Determinant of $I-2itA$

Question

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.

You've managed to mix two different meanings of $i$ in the same expression, I think. — Marc van Leeuwen, Apr 19 '17 at 03:35

Alijah Ahmed · Accepted Answer · 2017-04-18T18:41:18.943

3

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

edited Apr 18 '17 at 18:41

answered Apr 18 '17 at 18:29

Alijah Ahmed

11,609

This approach assumes that $A$ is diagnolizable... – Semiclassical Apr 18 '17 at 18:37
Since A is positive definite it has can be expressed in the form above by spectral decomposition. – tattybojangler Apr 18 '17 at 18:41
But thanks for the help! It makes perfect sense! – tattybojangler Apr 18 '17 at 18:41
@tattybojangler That's false. See for instance the matrix in this question: https://math.stackexchange.com/q/657621/137524 – Semiclassical Apr 18 '17 at 18:44
1

A positive definite matrix is symmetric and therefore has a spectral decomposition. – tattybojangler Apr 18 '17 at 18:47
1

Looking at some other sources, whether or not positive-definiteness assumes symmetric is a matter of convention. But if that's your context than that's fine. – Semiclassical Apr 18 '17 at 18:52

score 2 · Answer 2 · answered Apr 18 '17 at 18:55

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, $\det (\mathbf I-2it\mathbf A)=\prod_{i=1}^n(1-2it\lambda_i)$

score 0 · Answer 3 · answered Apr 18 '17 at 18:43

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.

Determinant of $I-2itA$

3 Answers3