The Unapologetic Mathematician

Mathematics for the interested outsider

Matrices III

Given two finite-dimensional vector spaces U and V, with bases \left\{e_i\right\} and \left\{f_j\right\} respectively, we know how to build a tensor product: use the basis \left\{e_i\otimes f_j\right\}.

But an important thing about the tensor product is that it’s a functor. That is, if we have linear transformations S:U\rightarrow U' and T:V\rightarrow V', then we get a linear transformation S\otimes T:U\otimes V\rightarrow U'\otimes V'. So what does this operation look like in terms of matrices?

First we have to remember exactly how we get the tensor product S\otimes T. Clearly we can consider the function S\times T:U\times V\rightarrow U'\times V'. Then we can compose with the bilinear function U'\times V'\rightarrow U'\otimes V' to get a bilinear function from U\times V to U'\otimes V'. By the universal property, this must factor uniquely through a linear function U\otimes V\rightarrow U'\otimes V'. It is this map we call S\otimes T.

We have to pick bases \left\{e_k'\right\} of U' and \left\{f_l'\right\} of V'. This gives us a matrix coefficients s_i^k for S and t_j^l for T. To calculate the matrix for S\otimes T we have to evaluate it on the basis elements e_i\otimes f_j of U\otimes V. By definition we find:

\left[S\otimes T\right](e_i\otimes f_j)=S(e_i)\otimes T(f_j)=\left(s_i^ke_k'\right)\otimes\left(t_j^lf_l'\right)=s_i^kt_j^le_k'\otimes f_l'

that is, the matrix coefficient between the index pair (i,j) and the index pair (k,l) is s_i^kt_j^l.

It’s not often taught anymore, but there is a name for this operation: the Kronecker product. If we write the matrices (as opposed to just their coefficients) \left(s_i^k\right) and \left(t_j^l\right), then we write the Kronecker product \left(s_i^k\right)\boxtimes\left(t_j^l\right)=\left(s_i^kt_j^l\right).

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May 26, 2008 - Posted by | Algebra, Linear Algebra

6 Comments »

  1. [...] Like we saw with the tensor product of vector spaces, the dual space construction turns out to be a functor. In [...]

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  2. [...] is one slightly touchy thing we need to be careful about: Kronecker products. When the upper index is a pair with and we have to pick an order on the set of such pairs. [...]

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  3. [...] the monoidal product on objects by multiplication — — and on morphisms by using the Kronecker product. That is, if we have an matrix and an matrix , then we get the Kronecker [...]

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  4. [...] wait, there’s more! The functor is linear over , so it’s a functor enriched over . The Kronecker product of matrices corresponds to the monoidal product of linear transformations, so the functor is [...]

    Pingback by The Category of Matrices III « The Unapologetic Mathematician | June 23, 2008 | Reply

  5. [...] can recognize this as a Kronecker product of two [...]

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  6. [...] want. And we know that when expressed in matrix form, the tensor product of linear maps becomes the Kronecker product of matrices. We write the character of as , that of as , and that of their tensor product as , [...]

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