Suppose that the system $$x'(t)=Ax(t)+Bu(t)$$ is controllable in $\mathbb{R}^n$, where $A$ is $n \times n$, $B$ is $ n \times m$ and $u(t)$ is $m \times 1$ Show that the system $$\left \{ \begin{array}{rclccc} x'(t) &=& Ax(t) &+& By(t)& \\ y'(t)&=&&&&u(t) \end{array} \right.$$ with the additional states $y\in\mathbb{R}^m$ and input $u\in\mathbb{R}^m$ is controllable in $\mathbb{R}^{n+m}$.
My attempt
If the first system is controllable, then the matrix $$M = \begin{pmatrix} B & AB & A^2B &\ldots & A^{n-1}B \end{pmatrix}$$ has rank M. Let the second system $$\begin{pmatrix}x' \\ y' \end{pmatrix} = \begin{pmatrix} A & B \\ 0 & 0\end{pmatrix}\begin{pmatrix} x \\ y\end{pmatrix}+\begin{pmatrix}0 \\ I \end{pmatrix}u(t)$$ Then, its Kalman matrix (aka controllability matrix) is $$\bar{M} = \begin{pmatrix} B & AB & A^2B & \ldots & A^{n+m-1}B \\ 0 & 0 & 0 & \ldots & 0\end{pmatrix}$$ The matrix $A$ is $n \times n$ and, by Cayley-Hamilton theorem, $$A^{n+l} = \sum_{k=0}^{n-1} \alpha_{k,l} A^{k}, l=0,1,...$$ I mean, $A^{n+l}$ is a linear combination of the first $n$ powers of $A$. This implies that we have $n$ linearly independent columns in $\bar{M}$ at the the terms until $A^{n-1}B$ (because the first system is controllable). But I don't see how to find the other $m$ columns. What am I doing wrong? Probably my controlability matrix is wrong, since it only can have $n$ l.i rows.
We could solve it for $y$ first and return to a system only in $x'$, substituting $u$ by some $\int u$. The Kalman matrix of first and second system would be the same and, by hypothesis, it would have $\text{rank }n$, correct? But why isn't my first approach correct?