# How to simplify symbolic expressions with KroneckerProduct

Having $X,Y$ being symbols for matrices, I was wondering if there is a way to simplify expressions like

KroneckerProduct[X, X] + KroneckerProduct[-X, X]


to give zero. Or

KroneckerProduct[2 X, 3 Y]


to produce something like:

6 KroneckerProduct[X, Y]


In general if $a,b$ are scalars, and $X,Y$ are matrices we have this mathematical identity:

KroneckerProduct[a X, b Y] == a*b KroneckerProduct[X, Y].

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• You can format inline code and code blocks by selecting the code and clicking the {} button above the edit window. The edit window help button ? is also useful for learning how to format your questions and answers. You may also find this this meta Q&A helpful – Michael E2 Aug 13 '15 at 0:46
• What sort of zero would you expect in the first example? – Michael E2 Aug 13 '15 at 0:49
• Thanks for you comments :) I'll try to get familiar ASAP. KroneckerProduct[-X, X]=-KroneckerProduct[X, X], so the sum is just like adding a matrix to its negative, which results in a zero matrix. – Milad Aug 13 '15 at 1:16

There is another option, using the relatively new tensor capabilities of Mathematica. This is pretty much copied from another answer by jose, but I don't need any assumptions here:

TensorExpand[KroneckerProduct[X, X] + KroneckerProduct[-X, X]]

(* ==> 0 *)

TensorExpand[KroneckerProduct[2 X, 3 Y]]

(* ==> 6 KroneckerProduct[X, Y] *)


There is a potential problem with this approach in that the first result is 0 rather than a matrix. However, Mathematica doesn't have a special symbol for the zero matrix of unspecified dimension. If you want to get a more "correct" output that keeps track of the product space in which the zero lives, it will be necessary to hand-craft the necessary algebraic rules and symbols.

• Great, this is what I needed. Thank you very much @Jens. Also, do you know how can I have KroneckerProduct[x, z] + KroneckerProduct[y, z]=KroneckerProduct[x+y, z] ? I looked through TensorReduce, and the other suggestions in the Help, but couldn't figure it out. I couldn't find a good reference for these. Thank you anyway. – Milad Aug 13 '15 at 1:49
• So I guess you tried TensorExpand[KroneckerProduct[x,z]+KroneckerProduct[y,z]==KroneckerProduct[x+y,z]] and got True, right? I guess what you want is a transformation rule that combines two products into one, i.e., a factorization. I think that will require defining a custom function. – Jens Aug 13 '15 at 6:21
• Here is a function that does the factorization: kroneckerFactor[expr_]:=expr//.{Plus[KroneckerProduct[x_,y_],KroneckerProduct[z_,y_]]:>KroneckerProduct[x+z,y],Plus[KroneckerProduct[y_,x_],KroneckerProduct[y_,z_]]:>KroneckerProduct[y,x+z]} – Jens Aug 13 '15 at 6:25
• I'd forgotten about the Tensor* family of MMA functions - +1 – ciao Aug 13 '15 at 6:31
• Exactly, that's what I meant. Thanks @Jens – Milad Aug 14 '15 at 1:23

av = Array[Subscript[a, ##] &, {2}];
bv = Array[Subscript[b, ##] &, {2}];
KroneckerProduct[av, bv] + KroneckerProduct[-av, bv]

{{0, 0}, {0, 0}}

• I was wondering if its possible to do this without specifying the matrix's dimensions and properties. This is just a part of a project, and everything involve is unknown matrices with some general properties. It would be perfect if using something close to Simplify I could produce: KroneckerProduct[aX, bY] = a*b KroneckerProduct[X, Y] – Milad Aug 13 '15 at 1:27

Some people (see The ubiquitous Kronecker product by Van Loan) have worked on finding two matrices $A, B$ of specified size whose tensor product $A \otimes B$ is closest (in a norm) to a given (larger) matrix $C$. That is, find $A, B$ which minimize $||C-A \otimes B||$. The algorithm is based on the SVD. There is a matlab implementation somewhere. It would be nice to see this algorithm implemented in Mathematica.

If the error is zero then the algorithm factorises - but If I recollect the article doesn't go into this. Is this more simple, unique? Who knows?

• This could perhaps be reworked into a separate question, but it isn't an answer to the current question. – Jens Aug 13 '15 at 6:12
• – pdmclean Aug 15 '15 at 13:15