Since $A$ is of rank $1$, we have $A = \begin{bmatrix}1 \\ u_2 \end{bmatrix}\begin{bmatrix}v_1 & v_2 \end{bmatrix} = \begin{bmatrix} v_1 & v_2\\ u_2 v_1 & u_2 v_2\end{bmatrix}$ The eigen values are $v_1 + u_2v_2$ and $0$. Given that the matrix is non-diagonalizable, a necessary condition is that the two eigenvalues must be equal. Hence, we have the other eigenvalue also to be zero i.e. $v_1 + u_2v_2 = 0 \implies u_2 = -\dfrac{v_1}{v_2}$. Hence, $A = \begin{bmatrix} v_1 & v_2\\ -\dfrac{v_1^2}{v_2} & -v_1\end{bmatrix}$ Hence, $A^2 = \begin{bmatrix}1 \\ -\dfrac{v_1}{v_2} \end{bmatrix}\begin{bmatrix}v_1 & v_2 \end{bmatrix} \times \begin{bmatrix}1 \\ -\dfrac{v_1}{v_2} \end{bmatrix}\begin{bmatrix}v_1 & v_2 \end{bmatrix}= \begin{bmatrix} 0 & 0\\ 0 & 0\end{bmatrix}$