The last step in an exercise is to prove the following condition. Given that $(A,B)$ is a controllable pair.
meaning: $\begin{cases}A\in\mathbb{R}^{n\times n} \\ B\in \mathbb{R}^{n\times m}\end{cases}$, rank$\left(\begin{bmatrix}B & AB & A^2B & \dotsc & A^{n-1}B\end{bmatrix}\right) = n$
What i now have to prove is that $(A,B Q_v^{\frac{1}{2}})$ is also a controllable pair.
meaning $\begin{cases}A\in\mathbb{R}^{n\times n} \\ B\in \mathbb{R}^{n\times m} \\ Q_v^{\frac{1}{2}}\in \mathbb{R}^{m\times m} \end{cases}$, rank$\left(\begin{bmatrix}BQ_v^{\frac{1}{2}} & ABQ_v^{\frac{1}{2}} & A^2BQ_v^{\frac{1}{2}} & \dotsc & A^{n-1}BQ_v^{\frac{1}{2}}\end{bmatrix}\right) = n$
I think it's fair to say that $$\begin{bmatrix}BQ_v^{\frac{1}{2}} & ABQ_v^{\frac{1}{2}} & A^2BQ_v^{\frac{1}{2}} & \dotsc & A^{n-1}BQ_v^{\frac{1}{2}}\end{bmatrix} = \begin{bmatrix}B & AB & A^2B & \dotsc & A^{n-1}B\end{bmatrix} \begin{bmatrix}Q_v^{\frac{1}{2}} \\ Q_v^{\frac{1}{2}} \\ \vdots \\Q_v^{\frac{1}{2}} \end{bmatrix}= \tilde{C} \tilde{Q}_v^{\frac{1}{2}}$$ Where $\tilde{C}\in\mathbb{R}^{n\times mn}, \tilde{Q}_v^{\frac{1}{2}} \in \mathbb{R}^{mn \times m}$I know that by Sylvester's inequality we have: $$ \text{rank}(\tilde{C}) + \text{rank}(\tilde{Q}_v^{\frac{1}{2}})-mn\leq \text{rank}(\tilde{C} \tilde{Q}_v^{\frac{1}{2}})$$
But I have no clue how to prove rank$(\tilde{C})$ = rank$(\tilde{C}\tilde{Q}_v^{\frac{1}{2}})$.
What's the most important about matrix $Q_v$ is that $Q_v$ is the variance matrix of a Gaussian white noise process. Which means that the diagonal of the matrix $Q_v$ is non-zero, but I don't know if I may assume that $Q_v \geq 0$ or $Q_v > 0$.