Consider the system $\displaystyle dX/dt=AX+Bu$, which has matrix form:

$\displaystyle d\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}/dt=\begin{pmatrix}1 & -1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}+\begin{pmatrix}1 \\ 0 \end{pmatrix}u$

for which we can show that $\displaystyle A=PJP^{-1}$, where

$\displaystyle A=\begin{pmatrix}1 & -1 \\ 1 & -1 \end{pmatrix}$, $\displaystyle P=\begin{pmatrix}1 & 0 \\ 1 & -1 \end{pmatrix}=P^{-1}$, and $\displaystyle J=\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}$.

Then $\displaystyle e^{At}=Pe^{Jt}P^{-1}=\begin{pmatrix}1 & 0 \\ 1 & -1 \end{pmatrix}\begin{pmatrix}1 & t \\ 0 & 1 \end{pmatrix}\begin{pmatrix}1 & 0 \\ 1 & -1 \end{pmatrix}=\begin{pmatrix}1+t & -t \\ t & 1-t \end{pmatrix}$

Hence $\displaystyle e^{At}B=\begin{pmatrix}1+t & -t \\ t & 1-t \end{pmatrix}\begin{pmatrix}1 \\ 0 \end{pmatrix}=\begin{pmatrix}1+t \\ t \end{pmatrix}$.

Suppose now that there is a vector $\displaystyle X_1=\begin{pmatrix}a \\ b \end{pmatrix}$

such that $\displaystyle X_1^{T}e^{At}B=\begin{pmatrix}a & b \end{pmatrix}\begin{pmatrix}1+t \\ t \end{pmatrix}=a(1+t)+bt=0$.

From the constant part we see that $\displaystyle a = 0$, while from the coefficient of $\displaystyle t$ it follows that $\displaystyle a+b=b=0$.

Since $\displaystyle X_1$ cannot be non-zero, we have shown that the above system is controllable.