$\newcommand\F{\mathbb{F}}\newcommand\R{\mathbb{R}}\newcommand\C{\mathbb{C}}\newcommand\Z{\mathbb{Z}}\newcommand\tr{\operatorname{trace}}\newcommand\End{\operatorname{End}}$

The trace of a sqaure matrix $A$ is defined to be the sum of the elements along the diagonal. A basic fact is that for any invertible matrix $M$, \begin{equation}\tag{*} \tr(M^{-1}AM) = \tr(A). \end{equation}

Let $V$ be an $n$-dimensional vector space over a field $\F$. The trace of a linear transformation $L: V \rightarrow V$ is usually defined as follows: Let $(v_1, \dots, v_n)$ be a basis of $V$. There exists a matrix $A$ such that $$ L(e_i) = A_i^je_j. $$ Then the trace of $L$ is defined to be $$ \tr(L) = \tr(A). $$ Fact (*) implies that this the right side remains the same, no matter which basis of $V$ is used.

A natural question is whether there is a way to define the trace of a linear transformation directly without using a basis or matrix. This would prove (*). One way is to use the universal property of the tensor product of two vector spaces.

Universal property of tensor product: Let $V$ and $W$ be vector spaces over a field $\F$. The tensor product $V\otimes W$ is a vector space with a bilinear map \begin{align*} B: V\times W &\rightarrow V\otimes W\\ (v,w) &\mapsto v\otimes w \end{align*} such that the following universal property holds: For any vector space $Z$ and bilinear map $$ b: V\times W \rightarrow Z, $$ there exists a unique linear map $$ \bar{b}: V\otimes W \rightarrow Z $$ such that $$ b = \bar{b}\circ B. $$

Space of linear transformations is tensor product: Let $\End(V)$ denote the space of linear transformations from $V$ to itself. The set of all rank $1$ linear transformations is the image of the following bilinear map \begin{align*} \phi: V \times V^* &\rightarrow \End(V), \end{align*} where for any $(v,\ell) \in V\times V^*$, the map $\phi(v,\ell): V \rightarrow V$ is the rank $1$ map such that for aany $w \in V$ to be $$ \phi(v, \ell)(w) = (\ell(w))v. $$ By the universal property above, this extends uniquely to a linear map $$ \bar\phi: V\otimes V^* \rightarrow \End(V). $$ It is straightforward to verify that $\bar\phi$ is an isomorphism.

Trace of linear transformation: The trace of any rank 1 linear transformation is given by the following natural bilinear function \begin{align*} e: V \times V^* &\rightarrow \F\\ (v,\ell) &\mapsto \ell(v) \end{align*} By the universal property above, this extends uniquely to a linear map $$ \bar{e}: V\otimes V^* \rightarrow \F. $$ The trace of a linear transformation $L \in \End(V)$ can now be defined to be $$ \tr(L) = \bar{e}\circ\bar{\phi}^{-1}(L). $$

Acknowledgement: This post was inspired by Levent Alpoge's Princeton PhD generals exam. Just look for "Harvard way".