Determinant of a modal matrix used in diagonalization
By Emily Wilson •
If A $\in \mathbb{R}^{n\times n}$ is diagonalized as $P^{-1}AP=\Lambda$ with P the modal matrix composed of the eigenvectors of $A$, is there a general way to determine the determinant of $P$, except that it is not equal to zero?
$\endgroup$1 Answer
$\begingroup$No. The problem begins with the fact that the eigenvectors themselves are not determined uniquely. Given $P$ you can multiply the first column by any non-zero scalar $c$ and obtain $P'$. Then you will still have $P'^{-1}AP' = \Lambda$ and $\det(P') = c \det(P)$. When you diagonalize $A$, the determinant of $P$ will depend on which eigenvectors you choose from each eigenspace and can be an arbitrary non-zero scalar.
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