This is regarding the compressed sensing problem. In the problem $y=X\beta$ where $y\in\mathbb{R}^{n}$, $X\in\mathbb{R}^{n\times p}$, and $\beta\in\mathbb{R}^{p}$, it is known that a random matrix $X$ where $X\sim_{\text{iid}}N(0,1/n)$ satisfies the restricted isometry property (RIP) condition with high probability, i.e.,
$$ \frac{\|X\beta\|_2^2}{\|\beta\|_2^2}\approx1. $$
See definition 1 of this note for a more formal definition.
My question: We have a matrix constructed from a diagonal of matrices: $$ X=\text{diag}(A,\dots,A) = \begin{bmatrix} A & \dots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \dots & A \end{bmatrix}, $$ where $A\in\mathbb{R}^{L\times L}$ and entries of $A$ are sampled independently from $N(0,1/n)$. The scaling of $L$ is $L\in\Theta(p)$, $n/p=\Theta(1)$, $L<n,p$. Does $X$ still satisfy the RIP condition with high probability?
My attempt: Let us denote $\beta_l\in\mathbb{R}^{L}$ as the $l$th component of $\beta$ if we were to partition $\beta$ into $p/L=\Theta(1)$ equal sized sub-vectors of length $L$. We have $$ \frac{\|X\beta\|_2^2}{\|\beta\|_2^2} =\left(\sum_{l=1}^{p/L}\|\beta_l\|_2^2\frac{\|A\beta_l\|_2^2}{\|\beta_l\|_2^2}\right)/\sum_{l=1}^{p/L}\|\beta_l\|_2^2 \approx 1, $$
since every component $\|A\beta_l\|_2^2/\|\beta_l\|_2^2 \approx 1$ because i.i.d. Gaussian matrices satisfy the RIP condition. This suggests that the diagonal Gaussian matrix might satisfy RIP. Is this right? If it is wrong, can you show me how to prove this?