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I am getting confused by the tensor product. I would appreciate some basic insight.

I consider $M_2\otimes M_3$. (Here $M_n$ denotes the complex $n\times n$ matrices.) The dimension of this space is 4*9 = 36. This is the same dimension that $M_6$ has. I can map a tensor element $\sum_i A_i\otimes B_i$ to an $M_6$ matrix via the (matrix) tensor (Kronecker) product. Since the dimensions are the same this should mean that the two spaces can be thought as the same. Am I right?

Now this means that for any matrix in $M_6$ it can be decomposed as a sum of Kronecker products of matrices from $M_2$ and $M_3$. How many terms are needed here (maximum)? Is the number of terms related to the tensor rank?

Assume you have a matrix $C$ in $M_6$ that has rank 3. How many terms are needed to express $C$ as $\sum_i A_i\otimes B_i$ thought to be in $M_2\otimes M_3$?

Or am I completely lost and asking nonsensical questions?

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    $\begingroup$ The number needed to represent a given matrix is between $0$ and $4$. Except in the rank $0$ case, this is not closely related to the matrix rank. $\endgroup$ Commented Aug 18 at 16:22
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    $\begingroup$ In fact $M_6$ is isomorphic to the tensor product $M_2 \otimes M_3$ of algebras, with the isomorphism given by Kronecker products, so we have much more than just that the dimensions match. $\endgroup$ Commented Aug 18 at 17:05
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    $\begingroup$ @WillSawin Do we have in general $min(n^2, m^2)$ for the maximum number of terms needed? $\endgroup$ Commented Aug 18 at 17:43
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    $\begingroup$ Yes, the number of terms could be called the tensor rank, though when a space has multiple relevant tensor product decompositions that terminology can be confusing. Certainly "may be possible" is true, for example the tensor product of a rank $2$ matrix in $M_2$ and a rank $3$ matrix in $M_3$ will be a rank $6$ matrix in $M_6$ with tensor rank $1$. $\min(n^2,m^2)$ is the maximum number of terms needed in general (since the problem is equivalent to the maximum rank of an $n^2\times m^2$ matrix). $\endgroup$ Commented Aug 18 at 18:47
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    $\begingroup$ I can certainly guess what M_n means (and I may or may not be right; there is no way to be sure). But I would vastly prefer if questions included definitions of their notation. $\endgroup$ Commented Aug 19 at 2:00

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