I am looking forward to an application of the universal property of the tensor product of vector spaces.

We DEFINE the determinant of a $n\times n$ matrix $A=(a_{ij})$ over a field $F$ by that for any fixed $i=1, 2, ..., n$, $$\det{A}=\sum_{j=1}^{n}(-1)^{i+j}a_{ij} |A_{ij}|,$$ where $|A_{ij}|$ is the minor of $A$ at position $(i, j)$.

The determinant is a multilinear form from $\overbrace{F^{n}\times F^n\times \cdots \times F^n}^{n \text{ times}}$ to $F$. By the universal property of tensor product of vector spaces, we have the following commutative diagram. $$ \begin{array}{rrl} {} \overbrace{F^{n}\times F^n\times \cdots \times F^n}^{n \text{ times}} & \longrightarrow & \overbrace{F^{n}\otimes F^n\otimes \cdots \otimes F^n}^{n \text{ times}} \\ & \det{}\searrow & \downarrow \theta \\ && F \end{array} $$ where $\theta$ is a linear transformation (functional).

Question. Can we prove the two important properties $\det{AB}=\det{A}\cdot \det{B}$ and $\det{A^t}=\det{A}$ via the linearity of $\theta$?

I want to use this question as a motivation of tensor product of vector spaces. So please don't use any concept which is deeper than tensor product, e.g., exterior product.

And avoiding prove it by the elementary method (as possible), e.g., rank and elementary matrix. Because I hope this question can show the power of tensor product. That is, it should be more elegent than the elementary proof.

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    $\begingroup$ This won't be enough—it's absolutely critical to observe that the determinant isn't just a multilinear form but a totally antisymmetric multilinear form. Hence, the universal object you really want to work with isn't the $n$-fold tensor product $\otimes^n F^n$ but the *$n$th exterior power* $\wedge^n F^n$, see, e.g., en.wikipedia.org/wiki/Exterior_algebra $\endgroup$ – Branimir Ćaćić Jun 18 '17 at 10:54
  • $\begingroup$ @BranimirĆaćić Thanks for your comment. But there is a textbook sates the theorem (Corollary 16, Section 10.4.(2), Foote's Abstract Algebra). Let $R$ be a commutative ring and let $M_1, ..., M_n$, $L$ be $R$-modules. Let $M_1\otimes M_2\otimes \cdots \otimes M_n$ denote any bracketing of the tensor product of these modules and let $\iota:M_1\times \cdots \times M_n\to M_1\otimes \cdots \otimes M_n$ be the map defined by $\iota(m_1, ..., m_n)=m_1\otimes \cdots\otimes m_n$. $\endgroup$ – bfhaha Jun 18 '17 at 11:54
  • $\begingroup$ Then if $\varphi:M_1\times \cdots \times M_n\to L$ is an $n$-multilinear map then there is a unique $R$-module homomorphism $\Psi:M_1\otimes \cdots \otimes M_n\to L$ such that $\varphi=\Psi\circ \iota$. Are there something I misunderstand? $\endgroup$ – bfhaha Jun 18 '17 at 11:54
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    $\begingroup$ That's absolutely true. The issue is that the usual tensor product is too crude a tool to prove the properties of the determinant that you want to prove in the sort of way that you want. The antisymmetry of the determinant is absolutely essential to its basic properties, so you need to use a refinement of the tensor algebra called the exterior algebra, which is the universal object specific to antisymmetric forms. $\endgroup$ – Branimir Ćaćić Jun 18 '17 at 12:09
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    $\begingroup$ See section 11.4 on determinants in Dummit and Foote. $\endgroup$ – Omnomnomnom Jun 18 '17 at 15:18

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