Showing that $\frac{\Vert Au\Vert_2^2}{\Vert u\Vert_2^2}\sim \chi_d^2$ I have for a matrix $A\in R^{d\times p}$ having i.i.d. standard normal entries and for fix $u \in R^p$, then we have that $\frac{\Vert Au\Vert_2^2}{\Vert u\Vert_2^2} \sim \chi_d^2$. If we denote $a_i$ being the $i$th row of $A$, then we have that $\frac{a_i^\top u}{\Vert u\Vert_2}\sim N(0,1)$, but I think I can use some independence of the rows, but I'm not sure. Can anyone help me to show that?

If you denote by $a_i$ the rows of $A$, then each element of $a_i$ has a standard normal distribution. Therefore,
$$
a_i^\top u = a_{i1}u_1 + \cdots +a_{ip}u_p \sim N(0, u^\top u)
$$
and thus
$$
\frac{a_i^\top u}{\sqrt{u^\top u}} \sim N(0,1).
$$
The desired result follows as per @Sycorax comment and by noting that
$$\Vert  Au\Vert_2^2 = (a_1^\top u)^2+\cdots+(a_d^\top u)^2,$$
for $u\neq 0_p$.
