Let $\frac{dx}{dt}=Ax(t)+Bu(t)$ be a control system and suppose that A has two or more linearly independent eigenvectors associated to the same eigenvalue. Show that the system is not controlable.
My attempt: I thought about using the Kalman rank condition but couldn't get to any solid result
No, The system is controllable iff the minimal polynomial of $A$ has degree $n$ and $B$ has a component in the direction of every eigenvector.
A counter-example is $$ A=\begin{bmatrix}1&1\\0&1\end{bmatrix} ,B=\begin{bmatrix}1\\1\end{bmatrix} $$ you can see that since $A$ is upper triangular matrix its eigenvalues are the diagonal elements which are $1$. Check the controllability matrix $R$ $$ R=[B\ \ |\ \ AB \ \ ]=\begin{bmatrix}1&2\\1&1\end{bmatrix} $$ Which has a rank of 2 and the system is controllable.
Edit: The controllablity condition is that the $Rank(R)=n=Rank(R^T)$ which means that the coulumn vectors of $R$ are linearly independent i.e. $$ (\sum_{i=0}^{n-1} \alpha_i A^i) B=0 \implies \alpha_i=0 \ \ \forall i $$ so if $A$ has a minimal polynomial of degree $n-1$ or less then you could find $\sum_{i=0}^{n-1} \alpha_i A^i=0$ with $\alpha_i \neq 0$ and so your system is uncontrollable. If $B$ doesn't have an component in one of the eigenvectors without loss of generality let this eigenvector be $v_{n}$ then choose $(\sum_{i=0}^{n-1} \alpha_i A^i) = \left(\Pi_{i=1}^{n-1}(A-\lambda_i I)\right)$ so $$ (\sum_{i=0}^{n-1} \alpha_i A^i) B=\left(\Pi_{i=1}^{n-1}(A-\lambda_i I)\right).\sum_{k=1}^{n-1}a_k v_k=\sum_{k=1}^{n-1} a_k \left(\Pi_{i=1,i\neq k}^{n-1}(A-\lambda_i I)\right).(A-\lambda_k I) v_k=0 $$ since $(A-\lambda_k I) v_k=0 \ \ \forall k$.