Theorem 6.2.4.

The trace and determinant of a \(2 \times 2\) matrix are invariant under a change of coordinates. That is, \(\det(T^{-1} A T) = \det(A)\) and \(\trace(T^{-1} A T) = \trace(A)\) for any \(2 \times 2\) matrix \(A\) and any invertible \(2 \times 2\) matrix \(T\text{.}\)

Proof.

It is straightforward to verify that \(\det(AB) = \det(A) \det(B)\) and \(\det(T^{-1}) = 1/\det(T)\) for \(2 \times 2\) matrices \(A\) and \(B\text{.}\) Therefore,
\begin{equation*} \det(T^{-1} A T) = \det(T^{-1}) \det(A) \det(T) = \frac{1}{\det(T)} \det(A) \det(T) = \det(A). \end{equation*}
A direct computation shows that \(\trace(AB) = \trace(BA)\text{.}\) Thus,
\begin{equation*} \trace(T^{-1} A T) = \trace (T^{-1} T A ) = \trace(A). \end{equation*}
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