Trace is a coefficient of the characteristic polynomial

Statement

Let $K$ be a commutative ring, let $n \ge 1$, and let $A$ be an $n \times n$ matrix over $K$. Then the coefficient of $t^{n-1}$ in the characteristic polynomial $p_A(t) = \det(t I_n - A)$ is $-\tr A$, where $\tr$ is the trace. So $p_A(t) = t^n - (\tr A)\, t^{n-1} + \cdots$.

Remarks

In the Leibniz expansion of $\det(t I_n - A)$, only the product of the diagonal entries $(t - a_{11}) \cdots (t - a_{nn})$ contributes to $t^{n-1}$. For $n = 2$, $p_A(t) = t^2 - (\tr A) t + \det A$. Together with the constant coefficient, it shows that $\tr$ and $\det$ are invariants of similarity.

Explanations

L1 human

The characteristic polynomial contains the trace as its second coefficient, with a minus sign. For a $2 \times 2$ matrix, the full polynomial is $t^2 - (\text{trace})\,t + \text{determinant}$. So the trace and the determinant are part of the information of the characteristic polynomial.