Q:

Convex Optimization: Gradient of $\log \det (X)$

In Boyd's CVX book, there is a step by step analysis of the gradient of so called log det function 

Three confusions:

Is the determinant for positive definite matrix exactly equivalent to the sum of eigen values is equal to the trace?
There is the claim that because $\Delta x$ is small (what does it mean by small), therefore $\lambda_i$ are small, is there any justification to this claim? Because after all we are computing the eigenvalues of $X^{-1/2}\Delta X X^{-1/2}$ not simply $\Delta X$
By first order approximation of $\log(1+\lambda_i) \approx \lambda_i$ I am assuming first order Mac series?

Thanks!

A:

If $A \in S_{++}$  then  $\det A$ is the product of the eigenvalues and $\log \det A$  is the sum of their logarithms.
If $\Delta X$ is small then so is $X^{-1/2} \Delta X X^{-1/2}$. This is just a multiplication by two fixed matrices after all.
You got the MacLaurin series wrong, $\log (1+x) = x - \frac{x^2}{2} + \dots$ for $|x| < 1$. After that correction the argument is OK.   

