Another basis independent property of linear operators is the trace.

Definition The trace of an \(n\times n\) matrix \(\mathbf{A}\), \(\tr\mathbf{A}\) is defined to be
\begin{equation}
\tr\mathbf{A}=\sum_{i=1}^nA_i^i=A_i^i.
\end{equation}

As \(\tr\mathbf{A}\mathbf{B}=(AB)^i_i=A^i_jB^j_i=(BA)^i_i=\tr\mathbf{B}\mathbf{A}\) it follows that \(\tr\mathbf{P}^{-1}\mathbf{A}\mathbf{P}=\tr\mathbf{A}\) so it makes sense to define the trace of any linear operator \(T:V\mapto V\), \(\tr T\), as the trace of any matrix representation of \(T\).

Working over an algebraically closed field \(K\), since any matrix \(\mathbf{A}\in\text{Mat}_n(K)\) is similar to an upper triangular matrix, we have \(\tr\mathbf{A}=\sum_{i=1}^n\lambda_i\) and \(\det\mathbf{A}=\prod_{i=1}^n\lambda_i\) (not all \(\lambda_i\) necessarily distinct), quantities which are in fact encoded as particular coefficients in the characteristic polynomial,
\begin{equation}
p_\mathbf{A}(x)=x^n-\tr\mathbf{A}x^{n-1}+e_2x^{n-2}-\dots+(-1)^{n-1}e_{n-1}x+(-1)^n\det\mathbf{A}.
\end{equation}
The coefficients \(e_1=\tr\mathbf{A},e_2,\dots,e_{n-1},e_n=\det\mathbf{A}\) are called the elementary symmetric functions.

There is a nice relationship between the trace and determinant. It can be shown that the matrix exponential,
\begin{equation}
\exp\mathbf{A}=\sum_{i=0}^\infty\frac{1}{i!}\mathbf{A}^i=\mathbf{I}_n+\mathbf{A}+\frac{1}{2!}\mathbf{A}^2+\dots,
\end{equation}
converges for any \(n\times n\) matrix over \(K\). Consider then, the function on \(\RR\) defined as,
\begin{equation*}
f(t)=\det\exp\mathbf{A}t.
\end{equation*}
Differentiating, we find that
\begin{equation*}
\frac{df(t)}{dt}=\tr\mathbf{A}f(t),
\end{equation*}
so that \(\ln f(t)=\tr\mathbf{A}t\) and in particular we have the following relationship between the determinant and trace,
\begin{equation}
\det\exp\mathbf{A}=\exp\tr\mathbf{A}.
\end{equation}