# $E^{*} \otimes E^{*}$ is canonically isomorphic to $Hom^2(E × E , \mathbb{R})$

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.

I have been following the textbook of Hoffman and Kunze.

• 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.
– Alex
Mar 4 at 21:53

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.

• 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.
– Jack
Mar 4 at 22:28
• 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 $\dim (E \times E)\dim \mathbb R = \dim E^2$. Mar 4 at 22:36

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.