Coordinate-free proof of $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$?

Solution 1:

$\newcommand{\tr}{\operatorname{tr}}$Here is an exterior algebra approach. Let $V$ be an $n$-dimensional vector space and let $\tau$ be a linear operator on $V$. The alternating multilinear map $$ (v_1,\dots,v_n) \mapsto \sum_{k=1}^n v_1 \wedge\cdots\wedge \tau v_k \wedge\cdots\wedge v_n $$ induces a unique linear operator $\psi: \bigwedge^n V \to \bigwedge^n V$. The trace $\tr(\tau)$ is defined as the unique number satisfying $\psi = \tr(\tau)\iota$, where $\iota$ is the identity. (This is possible because $\bigwedge^n V$ is one-dimensional.)

Let $\sigma$ be another linear operator. We compute \begin{align} (\tr\sigma)(\tr\tau) v_1 \wedge\cdots\wedge v_n &= \sum_{k=1}^n (\tr\sigma) v_1 \wedge\cdots\wedge \tau v_k \wedge\cdots\wedge v_n \\ &= \sum_{k=1}^n v_1 \wedge\cdots\wedge \sigma \tau v_k \wedge\cdots\wedge v_n \\ & \qquad + \sum_{k=1}^n \sum_{j \ne k} v_1 \wedge\cdots\wedge \sigma v_j \wedge \cdots \wedge \tau v_k \wedge\cdots\wedge v_n. \end{align}

Notice that the last sum is symmetric in $\sigma$ and $\tau$, and so is $(\tr\sigma)(\tr\tau) v_1 \wedge\cdots\wedge v_n$. Therefore $$ \sum_{k=1}^n v_1 \wedge\cdots\wedge \sigma \tau v_k \wedge\cdots\wedge v_n = \sum_{k=1}^n v_1 \wedge\cdots\wedge \tau \sigma v_k \wedge\cdots\wedge v_n, $$ i.e. $\tr(\sigma\tau)=\tr(\tau\sigma)$.

EDIT: To see that the trace is the sum of all eigenvalues, plug in your eigenvectors in the multilinear map defined at the beginning.

Solution 2:

The proof in Martin Brandenburg's answer may look scary but it is secretly about moving beads around on a string. You can see all of the relevant pictures in this blog post and in this blog post. The proof using pictures is the following:

In the first step $g$ gets slid down on the right and in the second step $g$ gets slid up on the left.

You can also find several proofs of the stronger result that $AB$ and $BA$ have the same characteristic polynomial in this blog post.

Solution 3:

The trace of an endomorphism $f : X \to X$ of a dualizable object $X$ in a monoidal category is the composition $1 \xrightarrow{\eta} X \otimes X^* \xrightarrow{f \otimes \mathrm{id}} X \otimes X^* \cong X^* \otimes X \xrightarrow{\epsilon} 1$. This coincides with the usual definition in the category of vector spaces. There is a more general categorical notion of trace, which then also applies to Hilbert spaces. Under suitable assumptions the formula $\mathrm{tr}(f \circ g)=\mathrm{tr}(g \circ f)$ holds. For more details, see the paper Traces in monoidal categories by Stolz and Teichner.

Solution 4:

The following is a simple combinatorial interpretation of this identity. Not exactly what you asked for, but still fun and relevant.

Suppose we have two sets $S,T$ with functions $g: S \to T$ and $f : T \to S$. Then $f\circ g : S \to S$ and $g\circ f: T \to T$ are endo-functions of $S$ and $T$ respectively. Now consider $\text{Fix}(f\circ g) \subseteq S$, the set of fixed points of $f\circ g$. It is easy to verify that

$$f|_{\text{Fix} (fg)}: \text{Fix} (fg) \to \text{Fix} (gf)$$

is a bijection, with inverse $g|_{\text{Fix} (gf)}$. Therefore, if $S,T$ are finite,

$$|\text{Fix} (fg)| = |\text{Fix} (gf)|.$$

But if $S,T$ are finite, we can represent $f$ as a $|S| \times |T|$ matrix and $g$ as a $|T| \times |S|$ matrix, each with $0$'s and $1$'s. (This matrix depends on an ordering of each set.) Then their products in either order represent the endo-functions $fg$ and $gf$. But it is obvious that for the matrix of an endo-function $h$, $|\text{Fix }h| = \text{Tr}(h)$ (irrespective of the order chosen). Thus, by the above, we $\text{Tr}(fg)=\text{Tr}(gf)$.

Solution 5:

Hint Compare the characteristic polynomials of $AB$ and $BA$.

The determinant (whence characteristic polynomials) admits basis-free definitions.

We have $$ \left(\matrix{I&A\\B&tI}\right)\left(\matrix{tI&-A\\0&I}\right)=\left(\matrix{tI&0\\*&tI-BA}\right) $$ and $$ \left(\matrix{I&A\\B&tI}\right)\left(\matrix{tI&0\\-B&I}\right)=\left(\matrix{tI-AB&*\\0&tI}\right). $$ Applying the determinant to these equations yields $$ t^m\det(tI-AB)=t^n\det(tI-BA). $$

Now over an algebraically closed field, we can define the eigenvalues of a linear operator as the zeros of its characteristic polynomial counted with multiplicities. The trace, which you defined as the sum of the latter, is $-1$ times the coefficient of degree $k-1$. So the formula above proves in particular that $\mathrm{tr}(AB)=\mathrm{tr}(BA)$.

Note I don't know how to prove without refering to any basis that the characteristic polynomial is actually a polynomial of degree $k$ with leading coefficient $1$. So I'm afraid this is a bit circular. Anyway, I don't think this is a very convenient way of defining the trace. For a viewpoint which is more useful when seeking infinite-dimensional generalizations, that other answer is probably more useful than what I just wrote above.