = [(B_(11)b_1 + B_(12)b_2),(B_(21)b_1 + B_(22)b_2)] `

Partie c) Passons maintenant aux trois produits tensoriels de Kronecker :

` [bbA] otimes [bbB] = [(A_(11),vdots, A_(12)),(ldots, ldots, ldots), (A_(21), vdots, A_(22))][(B_(11),B_(12)),(B_(21),B_(22))]`

` = [( A_(11)B_(11), A_(11)B_(12),vdots, A_(12)B_(11), A_(12)B_(12)), ( A_(11)B_(21), A_(11)B_(22),vdots, A_(12)B_(21), A_(12)B_(22) ), ( ldots, ldots, ldots, ldots, ldots ), ( A_(21)B_(11), A_(21)B_(12),vdots, A_(22)B_(11), A_(22)B_(12)), ( A_(21)B_(21), A_(21)B_(22),vdots, A_(22)B_(21), A_(22)B_(22) ) ]`

` (bba) otimes (bb"b") = ((a_1), (ldots),(a_2)) otimes ((b_1),(b_2)) = ((a_1 b_1), (a_1 b_2), (ldots), (a_2 b_1),(a_2 b_2)) `

` [bbA](bba) otimes [bbB](bb"b") = [(A_(11)a_1 + A_(12)a_2), (ldots), (A_(21)a_1 + A_(22)a_2)] otimes [(B_(11)b_1 + B_(12)b_2),(B_(21)b_1 + B_(22)b_2)] `

` = [ ( (A_(11)a_1 + A_(12)a_2)(B_(11)b_1 + B_(12)b_2) ), ( (A_(11)a_1 + A_(12)a_2)(B_(21)b_1 + B_(22)b_2) ), (ldots), ( (A_(21)a_1 + A_(22)a_2)(B_(11)b_1 + B_(12)b_2) ), ( (A_(21)a_1 + A_(22)a_2)(B_(21)b_1 + B_(22)b_2) ) ]`

Partie d) Vérification de la dimension des matrices obtenues :

On a bien :

` bbA otimes bbB `   de dimension 4 x 4

` bba otimes bb"b" `    de dimension 4 x 1

` bbA bba otimes bbB bb"b" `  de dimension 4 x 1  ERREUR à corriger dans l'énoncé de cette version du livre.

Partie e) Calcul de la partie gauche de l'équation (7.10)  ` ([bbA] otimes [bbB])([bba] otimes [bb"b"]) = [bbA] [bba] otimes [bbB] [bb"b"]` :

` ([bbA] otimes [bbB])([bba] otimes [bb"b"]) = [( A_(11)B_(11), A_(11)B_(12),vdots, A_(12)B_(11), A_(12)B_(12)), ( A_(11)B_(21), A_(11)B_(22),vdots, A_(12)B_(21), A_(12)B_(22) ), ( ldots, ldots, ldots, ldots, ldots ), ( A_(21)B_(11), A_(21)B_(12),vdots, A_(22)B_(11), A_(22)B_(12)), ( A_(21)B_(21), A_(21)B_(22),vdots, A_(22)B_(21), A_(22)B_(22) ) ] ((a_1 b_1), (a_1 b_2), (ldots), (a_2 b_1),(a_2 b_2)) `

` = [( A_(11)B_(11), A_(11)B_(12), A_(12)B_(11), A_(12)B_(12)), ( A_(11)B_(21), A_(11)B_(22), A_(12)B_(21), A_(12)B_(22) ), ( A_(21)B_(11), A_(21)B_(12), A_(22)B_(11), A_(22)B_(12)), ( A_(21)B_(21), A_(21)B_(22), A_(22)B_(21), A_(22)B_(22) ) ] ((a_1 b_1), (a_1 b_2), (a_2 b_1),(a_2 b_2)) `

` color (blue) ( = [( A_(11)B_(11)a_1 b_1 + A_(11)B_(12) a_1 b_2 + A_(12)B_(11) a_2 b_1 + A_(12)B_(12)a_2 b_2), ( A_(11)B_(21)a_1 b_1 + A_(11)B_(22) a_1 b_2 + A_(12)B_(21) a_2 b_1 + A_(12)B_(22)a_2 b_2), ( A_(21)B_(11)a_1 b_1 + A_(21)B_(12) a_1 b_2 + A_(22)B_(11) a_2 b_1 + A_(22)B_(12)a_2 b_2), ( A_(21)B_(21)a_1 b_1 + A_(21)B_(22) a_1 b_2 + A_(22)B_(21) a_2 b_1 + A_(22)B_(22)a_2 b_2) ] ) `

Partie f) Calcul de la partie droite :

` [bbA](bba) otimes [bbB](bb"b") = [ ( (A_(11)a_1 + A_(12)a_2)(B_(11)b_1 + B_(12)b_2) ), ( (A_(11)a_1 + A_(12)a_2)(B_(21)b_1 + B_(22)b_2) ), (ldots), ( (A_(21)a_1 + A_(22)a_2)(B_(11)b_1 + B_(12)b_2) ), ( (A_(21)a_1 + A_(22)a_2)(B_(21)b_1 + B_(22)b_2) ) ]`

` color (blue) ( = [ ( A_(11)a_1 B_(11)b_1 + A_(11)a_1 B_(12)b_2 + A_(12)a_2 B_(11)b_1 + A_(12)a_2 B_(12)b_2 ), ( A_(11)a_1 B_(21)b_1 + A_(11)a_1 B_(22)b_2 + A_(12)a_2 B_(21)b_1 + A_(12)a_2 B_(22)b_2 ), ( A_(21)a_1 B_(11)b_1 + A_(21)a_1 B_(12)b_2 + A_(22)a_2 B_(11)b_1 + A_(22)a_2 B_(12)b_2 ), ( A_(21)a_1 B_(21)b_1 + A_(21)a_1 B_(22)b_2 + A_(22)a_2 B_(21)b_1 + A_(22)a_2 B_(22)b_2 ) ] )`

Conclusion Comparaison de la partie droite et de la partie gauche :

La deuxième matrice (la partie droite) est bien strictement égale à la première (la partie gauche) simplement en remettant en ordre chaque terme sous la forme ` bbA bbB bba bb"b" ` au lieu de ` bbA bba bbB bb"b" `. Et ce sans changer l'ordre des termes.

Ce que l'on voulait vérifier (avec peine quand même ! )