Notes:Concrete tensor product

Notes

Let $(V_1,\mathbb{F})$ and $(V_2,\mathbb{F})$ both be vector spaces over the field $\mathbb{F}$. Let $V_i^*:\eq L(V_i,\mathbb{F})$ - the dual vector space. We wish to show that:

• $V_1\otimes V_2\cong L(V_1,V_2;\mathbb{F})$, with some intermediate steps.

Concrete tensor "operation"

This is where I am struggling - I need to make this formal.

• Define $\otimes_{V_1,V_2}:V^*_1\times V^*_2\rightarrow L(V_1,V_2;\mathbb{F})$ by $$\otimes_{V_1,V_2}:(\alpha,\beta)\mapsto\left(\begin{array}{cc}\big(:V_1\times V_2\rightarrow\mathbb{F}\big)\\\big(:(u,v)\mapsto \alpha(u)\beta(v)\big)\end{array}\right)$$ where $\alpha(u)\cdot\beta(v)$ is just the multiplication operation of the field $\mathbb{F}$

Is this a surjection?

• Let $f\in L(V_1,V_2;\mathbb{F})$ be given.
  • We claim: $\exists (\alpha,\beta)\in V^*_1\times V^*_2[\alpha\otimes_{V_1,V_2}\beta\eq f]$ - really I cannot see how to pick $(\alpha,\beta)$ and go from there. So we must construct it....
    • If $f\eq g$ for some functions of the form $(:X\rightarrow Y)$ then $\forall x\in X[f(x)\eq g(x)]$ - we must do this.
    • Let $(u,v)\in V_1\times V_2$ be given - we need to show $(\alpha\otimes\beta)(u,v)\eq f(u,v)$ (using $\otimes$ as short for $\otimes_{V_1,V_2}$)
      • Note that $$u\eq\sum^{n_1}_{c_1\eq 1}u_{c_1}e^{(1)}_{c_1}$$ and $$v\eq\sum^{n_2}_{c_2\eq 1}v_{c_2}e^{(2)}_{c_2}$$ for $n_i:\eq\text{Dim}(V_i)$
      • Then $$f(u,v)\eq f\left(\sum^{n_1}_{c_1\eq 1}u_{c_1}e^{(1)}_{c_1},\sum^{n_2}_{c_2\eq 1}v_{c_2}e^{(2)}_{c_2}\right)$$

        $$\eq\sum^{n_1}_{c_1\eq 1}u_{c_1} f\left(e^{(1)}_{c_1},\sum^{n_2}_{c_2\eq 1}v_{c_2}e^{(2)}_{c_2}\right)$$

        $$\eq\sum^{n_1}_{c_1\eq 1}u_{c_1}\sum^{n_2}_{c_2\eq 1} v_{c_2} f(e^{(1)}_{c_1},e^{(2)}_{c_2})$$

        $$\eq\sum^{n_1}_{c_1\eq 1}\sum^{n_2}_{c_2\eq 1}u_{c_1}v_{c_2}f(e^{(1)}_{c_1},e^{(2)}_{c_2})$$

      • and $$(\alpha\otimes\beta)(u,v)\eq\alpha(u)\beta(v)$$

        $$\eq\alpha\left(\sum^{n_1}_{c_1\eq 1}u_{c_1}e^{(1)}_{c_1}\right)\beta\left(\sum^{n_2}_{c_2\eq 1}v_{c_2}e^{(2)}_{c_2}\right)$$

        $$\eq\left(\sum^{n_1}_{c_1\eq 1}u_{c_1}\alpha(e^{(1)}_{c_1})\right)\beta\left(\sum^{n_2}_{c_2\eq 1}v_{c_2}e^{(2)}_{c_2}\right)$$

        $$\eq\sum^{n_1}_{c_1\eq 1}u_{c_1}\alpha(e^{(1)}_{c_1})\beta\left(\sum^{n_2}_{c_2\eq 1}v_{c_2}e^{(2)}_{c_2}\right)$$

        $$\eq\sum^{n_1}_{c_1\eq 1}u_{c_1}\alpha(e^{(1)}_{c_1})\left(\sum^{n_2}_{c_2\eq 1}v_{c_2}\beta(e^{(2)}_{c_2})\right)$$

        $$\eq\sum^{n_1}_{c_1\eq 1}\sum^{n_2}_{c_2\eq 1}u_{c_1}v_{c_2}\alpha(e^{(1)}_{c_1})\beta(e^{(2)}_{c_2})$$

      • Observe: $$\sum^{n_1}_{c_1\eq 1}\sum^{n_2}_{c_2\eq 1}u_{c_1}v_{c_2}f(e^{(1)}_{c_1},e^{(2)}_{c_2})\eq\sum^{n_1}_{c_1\eq 1}\sum^{n_2}_{c_2\eq 1}u_{c_1}v_{c_2}\alpha(e^{(1)}_{c_1})\beta(e^{(2)}_{c_2})$$ is what we require