Trace and Determinant
Trace
By definition the trace of square matrix
\[
A =
\begin{pmatrix}
a_{11} & a_{12} & \dots & a_{1n} \\
a_{21} & a_{22} & \dots & a_{2n} \\
\vdots & & \ddots & \\
a_{n1} & a_{n2} & \dots & a_{nn} \\
\end{pmatrix}
\]
is the sum of its diagonal elements, i.e.
\[
\Tr(A) = a_{11}+\cdots+a_{nn} = \sum_{i=1}^n a_{ii}
\]
Fact. $\Tr (AB) = \Tr(BA)$.
Fact. $\Tr\bigl(T^{-1}AT\bigr) = \Tr A$
Fact. $\Tr A = \lambda_1 + \cdots + \lambda_n$, i.e. the trace
of a matrix is the sum of its eigenvalues.
In the special case where $A$ is a $3\times3$ matrix with one real eigenvalue
$\lambda$, and two complex eigenvalues $\mu\pm i\omega$, we find that
\[
\Tr A
= \lambda + ( \mu + i\omega ) +( \mu -i\omega ) = \lambda+2\mu.
\]
Thus if we know the real eigenvalue and the trace of $A$ then we can also find
the real part of the complex eigenvalue.
Expanding determinants
Recall that determinants are additive in the sense that
\[
\left|
\begin{array}{cccc}
b_1+c_1 & b_2+c_2 & \cdots & b_n+c_n \\
a_{21}& a_{22}& & a_{2n} \\
& & \ddots &\\
a_{n1}& a_{n2}& & a_{nn} \\
\end{array}
\right |
=
\left|
\begin{array}{cccc}
b_1 & b_2 & \cdots & b_n \\
a_{21}& a_{22}& & a_{2n} \\
& & \ddots &\\
a_{n1}& a_{n2}& & a_{nn} \\
\end{array}
\right |
+
\left|
\begin{array}{cccc}
c_1 & c_2 & \cdots & c_n \\
a_{21}& a_{22}& & a_{2n} \\
& & \ddots &\\
a_{n1}& a_{n2}& & a_{nn} \\
\end{array}
\right |
\]
always holds.
The determinant
\[
D(x) = \left|
\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1n} +x \\
a_{21}& a_{22}& & a_{2n} \\
& & \ddots &\\
a_{n1}& a_{n2}& & a_{nn} \\
\end{array}
\right |
\]
defines a function of $x$. Even though $D(x)$ is defined in terms of a large
determinant, its dependence on $x$ is not very complicated, namely,
\begin{align*}
D(x)&= \left|
\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1n} +x \\
a_{21}& a_{22}& & a_{2n} \\
& & \ddots &\\
a_{n1}& a_{n2}& & a_{nn} \\
\end{array}
\right | \\
&=
\left|
\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21}& a_{22}& & a_{2n} \\
& & \ddots &\\
a_{n1}& a_{n2}& & a_{nn} \\
\end{array}
\right |
+
\left|
\begin{array}{cccc}
0& 0 & & x\\
a_{21}& a_{22} & & a_{2n} \\
& & \ddots &\\
a_{n1}& a_{n2} && a_{nn} \\
\end{array}
\right |
\\[1ex]
&=
\left|
\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21}& a_{22}& & a_{2n} \\
& & \ddots &\\
a_{n1}& a_{n2}& & a_{nn} \\
\end{array}
\right |
+(-1)^{n}x\;
\left|
\begin{array}{cccc}
a_{21}& & a_{2,n-1} \\
& \ddots &\\
a_{n1}& & a_{n,n-1} \\
\end{array}
\right |
\end{align*}
So $D(x)$ is just a linear function of $x$:
\[
D(x) = px+q,
\]
where
\[
p=\left|
\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21}& a_{22}& & a_{2n} \\
& & \ddots &\\
a_{n1}& a_{n2}& & a_{nn} \\
\end{array}
\right |,\qquad
q=(-1)^n
\left|
\begin{array}{cccc}
a_{21}& & a_{2,n-1} \\
& \ddots &\\
a_{n1}& & a_{n,n-1} \\
\end{array}
\right |.
\]