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.

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 $\det{(AB)}=\det(BA)$) $$\begin{align}\det (I_n-2itA)&=\det(UU^{-1}-2U\Lambda U^{-1})\\\&=\det( \color{red}{U}\color{blue}{(I-2it\Lambda)U^{-1}})\\\&=\det( \color{blue}{(I-2it\Lambda)U^{-1}}\color{red}{U})\\\&=\det(I-2it\Lambda)\\\&=\prod_{1=i}^n(1-2it\lambda_i)\end{align}$$