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!
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. $$\eqalign{ Q &= \sum_k \sigma_k u_k v_k^T = \sum_k a_k b_k^T = \sum_k A_k \\ }$$