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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    – Michael E2
    Aug 13, 2015 at 0:46
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    – Michael E2
    Aug 13, 2015 at 0:46
  • $\begingroup$ What sort of zero would you expect in the first example? $\endgroup$
    – Michael E2
    Aug 13, 2015 at 0:49
  • $\begingroup$ 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. $\endgroup$
    – Milad
    Aug 13, 2015 at 1:16

3 Answers 3


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.

  • $\begingroup$ 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. $\endgroup$
    – Milad
    Aug 13, 2015 at 1:49
  • $\begingroup$ 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. $\endgroup$
    – Jens
    Aug 13, 2015 at 6:21
  • $\begingroup$ 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]} $\endgroup$
    – Jens
    Aug 13, 2015 at 6:25
  • $\begingroup$ I'd forgotten about the Tensor* family of MMA functions - +1 $\endgroup$
    – ciao
    Aug 13, 2015 at 6:31
  • $\begingroup$ Exactly, that's what I meant. Thanks @Jens $\endgroup$
    – Milad
    Aug 14, 2015 at 1:23

How about:

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

{{0, 0}, {0, 0}}
  • $\begingroup$ 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] $\endgroup$
    – Milad
    Aug 13, 2015 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?


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