$\mathbb{E}[Z \, Z^T] = I$ with $Z$ gaussian standard vector and $I$ identity matrix

I'm wondering why $\mathbb{E}[Z \, Z^T] = I$ with $Z$ gaussian standard vector and $I$ identity matrix.

If I expand the expectation, I get:

$$ \mathbb{E}[Z \, Z^T] = \mathbb{E} [Z_1^2 + \quad \ldots \quad + Z_d^2] $$

Solution 1:

If $Z=(z_1,z_2,...,z_d)$ is s.t. $z_k\sim \mathcal{N}(0,1),k\leq d$ IID then $$ZZ'=\begin{bmatrix} z_1^2&z_1z_2&...&z_1z_d\\ z_1z_2&z_2^2&...&z_2z_d\\ ...&...&...&...\\ z_1z_d&z_2z_d&...&z_d^2 \end{bmatrix}$$ and $$\implies E[ZZ']=\begin{bmatrix} E[z_1^2]&E[z_1z_2]&...&E[z_1z_d]\\ E[z_1z_2]&E[z_2^2]&...&E[z_2z_d]\\ ...&...&...&...\\ E[z_1z_d]&E[z_2z_d]&...&E[z_d^2] \end{bmatrix}=\begin{bmatrix} 1&0&...&0\\ 0&1&...&0\\ ...&...&...&...\\ 0&0&...&1 \end{bmatrix}=I_d$$