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I'm trying to work out some examples of applying the tensor product in some concrete cases to get a better understanding of it. Within this context, let $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ be a linear map with matrix $A$ and let $g:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ be a linear map with matrix $B$. It follows almost immediately from the universal property of the tensor product that there exists a unique linear map $$ f \otimes g: \mathbb{R}^2 \otimes \mathbb{R}^2 \rightarrow \mathbb{R}^2 \otimes \mathbb{R}^2 $$ such that $$ (f \otimes g)(u \otimes v) = f(u) \otimes g(v). $$ Suppose we take a basis for $\mathbb{R}^2$, say the canonical one, $(e_1, e_2)$, and let $u = a^ie_i$ and $v = b^je_j$. Then, using the linearity of $f$ and $g$ and multilinearity of $\otimes$ we arrive at the general expression $$ (f \otimes g)(u \otimes v) = a^ib^j f(e_i) \otimes g(e_j) $$ where the summation convention is in force. Since $f \otimes g$ is a linear map we ought to be able to represent it by a matrix $A \otimes B$ whose columns represent the action of $f \otimes g$ a basis of $\mathbb{R}^2 \otimes \mathbb{R}^2$.

Now, since $\{e_i \otimes e_j| i,j = 1, 2 \}$ is a basis for $\mathbb{R}^2 \otimes \mathbb{R}^2$ the resulting matrix with respect to this basis should have columns that are given by $(f \otimes g)(e_i \otimes e_j)$ where $i,j=1, 2$. The first column of this matrix, for example, is the vector obtained by $$ (f \otimes g)(e_1 \otimes e_1) = f(e_1) \otimes g(e_1) $$ which is the tensor product of the first column of $A$ and the first column of $B$ since $A$ and $B$ respectively represent $f$ and $g$.

This is where I am stuck. I understand $f(e_1) \otimes g(e_1)$ is essentially an equivalence class that is obtained in the existence proof of the tensor product but I'm not so sure how one works this out concretely based on the definition of the tensor product and its universal construction.

So my question is, how does one concretely compute $f(e_1) \otimes g(e_1)$?

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$f$ has matrix $A$ with respect to $\{e_1, e_2\}$. Hence $f(e_1) = a^i_1e_i$, analogously $g(e_1) = b^j_1e_j$. By bilinearity of $\otimes$, therefore \[ f(e_1) \otimes g(e_1) = a^i_1b^j_1 (e_i \otimes e_j) \] So if say, you decide that $(e_1 \otimes e_1, e_1 \otimes e_2, e_2 \otimes e_1, e_2 \otimes e_2)$ is your ordered basis of $\mathbb R^2 \otimes \mathbb R^2$, then the first column of your matrix is $(a_1^1b_1^1, a_1^1b_1^2, a_1^2b_1^1, a_1^2b_1^2)^t$.

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  • $\begingroup$ Ah, I see...makes perfect sense. $\endgroup$ – ItsNotObvious Jun 18 '12 at 20:06
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Check out http://en.wikipedia.org/wiki/Kronecker_product

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