[proofplan]
The trace of an $n \times n$ matrix is the sum of its diagonal entries. The proof rests on a single algebraic property — the **cyclic invariance** of the trace, $\operatorname{tr}(BC) = \operatorname{tr}(CB)$ — which follows by interchanging the order of summation in the double sum defining the trace of a product. Applying this identity to the product $X A X^{-1}$ with the bracketing $(X A)(X^{-1})$ moves the $X^{-1}$ factor to the front, giving $\operatorname{tr}(X^{-1} X A) = \operatorname{tr}(A)$.
[/proofplan]
[step:Establish the cyclic property of the trace for two matrices]
Let $B, C \in \operatorname{Mat}_n(\mathbb{F})$. The trace
\begin{align*}
\operatorname{tr}: \operatorname{Mat}_n(\mathbb{F}) &\to \mathbb{F} \\
M &\mapsto \sum_{i=1}^n M_{ii}
\end{align*}
is by definition the sum of diagonal entries. Using the matrix product formula $(BC)_{ii} = \sum_{j=1}^n B_{ij} C_{ji}$:
\begin{align*}
\operatorname{tr}(BC) = \sum_{i=1}^n (BC)_{ii} = \sum_{i=1}^n \sum_{j=1}^n B_{ij} C_{ji} = \sum_{j=1}^n \sum_{i=1}^n C_{ji} B_{ij} = \sum_{j=1}^n (CB)_{jj} = \operatorname{tr}(CB).
\end{align*}
The interchange of the two finite sums is valid because addition in $\mathbb{F}$ is commutative and associative.
[guided]
We start by proving the most basic identity available — that the trace does not see the order of a product of two matrices. The trace is defined entry by entry as
\begin{align*}
\operatorname{tr}: \operatorname{Mat}_n(\mathbb{F}) &\to \mathbb{F} \\
M &\mapsto \sum_{i=1}^n M_{ii},
\end{align*}
so the only tool we have is direct computation. Recall the formula for matrix multiplication: $(BC)_{ii} = \sum_{j=1}^n B_{ij} C_{ji}$. Substituting:
\begin{align*}
\operatorname{tr}(BC) = \sum_{i=1}^n (BC)_{ii} = \sum_{i=1}^n \sum_{j=1}^n B_{ij} C_{ji}.
\end{align*}
Now we ask: can we swap the roles of $B$ and $C$? The double sum runs over all pairs $(i, j) \in \{1, \ldots, n\}^2$, and addition in $\mathbb{F}$ is commutative and associative, so we may reorder the terms. Interchanging the order of summation and relabelling:
\begin{align*}
\sum_{i=1}^n \sum_{j=1}^n B_{ij} C_{ji} = \sum_{j=1}^n \sum_{i=1}^n C_{ji} B_{ij} = \sum_{j=1}^n (CB)_{jj} = \operatorname{tr}(CB).
\end{align*}
This is the **cyclic property** for two factors: $\operatorname{tr}(BC) = \operatorname{tr}(CB)$. Note that we did **not** use commutativity of matrix multiplication — only commutativity of scalar addition. The matrices $B$ and $C$ themselves need not commute.
[/guided]
[/step]
[step:Apply the cyclic property to $XAX^{-1}$ with bracketing $(XA)(X^{-1})$]
Set $B = XA$ and $C = X^{-1}$ in the cyclic identity from the previous step:
\begin{align*}
\operatorname{tr}(XAX^{-1}) = \operatorname{tr}\bigl((XA)\, X^{-1}\bigr) = \operatorname{tr}\bigl(X^{-1}\, (XA)\bigr) = \operatorname{tr}(X^{-1}XA).
\end{align*}
Since $X \in \operatorname{GL}_n(\mathbb{F})$, the matrix $X^{-1}$ exists and satisfies $X^{-1}X = I_n$. By associativity of matrix multiplication, $X^{-1}(XA) = (X^{-1}X)A = I_n A = A$. Therefore
\begin{align*}
\operatorname{tr}(XAX^{-1}) = \operatorname{tr}(A),
\end{align*}
which is the desired identity.
[guided]
We now apply the cyclic property to the matrix of interest. The expression $XAX^{-1}$ is a triple product, so we need to choose a bracketing to fit the form $\operatorname{tr}(BC)$. The strategic choice is $B := XA$, $C := X^{-1}$, because then the swap $BC \to CB$ moves the $X^{-1}$ to the **front**, where it can cancel against the $X$ in $XA$.
Substituting into the cyclic identity:
\begin{align*}
\operatorname{tr}(XAX^{-1}) = \operatorname{tr}\bigl((XA)\, X^{-1}\bigr) = \operatorname{tr}\bigl(X^{-1}\, (XA)\bigr).
\end{align*}
The hypothesis $X \in \operatorname{GL}_n(\mathbb{F})$ guarantees that $X^{-1}$ exists and that $X^{-1}X = I_n$ — this is the only place where invertibility of $X$ is used. Now associativity of matrix multiplication lets us regroup:
\begin{align*}
X^{-1}(XA) = (X^{-1}X)A = I_n A = A.
\end{align*}
Substituting back:
\begin{align*}
\operatorname{tr}(XAX^{-1}) = \operatorname{tr}(A).
\end{align*}
This completes the proof. The same argument shows that the trace is invariant under conjugation by **any** invertible matrix, not just elements of $\operatorname{GL}_n(\mathbb{F})$ acting on $\operatorname{Mat}_n(\mathbb{F})$ — the hypothesis $A \in \operatorname{GL}_n(\mathbb{F})$ is in fact stronger than necessary; the identity $\operatorname{tr}(XAX^{-1}) = \operatorname{tr}(A)$ holds for any $A \in \operatorname{Mat}_n(\mathbb{F})$ and any invertible $X$.
[/guided]
[/step]
