How to show $\text{Tr}(M\log N)=\sum_{i,j}^n\lambda_i\log(\tilde{\lambda_j})(u_i^{\top}\tilde{u}_j)^2$? The above question is the equation $(2.4)$ of the following paper:
MATRIX EXPONENTIATED GRADIENT UPDATES.
Let $M$ and $N$ be two $n \times n$ positive definite matrices where $M=U\Lambda U^{\top}$, $N=\tilde{U}\tilde{\Lambda} \tilde{U}^{\top} 
$ and $(\lambda_i,v_i)$ are eigenpairs of $M$, likewise for $N$.
How to show the following
$$\text{Tr}(M\log N)=\sum_{i,j}\lambda_i\log(\tilde{\lambda_j})(u_i^{\top}\tilde{u}_j)^2$$
First I do not know what $i,j$ mean in summation, and how we have it using two summations. Second how to get that.
My try:
\begin{align}
\text{Tr}(M\log N) &=
\text{Tr}(U\Lambda U^{\top}
\tilde{U}\log(\tilde{\Lambda})\tilde{U}^{\top}) \\
& = \text{Tr}(\Lambda U^{\top}
\tilde{U}\log(\tilde{\Lambda})\tilde{U}^{\top}U)
\end{align}
How can I proceed using matrix calculus to get the result not by expanding? what is the hidden trick?
 A: The first version of the formula with one index is false. Take $M=\begin{pmatrix}50&37\\37&61\end{pmatrix},N=diag(2,1)$.
A: To easy the notations, let us take
$$
\kappa_j=\log\tilde{\lambda}_j
$$
and
$$
v_j=\tilde{u}_j.
$$
With these notations, it suffices to show
$$
\text{tr}\left(M\log N\right)=\sum_{i,j=1}^n\lambda_i\kappa_j\left(u_i^{\top}v_j\right)^2.
$$
Now, let us begin our proof. Using
\begin{align}
M&=\sum_{j=1}^n\lambda_ju_ju_j^{\top},\\
N&=\sum_{j=1}^n\kappa_jv_jv_j^{\top},
\end{align}
we have
\begin{align}
\text{tr}\left(M\log N\right)&=\text{tr}\left(\sum_{i=1}^n\lambda_iu_iu_i^{\top}\sum_{j=1}^n\kappa_jv_jv_j^{\top}\right)\\
&=\text{tr}\left(\sum_{i,j=1}^n\lambda_i\kappa_ju_iu_i^{\top}v_jv_j^{\top}\right)\\
&=\sum_{i,j=1}^n\lambda_i\kappa_j\text{tr}\left(u_iu_i^{\top}v_jv_j^{\top}\right)\\
&=\sum_{i,j=1}^n\lambda_i\kappa_j\text{tr}\left(v_j^{\top}u_iu_i^{\top}v_j\right)\\
&=\sum_{i,j=1}^n\lambda_i\kappa_j\left(v_j^{\top}u_i\right)\left(u_i^{\top}v_j\right)\\
&=\sum_{i,j=1}^n\lambda_i\kappa_j\left(u_i^{\top}v_j\right)^2.
\end{align}
This completes the proof.
I would like to thank @loupblanc for kind advices.
