A is 5×5 matrix, all of whose entries are 1 A is 5×5 matrix, all of whose entries are 1, then
(a)  A is not diagonalizable  (b)  A is idempotent  (c)   A is nilpotent
(d)   The minimal polynomial and the characteristics polynomial of A are not equal.
by the given condition matrix must be singular and hence it is not diagonalizable.
am i correct.
 A: $A$ is symmetric (or selfadjoint, if your matrices are complex), so it is diagonalizable. 
It is not idempotent, because $A^2=5A$. 
It is not nilpotent, because $A^n=5^{n-1}A$.
The minimal polynomial of $A$ is $p_m(t)=t(t-5)$, while the characteristic polynomial is $p_c(t)=t^{4}(t-5)$. So they are different. 
The characteristic polynomial can be obtained this way: we have
$$
A=\begin{bmatrix}1&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 1&0&\cdots&0
\end{bmatrix}\,\begin{bmatrix}1&\cdots&1\\0&\cdots&0\\ &\ddots\\0&\cdots&0
\end{bmatrix},
$$
so its spectrum agrees with that of 
$$
\begin{bmatrix}1&\cdots&1\\0&\cdots&0\\ &\ddots\\0&\cdots&0
\end{bmatrix}\,\begin{bmatrix}1&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 1&0&\cdots&0
\end{bmatrix}=\begin{bmatrix}5&0&\cdots&0\\0&0&\cdots&0\\ & & \ddots \\0& & \cdots &0
\end{bmatrix}.
$$
The eigenvalues are then $5,0,0,0,0$, and so $p_c(t)=t^4(t-5)$. As $A^5-5A=0$, the minimal polynomial is at most $t(t-5)$. But it also requires $0$ and $5$ as roots, so $p_m(t)=t(t-5)$.
