Variance identity for i.i.d mean-zero random vector. Let $Z_1, \ldots, Z_k$ independent mean-zero random vectors in $\mathbb{R}^n$. Show that
\begin{equation}
\mathbb{E} \left\| \sum_{j=1}^k Z_j \right\|_2^2 =
\sum_{j=1}^k \mathbb{E} \left\|  Z_j \right\|_2^2
\end{equation}
Answer in correct place. Thank you all for the confirmation.

As we know:
\begin{equation}
\left\| Z \right\|_2^2 = \langle Z, Z \rangle = Z^T Z = \sum_{i=1}^n  z_i z_i
\end{equation}
with $z_i$ being the $i$-th component of $Z$ vector. We can rewrite the left hand side as
\begin{align}
\mathbb{E} \left\| \sum_{j=1}^k Z_j \right\|_2^2 &= \mathbb{E} \left\langle \sum_{j=1}^k Z_j,  \sum_{j=1}^k Z_j \right\rangle \\
&= \mathbb{E} \sum_{i=1}^n \sum_{j=1}^k {Z_j}_i {Z_j}_i \\
&= \mathbb{E} \sum_{j=1}^k \sum_{i=1}^n  {Z_j}_i {Z_j}_i \\
\text{assuming mean 0} \\
&= \sum_{j=1}^k \left( \mathbb{E} \sum_{i=1}^n  {Z_j}_i {Z_j}_i  \right) \\
&= \sum_{j=1}^k  \mathbb{E} \langle  Z_j, Z_j \rangle  \\
&= \sum_{j=1}^k  \mathbb{E} \left\| Z_j \right\|^2_2 \\
\end{align}
