Relationship between eigen values and state feedback
My question is that even if we have control over eigen values of this new vector how does that help us control the system i.e. to bring our system to our desired state?


If you consider your system as just the plant, then all you can achieve by state feedback control is stabilization (if the system is stabilizable) with pole-placement (if the system is controlable). However, by augmenting the system with a controller that follows the Internal Model Principle, you can use the same state-feedback design method and track a given reference signal.

Demonstration: plant system is G, with controller system F. 
$$
G:\begin{cases}
\dot{x}(t) = Ax(t)+Bu(t)\\
y(t) = Cx(t)
\end{cases}
$$

$$
F:\begin{cases}
\dot{x}_c(t) = A_cx_c(t)+B_ce(t)\\
\end{cases}
$$

\$y(t)\$ is the linear combination of states that must track reference signal \$r(t)\$. Both \$y(t)\$ and \$r(t)\$ may be vector signals (multiple-input-multiple-output case).

When closing the feedback loop, error signal is \$e(t) = r(t)-y(t)\$. The augmented system for state vector \$\bar{x}(t) = \begin{bmatrix}x(t)\\x_c(t)\end{bmatrix}\$ then follows:

$$
\begin{cases}
\dot{\bar{x}}(t) = \begin{bmatrix}
A &amp; 0\\
-B_cC &amp; A_c
\end{bmatrix}\bar{x}(t) + \begin{bmatrix}
B\\
0
\end{bmatrix}u(t)+\begin{bmatrix}
0\\
B_c
\end{bmatrix}r(t) = \bar{A}\bar{x}(t)+\bar{B}u(t) + \bar{B}_rr(t)\\\\
y(t) = \begin{bmatrix}C&amp;0\end{bmatrix}\bar{x}(t)=\bar{C}\bar{x}(t)
\end{cases}
$$

Now apply state-feedback \$u(t) = K\bar{x}(t)\$:

$$
\dot{\bar{x}}(t) = (\bar{A}+\bar{B}K)\bar{x}(t) + \bar{B}_rr(t)\\
$$

If \$A\$ or \$A_c\$ follow the Internal Model Principle (replicate the modeled dynamics of the reference signal), then the output \$y(t)\$ will track the reference \$r(t)\$. The transfer function from \$r(t)\$ to \$y(t)\$ is \$T(s)\$.

$$
T(s) = \bar{C}(sI-(\bar{A}+\bar{B}K))^{-1}\bar{B}_r
$$