Representation of negative Quantum entropy in terms of eigenvalues, i.e., $\text{Tr}(M\log M -M)=\sum_{i=1}^{n}(\lambda_i\log(\lambda_i)-\lambda_i)$?

Question

Negative Quantum entropy or Negative Von Nuemann entropy is defined as $f(M)=\text{Tr}(M\log M -M)$.

Where $M$ is a positive definite matrix in $\mathbb{S}_+^n$, $\log$ is natural matrix logarithm for which $\log(M)$ is defined as $\log(M)=\sum_{i=1}^{n}\log(\lambda_i)v_iv_i^T$ where $(\lambda_i,v_i)$ are eigenpairs of $M$.

Show $f(M)=\text{Tr}(M\log M -M)=\sum_{i=1}^{n}(\lambda_i\log(\lambda_i)-\lambda_i)$.

Answer 1

Since $M\in\mathbb{S}_+^n$, there must exist an orthogonal matrix $U$ and a diagonal matrix $\Lambda=\text{diag}\left\{\lambda_1,\lambda_2,...,\lambda_n\right\}$, with each $\lambda_j>0$, such that $$ M=U\Lambda U^{\top}. $$ Hence, using the definition of $\log M$, \begin{align} M\log M-M&=\left(U\Lambda U^{\top}\right)\left(U\log\Lambda\,U^{\top}\right)-U\Lambda U^{\top}\\ &=U\left(\Lambda\log\Lambda-\Lambda\right)U^{\top}. \end{align} Consequently, \begin{align} f(M)&=\text{tr}\left(M\log M-M\right)\\ &=\text{tr}\left(U\left(\Lambda\log\Lambda-\Lambda\right)U^{\top}\right)\\ &=\text{tr}\left(\left(\Lambda\log\Lambda-\Lambda\right)U^{\top}U\right)\\ &=\text{tr}\left(\Lambda\log\Lambda-\Lambda\right)\\ &=\sum_{j=1}^n\lambda_j\left(\log\lambda_j-1\right). \end{align}