Determinant of $I-2itA$

Suppose A is an $n \times n$ positive defnite definite matrix.

I want to show that

$$\det(I_{n}-2itA)=\prod_{i=1}^{n} (1-2it\lambda_{i})$$

where $\lambda_{i}$ are the eigenvalues of A.

I can’t seem to be able to show this! Thanks for your help.

Answer

For the case where $A$ is diagonalisable, we can consider the eigendecomposition of matrix $A$, as $U\Lambda U^{-1}$, where matrix $\Lambda$ is a diagonal matrix whose entries are the eigenvalues of $A$.

We thus have (the coloured parts below is the property that $\det{(AB)}=\det(BA)$)
$$\begin{align}\det (I_n-2itA)&=\det(UU^{-1}-2U\Lambda U^{-1})\\&=\det( \color{red}{U}\color{blue}{(I-2it\Lambda)U^{-1}})\\&=\det( \color{blue}{(I-2it\Lambda)U^{-1}}\color{red}{U})\\&=\det(I-2it\Lambda)\\&=\prod_{1=i}^n(1-2it\lambda_i)\end{align}$$

Attribution
Source : Link , Question Author : tattybojangler , Answer Author : Alijah Ahmed

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