$\Vert Z\Vert_2^2\overset{\text{d}}{\underset{\text{}}{=}}\sum_{k=1}^d\lambda_kw_k^2,\,$ where $w_k\sim N(0,1)$ i.i.d. 
Let $\Sigma\in \mathbb R^{d,d}$ positive semi definit and $Z\sim N(0,\Sigma)$ a random vector in $\mathbb R^d$. 
  I want to show that $\Vert Z\Vert_2^2\overset{\text{d}}{\underset{\text{}}{=}}\sum_{k=1}^d\lambda_kw_k^2,\,$ where $w_k\sim N(0,1)$ i.i.d. random variables and $\lambda_i \,,i=1,\dots d$ are the eigenvalues of $\Sigma$.

My thoughts:
Since $\Sigma$ is postiv semi defint all its eigenvalues are non negative. By definition of $Z=(Z_1,\dots,Z_d)$ where each $Z_i \sim N(0,\sigma_i^2)$ $$\Vert Z\Vert_2^2= Z_1^2+\dots + Z_d^2=\sigma_1^2 \frac{Z_1^2}{\sigma_1^2}+\dots + \sigma_d^2 \frac{Z_d^2}{\sigma_d^2}\overset{\text{d}}{\underset{\text{}}{=}}\chi_1^2\sigma_1^2+\dots + \chi_1^2\sigma_d^2$$I do not see how to continue here, clearly each $w_i^2\sim \chi_1^2$, but now?

Let $W\sim N(0,I)$ and use eigendecomposition (see comment above) to write $\Sigma = Q\Lambda Q’$. Then $Z’Z \sim W’ \Sigma W \sim (Q’W)’ \Lambda (Q’W)\sim W’\Lambda W$, where the last equality in distribution follows from $Var(Q’W) = Q’Q=I$.
