I am trying to learn some matrix differentiation, and came across example of calculating the derivative of $$f(X)=\log\det(X)$$ where the $X$ is a symmetric positive definite matrix. I came to the underlined step, but I couldn't make out how this step is derived. What's the theory or theorems underlying this step? I tried to look into the determinant formulas to work it out, but sadly, failed. So, would anyone please be kind enough to explain to me?
The first step is to use the fact that the determinant of a matrix is given by the product of its eigenvalues so if $A$ is any matrix and its eigenvalues are $\alpha_1,...,\alpha_n$ then $$ \det A = \prod_{i=1}^n \alpha_i. $$ Using the fact that the log of a product is the sum of the logs it follows $$ \log \det A = \log \left ( \prod_{i=1}^n \alpha_i \right )= \sum_{i=1}^n \log(\alpha_i) $$ The rest of the work is applying the fact that for any matrix $A$ with eigenvalues $\alpha_1,...,\alpha_n$ the matrix $I+A$ has eigenvalues $1+\alpha_1,...,1+\alpha_n$. To see this let $v_i$ be the $i$th eigenvector of $A$, that is $Av_i = \alpha_iv_i$. Then $$ (I+A)v_i = Iv_i + Av_i = 1v_i + \alpha_iv_i = (1+\alpha_i)v_i. $$ Combining this with the identity for $\log \det$ gives for any matrix $A$ that $$\log \det (I + A) = \sum_{i=1}^n \log(1 + \alpha_i).$$ The line in question is just applying this formula with $A = X^{-1/2}\Delta X X^{-1/2}$ since they assume $\lambda_1,...,\lambda_n$ are the eigenvalues of $X^{-1/2}\Delta X X^{-1/2}$.