Log-Determinant Concavity Proof

Note that when $A,B$ are square matrices of the same size, $\det AB=\det BA=\det A\det B~$. We use that here: $$\begin{aligned} \log\det(Z^{1/2}(I+tZ^{-1/2}VZ^{-1/2})Z^{1/2}) &= \log\det((I+tZ^{-1/2}VZ^{-1/2})Z) \\ &= \log\left(\det(I+tZ^{-1/2}VZ^{-1/2})\det Z\right) \\ &= \log\det(I+tZ^{-1/2}VZ^{-1/2}) + \log\det Z. \end{aligned}$$ Let us define $\tilde{V} = Z^{-1/2}VZ^{1/2}$, and let $\tilde{V}=Q\Lambda Q^T$ be the Schur decomposition of $\tilde{V}$: $Q^TQ=QQ^T=I$ and $\Lambda=\mathop{\textrm{diag}}(\lambda_1,\lambda_2,\dots,\lambda_n)$. $$\begin{aligned} \log\det(I+t\tilde{V}) &= \log\det(I+tQ\Lambda Q^T) = \log\det(QQ^T+tQ\Lambda Q^T) \\ &= \log\det Q(I+t\Lambda)Q^T = \log\det Q^TQ(I+t\Lambda) \\ &= \log\det(I+t\Lambda) = \log \prod_i (1+t\lambda_i)\\ &= \sum_i \log(1+t\lambda_i). \end{aligned}$$ The last two steps depend on the fact that $I+t\Lambda$ is diagonal, so its determinant is just the product of its diagonal elements.