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In the book of A Course in Algebra by E.B Winberg, at page $301$ it states that

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However, it does not explain how did we derive 8.27, so my first question is how can we prove 8.27 ?

Secondly, in this example we are considering the tensor product of two set of linear operators, of two vector spaces, so my question is if they were homomorphisms, i.e codomain was a vector space instead of the corresponding field of the vector space, what would be the tensor product of the matrix representations of the maps ?

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  • $\begingroup$ Look at the book "Abstract Algebra" by Dummit and Foote, Chapter 11, Section 2 (in my copy). They defined "tensor product" of two matrices, which is same as tensor product of two linear maps between finite densional vector spaces. $\endgroup$ – Krish Sep 7 '17 at 13:55
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Let's decompose $\mathcal{A}$ and $\mathcal{B}$ w.r.t. $\{e_1,e_2,...,e_n\}$ and $\{f_1,f_2,...,f_m\}$, respectively.

$\begin{equation}\mathcal{A}=\\ e^*_1\otimes(a_{11}e_1+a_{12}e_2+...+a_{1n}e_n)+\\ e^*_2\otimes(a_{21}e_1+a_{22}e_2+...+a_{2n}e_n)+\\ ...+\\ e^*_n\otimes(a_{n1}e_1+a_{n2}e_2+...+a_{nn}e_n) \end{equation}$

where $\{e^*_1,e^*_2,...,e^*_n\}$ is the dual base of $V^*$ w.r.t. $\{e_1,e_2,...,e_n\}$ and $\otimes$ is understood to be $\otimes_{L(V)}$ the tensor product defined in $V^*\bigotimes V\cong L(V)$

$\begin{equation}\mathcal{B}=\\ f^*_1\otimes(b_{11}f_1+b_{12}f_2+...+b_{1m}f_m)+\\ f^*_2\otimes(b_{21}f_1+b_{22}f_2+...+b_{2m}f_m)+\\ ...+\\ f^*_m\otimes(b_{m1}f_1+b_{m2}f_2+...+b_{mm}f_m) \end{equation}$

where $\{f^*_1,f^*_2,...,f^*_m\}$ is the dual base of $W^*$ w.r.t. $\{f_1,f_2,...,f_m\}$ and $\otimes$ is understood to be $\otimes_{L(W)}$ is the tensor product defined in $W^*\bigotimes W\cong L(W)$

Then their tensor product is:

$\mathcal{A}\otimes\mathcal{B}=\displaystyle\sum_{i,j,k,l}(e^*_i\otimes_{L(V)} a_{ij}e_j)\otimes(f^*_k\otimes_{L(W)} b_{kl}f_l)$

where $\otimes$ is to be understood $\otimes_{L(V)\bigotimes L(W)}$

Now observe that canonically $L(V)\bigotimes L(W)\cong(V^*\bigotimes V)\bigotimes(W^*\bigotimes W)\cong(V^*\bigotimes W^*)\bigotimes(V\bigotimes W)\cong L(V\bigotimes W)$.

So choosing in $V\bigotimes W$ the base $\{e_1\otimes f_1,e_2\otimes f_1, ...,e_n\otimes f_1,$ $e_1\otimes f_2,e_2\otimes f_2, ...,e_n\otimes f_2, ..., e_1\otimes f_m,e_2\otimes f_m, ...,e_n\otimes f_m\}$ means to choose as isomorphism that which imposes such contraints

$$(e^*_i\otimes_{L(V)} e_j)\otimes_{L(V)\bigotimes L(W)}(f^*_k\otimes_{L(W)} f_l)=(e^*_i\otimes_{V^*\bigotimes W^*} f^*_k)\otimes_{L(V\bigotimes W)}(e_j\otimes_{V\bigotimes W} f_l)$$

So it follows: $\mathcal{A}\otimes\mathcal{B}=\displaystyle\sum_{i,j,k,l}(e^*_i\otimes_{V^*\bigotimes W^*} f^*_k)\otimes_{L(V\bigotimes W)}(a_{ij}b_{kl}e_j\otimes_{V\bigotimes W} f_l)$

Its matrix representation in the ordered base $\{e_1\otimes f_1,e_2\otimes f_1, ...,e_n\otimes f_1,$ $e_1\otimes f_2,e_2\otimes f_2, ...,e_n\otimes f_2, ..., e_1\otimes f_m,e_2\otimes f_m, ...,e_n\otimes f_m\}$ is then: $$\begin{pmatrix}a_{ij}b_{kl}\end{pmatrix}$$ that as you can see, given the order of the base, can be block-partitioned in the matrix presented in the book.

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