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This question was asked in exercise of my linear algebra course and I am not able to prove part (ii) of it.

Let $E$ be a finite dimensional vector space ( over $\mathbb{R}$).

(I) What is the defining characteristic of the tensor product $ E\otimes E$.

(II) Prove that $ E^* \otimes E^*$ is canonically isomorphic to $Hom^2( E \times E; \mathbb{R})$ .

Ist part I have done.

$Hom^2( E × E , \mathbb{R}) $ is the space of bilinear maps $ f: E× E \to \mathbb{R}$.

I am not able to construct a map which could act as isomorphism.

Can you please help me with this?

I have been following the textbook of Hoffman and Kunze.

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    $\begingroup$ Maybe include your solution of Question 1. Then prove that $Hom^2(E\times E,\mathbb R)$ satisfies this defining characteristic. By using this universal property you maybe dont have to write down an isomorphism explicitely. $\endgroup$
    – Alex
    Mar 4 at 21:53

2 Answers 2

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Take the map $f\colon E^* \times E^* \to \mathrm{Hom}(E\times E;\mathbb R)$ mapping $(\alpha,\beta)$ to a homomorphism $f(\alpha,\beta)$ acting on $(v,w) \in E \times E$ as $f(\alpha,\beta)(v,w) := \alpha(v)\beta(w)$. This is bilinear by definition, so it extends to a linear map $\tilde{f}\colon E^* \otimes E^* \to \mathrm{Hom}(E \times E; \mathbb R)$ such that $\tilde{f}(\alpha \otimes \beta) = f(\alpha,\beta)$. Its kernel is trivial, and $\dim(E^* \otimes E^*) = \dim \mathrm{Hom}(E\times E;\mathbb R)$, so $\tilde{f}$ is an isomorphism.

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  • $\begingroup$ can you please tell how kernal of $\tilde{f}$ is trivial ? Also how is dimension $( E^* \times E^* ) = $ dimension of $Hom^2 ( E \times E , \mathbb{R}) $. If I am not wrong, dimension of $E^* \otimes E^{*}) = n^2$. But how to compute dimension of $ Hom( E× E; \mathbb{R})$? Please help. $\endgroup$
    – Jack
    Mar 4 at 22:28
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    $\begingroup$ As for injectivity, you want $f(\alpha,\beta)$ to vanish when applied to any pair $(v,w)$. Then necessarily either $\alpha=0$ or $\beta = 0$, and hence in both cases $\alpha \otimes \beta = 0$. The kernel of $\tilde{f}$ is then trivial. The dimension of $\mathrm{Hom}(E \times E; \mathbb R)$ is just $\dim (E \times E)\dim \mathbb R = \dim E^2$. $\endgroup$
    – Gibbs
    Mar 4 at 22:36
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I would solve part II by using part I. I assume your answer to part I was something like this:

The tensor product $T:=E\otimes E$ is uniquely determied by the following property:

There exists a bilinear map $\otimes:\, E\times E \to T$ such that for every bilinear map $f:E\times E\to Z$ there exists a unique linear map $\tilde f:\,T\to Z$ such that $f=\tilde f\circ \otimes$.

Lets prove that $Hom^2(E\times E,\mathbb R)$ satsifies the universal property of $E^\ast \otimes E^\ast$. We define $\otimes:E^\ast \times E^\ast\to Hom^2(E\times E,\mathbb R)$ by $\otimes(\alpha,\beta)(v,w)=\alpha(v)\beta(w)$. This map is bilinear and if $f:E^\ast \times E^\ast\to Z$ is any bilinear map then we define $\tilde f:\,Hom^2(E\times E,\mathbb R)\to Z$ by $\tilde f(g)=f(g\circ\iota_1,g\circ \iota_2)$ where $\iota_i:E\to E\times E$ is the inclusion of $E$ into the $i$-th factor ($i=1,2$). Then $f=\tilde f\circ\otimes$ and this in fact defines $\tilde f$ uniquely.

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