Let $A$ be a $2\times2$ real square matrix of rank $1$. If $A$ is not diagonalizable, then which of the following is true Let $A$ be a $2\times2$ real square matrix of rank $1$. If $A$ is not diagonalizable, then which of the following is true.
(a)  $A$ is nilpotent
(b)  $A$ is not nilpotent
(c) the characteristic polynomial of $A$ is linear.
(d)  $A$ has a non-zero eigenvalue.   
I can say that d is false.
 A: Think about the Jordan normal form. Since it is not diagonalizable, it must be a 2-block. Since it has rank 1, the eigenvalues must be 0. Hence, $A^2=0$, so the matrix is nilpotent, it has minimal polynomial = characteristic polynomial which is $A^2=0$.
Hence only (a) is true.
A: 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,
$$A^2 = \begin{bmatrix}1 \\ -\dfrac{v_1}{v_2} \end{bmatrix}\begin{bmatrix}v_1 & v_2 \end{bmatrix} \times  \begin{bmatrix}1 \\ -\dfrac{v_1}{v_2} \end{bmatrix}\begin{bmatrix}v_1 & v_2 \end{bmatrix}= \begin{bmatrix} 0 & 0\\ 0 & 0\end{bmatrix}$$
A: The characteristic polynomial of $A$ is $\chi_A(x)=x^2+bx+c$ for some $b,c$. 
Since $A$ has rank $1\Rightarrow \det A=0 \Rightarrow c=0$. Thus $\chi_A(x)=x(x+b)$.
Since $A$ is non-diagonalizable it has only one eigenvalue $\Rightarrow b=0 \Rightarrow\chi_A(x)=x^2 \ldots$  
A: Hint: Consider $$\left(\begin{array}{cc}0&1\\ 0&0\end{array}\right).$$
