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Here’s another interesting looking twitter math post, this time about 2×2 matrix eigenvalues:
Theorem 1.1: Eigenvalues of a 2×2 matrix.
\begin{equation*}
m \pm \sqrt{ m^2 – p }.
\end{equation*}
This is also not hard to verify.
Start proof:
Let
\begin{equation}\label{eqn:2x2eigen:20}
A =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix},
\end{equation}
where we are looking for \( \lambda \) that satisfies the usual zero determinant condition
\begin{equation}\label{eqn:2x2eigen:40}
\begin{aligned}
0
&= \Abs{ A – \lambda I } \\
&=
\begin{vmatrix}
a – \lambda & b \\
c & d – \lambda
\end{vmatrix} \\
&=
\lr{ a – \lambda } \lr{ d – \lambda } – b c \\
&=
a d – b c – \lambda \lr{ a + d } + \lambda^2 \\
&=
\mathrm{Det}{A} – \lambda \mathrm{Tr}{A} + \lambda^2 \\
&=
\lr{ \lambda – \frac{\mathrm{Tr}{A}}{2} }^2 + \mathrm{Det}{A} – \lr{ \frac{\mathrm{Tr}{A}}{2}}^2,
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:2x2eigen:n}
\lambda = \frac{\mathrm{Tr}{A}}{2} \pm \sqrt{ \lr{ \frac{\mathrm{Tr}{A}}{2}}^2 – \mathrm{Det}{A} }.
\end{equation}
substitution of the variables in the problem statement finishes the proof.
End proof.
Clearly the higher dimensional characteristic equation will also have both a trace and determinant dependency as well, but the cross terms will be messier (and nobody wants to solve cubic or higher equations by hand anyways.)