Diagonal Matrices and the Hadamard Product I am trying to show that
$$ \begin{bmatrix} \Sigma{11} \\ \vdots \\ \Sigma{nn}  \end{bmatrix} =
(\boldsymbol{\mathbf{Q}} \odot \mathbf{Q})\boldsymbol{\lambda} $$
Where $$\boldsymbol{\Sigma} =\bf{QDQ^{^{T}}}$$  and   $$\mathbf{D}=diag(\boldsymbol{\lambda}) $$ 
with the eigenvalues of Σ.
What rules or properties can I use to start tackling this problem?
Thanks in advance!
 A: The third order tensor ${\cal H}$ with components
$$\eqalign{
{\cal H}_{ijk}  &= \begin{cases} 1 &\text{if }(i=j=k) \\ 0 & \text{otherwise}\end{cases}
}$$
has some very useful properties.
It can be used to write both of the diag-operators and the Hadamard product of two vectors.
$$\eqalign{
{\rm Diag}(a) &= {\cal H}\cdot a &= ({\rm diagonal\,matrix\,from\,a\,vector}) \\
{\rm diag}(B) &= {\cal H}:B  &= ({\rm vector\,from\,diagonal\,of\,a\,matrix}) \\
x\odot y &= x^T\cdot{\cal H}\cdot y &= ({\rm hadamard\,product})\\
x\odot y  &= {\cal H}:xy^T &= ({\rm ditto})\\
}$$
where $(\cdot)$ and $(:)$ represent the dot product and double-dot product.
Define two rank-one matrices 
$${
A = ab^T,\; F=fg^T
\quad\implies\;
A^T = ba^T,\; F^T=gf^T}$$
The Hadamard product of these matrices is
$$\eqalign{
A\odot F^T
 &= (ab^T)\odot(gf^T) \\
 &= (a\odot g)\,(b\odot f)^T \\
}$$
Given a vector $v$, expand the following expression.
$$\eqalign{
x &= {\rm diag}\Big(A\cdot{\rm Diag}(v)\cdot F\Big) \\
  &= {\cal H}:\Big((ab^T)\cdot({\cal H}\cdot v)\cdot(fg^T)\Big) \\
  &= {\cal H}:\Big(ag^T(b^T\cdot{\cal H}\cdot f)\cdot v\Big) \\
  &= ({\cal H}:ag^T)\,(b\odot f)^Tv \\
  &= (a\odot g)\,(b\odot f)^Tv \\
  &= \big(A\odot F^T\big)\,v \\
}$$
So this confirms the result for rank-one matrices.
I'll leave it to you to work out the case for full-rank matrices.
But note that using the SVD allows you to write any matrix as a sum of rank-one matrices, e.g. 
$$\eqalign{
Q
 &= \sum_k \sigma_k u_k v_k^T
  = \sum_k a_k b_k^T
  = \sum_k A_k
\\
}$$
