Lemma: $\forall A\in R^{n\times n}$ and a diagonal matrix $\forall W\in R^{n\times n}$ with $ w_{11}\geq w_{22}\geq ...\geq w_{_{nn}} >0$. The singular value decomposition of A denoted by: $A=XM Y^{T}$, we have:
$tr\left ( WU^{T}AV\right )\leq tr\left ( WX^TAY\right )$
where $U^{T}U=I; V^{T}V=I;$; I is identity matrix. When U=X and V=Y, the $tr\left ( WU^{T}AV\right )$ reaches maximum.
Please help me to prove the above lemma. Thank you very much.
Your Lemma is wrong. Consider the following counter example:
$W = \begin{bmatrix} 10 & 0 \\ 0 & 1 \end{bmatrix}$, $A = \begin{bmatrix} 1 & 0 \\ 0 & 10 \end{bmatrix}$, $U = V=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, $ X = Y = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ leads to
which is a contradiction to $tr\left ( WU^{T}AV\right )\leq tr\left ( WX^TAY\right )$.