how to control state space when it is not controllable?
Controlability means that a state-space feedback is able to assign closed-loop eigenvalues of a linear system to anywhere in the complex plane. However, the stabilizability of a system is a weaker notion, requiring only that the closed-loop eigenvalues all end up on the left half-plane.

Consider \$A\$ is of order \$n\$. For all \$\lambda \in \mathbb{C}\$ that are eigenvalues of \$A\$ with positive real parts (\$\Re (\lambda )\geq 0\$):

\$\text{rank}\begin{bmatrix}\lambda I-A&amp;B\end{bmatrix} = n\$ \$\iff\$ the system is stabilizable

Example:
$$
A = \begin{bmatrix} 1 &amp; 1 \\ 4 &amp; -2\end{bmatrix} \qquad B = \begin{bmatrix}1\\1\end{bmatrix}
$$

This system is not controllable. Its controllability matrix, with rank 1, is
$$
Co = \begin{bmatrix} B &amp; AB \end{bmatrix} = \begin{bmatrix}1 &amp; 2 \\ 1 &amp; 2\end{bmatrix}
$$
However, this system is stabilizable. There are 2 eigenvalues of \$A\$: \$\lambda=-3\$ and \$\lambda=2\$.
$$
\text{rank}\begin{bmatrix}2I-A&amp;B\end{bmatrix} = 
\text{rank}\begin{bmatrix}1&amp;-1&amp;1\\-4&amp; 4 &amp; 1\end{bmatrix} = 2
$$

One way of stabilizing this system, using state-feedback with pole placement ideas, is to keep the stable (yet uncontrollable) eigenvalue \$\lambda=-3\$ in its place for the closed-loop system. If one wishes for the closed-loop eigenvalues of 
\$\lambda=-3\$ and \$\lambda=-5\$, then it follows the characteristic equation:
$$(s+3)(s+5) = s^2+8s+15$$

Gain calculation:
$$K = \begin{bmatrix}k_1 &amp; k_2\end{bmatrix}$$
$$|sI-(A-BK)| = \text{det}\begin{bmatrix}s-1+k_1&amp;-1+k_2\\-4+k_1&amp;s+2+k_2\end{bmatrix}=$$
$$ = s^2+s(k_1+k_2+1)+3k_1+3k_2-6$$
$$\begin{cases}
k_1+k_2+1 = 8\\
3k_1+3k_2-6 = 15
\end{cases}
$$
This system has infinite possible solutions. Assuming \$k_1 = 0\$, then \$K=\begin{bmatrix}0&amp;7\end{bmatrix}\$ achieves the desired closed-loop eigenvalues.