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Let $\mathbb P_n$ be the space of all $n \times n$ self-adjoint positive definite matrices. Consider the function $\varphi: \mathbb P_n \longrightarrow \mathbb R$ defined by $$\varphi (A) = \text {tr}\ (A \log A).$$ Show that $\varphi$ is operator convex.

I need to show that for any $A,B \in \mathbb P_n$ $$(1-t) \varphi (A) + t \varphi (B) \geq \varphi ((1-t) A + t B)$$ for all $t \in [0,1].$

I can able to show the result when $A$ and $B$ are both diagonal by using the fact that the function $x \mapsto x \log x$ is convex on $(0,\infty).$ Now for any arbitrary $A$ and $B$ there exist unitary matrices $U$ and $V$ such that $U^* A U = D$ and $V^* B V = D',$ where $D$ and $D'$ are both diagonal matrices. Now it is easy to see that $\varphi (A) = \varphi (D)$ and $\varphi (B) = \varphi (D').$ Hence we have $$\begin{align*} (1-t) \varphi (A) + t \varphi (B) & = (1-t) \varphi (D) + t \varphi (D') \\ & \geq \varphi ((1-t) D + t D') \end{align*}$$ If we can somehow show that $$\varphi ((1-t) D + t D') = \varphi ((1-t) A + t B)$$ then we are through. If $A$ and $B$ commute then it is true since then $A$ and $B$ are simultaneously diagonalizable. But how can it shown if $A$ and $B$ don't commute? Could anyone suggest something needful?

Thanks a bunch!

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