Let $M$ be a square matrix of order $2$ with rank $1.$ Then Let $M$ be a square matrix of order $2$ with rank $1.$ Then
(i) diagonalizable and nonsingular.
(ii) digonalizable and nilpotent
(iii) neither diagonalizable nor nilpotent.
(iv) either diagonalizable or nilpotent but not both.
$1$ false , let char poly be $x^2+bx+c=0$ since $x=0$ is a solution so $c=0$, so charpoly is $x(x+b)=0$ if $b=0$ then our matrix is Nilpotent and hence non diagonalizable as then charpoly and minpoly is $x^2=0$ am I okay with my logic? but what would happen if $b\ne 0$? thank you for help.
 A: Think of the Jordan form of such a matrix. If, in your notation, $b = 0$, so that the characteristic polynomial is $x^2$, you still have two possibilities for the minimal polynomial, that is either $x$ or $x^2$, which correspond to Jordan forms
$$
\begin{bmatrix}
0&0\\
0&0
\end{bmatrix},
\qquad
\begin{bmatrix}
0&1\\
0&0
\end{bmatrix}.
$$
But the zero matrix has rank $0$, so it has to be excluded.
If $b \ne 0$, then the characteristic polynomial equals the minimal polynomial, and the Jordan form is
$$
\begin{bmatrix}
0&0\\
0&-b
\end{bmatrix}.
$$
By looking at these possibilities, you should see the correct answer now.
A: The decomposition of $M$ in Jordan canonical form will be of the form 
$$
\begin{bmatrix}
0 & 1 \\
0 & 0 \\
\end{bmatrix}
$$
or 
$$
\begin{bmatrix}
0 & 0 \\
0 & 0 \\
\end{bmatrix}
$$
if $b = 0$, or 
$$
\begin{bmatrix}
0 & 0 \\
0 & -b \\
\end{bmatrix}
$$
where $-b$ is some non-zero eigenvalue when the characteristic polynomial is of the form $x(x+b)$. Therefore $M$ is either diagonalizable or nilpotent but not both (when it is both, it is the zero matrix which does not have rank $1$). 
Hope that helps,
