Is the system's pole movable?
I would have liked to put this answer as a comment, but I don't have enough reputation to do so.



First a small remark. There is a problem with the dimensions of the matrices in the question: If you plug the expression for \$u\$ in the state equation, then you multiply a \$2\times1\$ matrix with a \$2\times1\$ matrix:
$$
\begin{aligned}
\dot{x}
&amp; = A x + B u \\
&amp; = A x + \begin{bmatrix} 0 \\ 1 \end{bmatrix} (\begin{bmatrix} k_1 &amp; k_2 \end{bmatrix} x + \begin{bmatrix} 0 \\ 1 \end{bmatrix} u')
\end{aligned}
$$
and the problem occurs in the product \$\begin{bmatrix} 0 \\ 1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix}\$.

Typically, a state feedback is formulated as \$u = - K x + u'\$ where \$u'\$ is a reference signal. Here, you get \$u = - \begin{bmatrix} k_1 &amp; k_2 \end{bmatrix} x + u'\$, but you can of course take the minus sign into account in the coefficients \$k_1\$ and \$k_2\$. Considering \$u = \begin{bmatrix} k_1 &amp; k_2 \end{bmatrix} x + u'\$, the matrix \$A'\$ is correct.



Your analysis is correct.
The open-loop system has two eigenvalues in the origin. With the state-feedback controller, you can only control one pole of this open-loop system, since the system is not controllable. You can easily see this in the \$A\$ and \$B\$ matrix: neither of the states, nor the input affect the evolution of the first state (the top rows of the matrices \$A\$ and \$B\$ are zero). This means that \$\dot{x}_1=0\$ no matter what and so \$x_1\$ stays at its initial value.

If the top element of the matrix \$B\$ was non-zero, then the first state equation would be \$\dot{x}_1=b_1 u\$, which corresponds to an integrator (as mentioned in Chus comment). Nevertheless, since the top element of \$B\$ is zero here, the integrator will start at the initial value of \$x_1\$ and keep integrating zeros.

You can also see that one pole is not moveable from the rank of the controllability matrix. Here, \$rank(\begin{bmatrix}B &amp; AB\end{bmatrix}) = rank(\begin{bmatrix}0 &amp; 0 \\ 1 &amp; 0\end{bmatrix}) = 1\$. This means that you can only move one pole. It also means that the states \$x=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\$ that you can reach from a zero initial state live in a one-dimensional space. In fact, those states that you can reach are linear combinations of the columns of the controllability matrix, here \$\alpha \begin{bmatrix} 0 \\ 1 \end{bmatrix}\$, where \$\alpha\$ can be any number you like. And again, you can see that you cannot control the first state.

In conclusion, one pole in the origin is controllable (with \$k_2\$) and the other pole in the origin is not controllable.