Question: Given two matrices $A$ and $C$ of order $n\times n$ and $m\times n$ respectively. We define the following matrix by block:
$$
D=\left( \begin{matrix}
C\\
CA\\
:\\
:\\
CA^{q-1}
\end{matrix}\right)
$$
where $q\in\mathbb N^\*$.

My question is: is there any relation between $q$ and the rank of the matrix $D$? if so for what value of $q$ we have $rank(D)=n$ ? Thank you for your time.

Answer: 
\begin{align\*}
\sum\_{k=1}^M k^2(D\_k-D\_{k+1})
&= \sum\_{k=1}^M k^2 D\_k - \sum\_{k=1}^M k^2 D\_{k+1} \\
&= \sum\_{k=1}^M k^2 D\_k - \sum\_{k=2}^{M+1} (k-1)^2 D\_k \\
&= D\_1 + \sum\_{k=2}^M (k^2-(k-1)^2) D\_k - M^2 D\_{M+1} \\
&= D\_1 + \sum\_{k=2}^M (2k-1) D\_k - M^2 D\_{M+1} \\
&= \sum\_{k=1}^M (2k-1) D\_k - M^2 D\_{M+1} \\
&\le \sum\_{k=1}^M (2k-1) D\_k
\end{align\*}
\begin{align\*}
\frac{1}{M^2}\sum\_{k \leq M} \sigma^2\_k &\le \frac{1}{M^2} \sum\_{k=1}^M (2k-1) D\_k \\ 
&=\frac{1}{M^2} \left(D\_1 + 3D\_2 + 5D\_3 + 7D\_4 + \ldots \right)\\
&=\frac{1}{M^2} \left(\sum\_{n \geq 1}\frac{\sigma^2\_n}{n^2} + 3\sum\_{n \geq 2}\frac{\sigma^2\_n}{n^2} + 5\sum\_{n \geq 3}\frac{\sigma^2\_n}{n^2}+7\sum\_{n \geq 4}\frac{\sigma^2\_n}{n^2} + \ldots\right)\\ 
&=\frac{1}{M^2} \left\{\left(\frac{\sigma\_1^2}{1^2} + \frac{\sigma\_2^2}{2^2} + \ldots \right) + 3\left(\frac{\sigma\_2^2}{2^2} + \frac{\sigma\_3^2}{3^2} + \ldots \right) + 5\left(\frac{\sigma\_3^2}{3^2} + \frac{\sigma\_4^2}{4^2} + \ldots \right) + 7\left(\frac{\sigma\_4^2}{4^2} + \ldots \right) + \ldots\right\}\\
&=\frac{1}{M^2} \left\{\frac{\sigma\_1^2}{1^2} + \frac{4\sigma\_2^2}{2^2} + \frac{9\sigma\_3^2}{3^2} + \frac{16\sigma\_4^2}{4^2} + \ldots \right\}\\
&= \frac{1}{M^2} \left(\sigma\_1^2 + \sigma\_2^2 + \sigma\_3^2 + \sigma\_4^2 + \ldots\right)\\
&= \frac{1}{M^2}\sigma\_1^2 + \frac{1}{M^2}\sigma\_2^2 + \frac{1}{M^2}\sigma\_3^2 + \frac{1}{M^2}\sigma\_4^2 + \ldots 
\end{align\*}

Applying limit

\begin{align\*}
\lim\_{M \rightarrow \infty} \frac{1}{M^2}\sum\_{k=1}^M \sigma^2\_k &\le \lim\_{M \rightarrow \infty} \left(\frac{1}{M^2}\sigma\_1^2 + \frac{1}{M^2}\sigma\_2^2 + \frac{1}{M^2}\sigma\_3^2 + \frac{1}{M^2}\sigma\_4^2 + \ldots \right) \\
&= 0 + 0 + 0 + 0 + \ldots \\
&= 0.
\end{align\*}

Therefore

$$\lim\_{M \rightarrow \infty} \frac{1}{M^2}\sum\_{k \leq M} \sigma^2\_k = 0.$$