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. 
 A: 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 $\det{(AB)}=\det(BA)$)
$$\begin{align}\det (I_n-2itA)&=\det(UU^{-1}-2U\Lambda U^{-1})\\&=\det( \color{red}{U}\color{blue}{(I-2it\Lambda)U^{-1}})\\&=\det( \color{blue}{(I-2it\Lambda)U^{-1}}\color{red}{U})\\&=\det(I-2it\Lambda)\\&=\prod_{1=i}^n(1-2it\lambda_i)\end{align}$$ 
A: You have $\det \mathbf{A}=\prod_{i=1}^{n}\lambda_i$ and $\mathbf{Ax_i}=\lambda_i \bf x_i$ $\quad(i=1,2,...,n)$
So, $(\mathbf I-2it\mathbf A)\mathbf x_i=\mathbf x_i-2it(\lambda_i \mathbf x_i)=(1-2\lambda_i it)\mathbf x_i$
$\Rightarrow 1-2\lambda_i it$ is an eigenvalue of $\mathbf I-2it\mathbf A$ corresponding to the eigenvector $\mathbf x_i$
Thus, $\det (\mathbf I-2it\mathbf A)=\prod_{i=1}^n(1-2it\lambda_i)$
A: When you say "$\lambda_i$ are the eigenvalues of $A$", you probably mean to say that $\prod_{i=1}^n(X-\lambda_i)$ is the characteristic polynomial $\det(XI_n-A)$. Substitute $X:=(2\mathbf it)^{-1}$ and multiply the resulting expression by $(2\mathbf it)^n$ to obtain $\det(I_n-2\mathbf itA)=\prod_{i=1}^n(1-2\mathbf it\lambda_i)$. (Here $\mathbf i^2$ is $-1$, but $i^2$ is not.)
Note that as long as that is the characteristic polynomial, it does not matter whether $A$ is diagonalisable.
