$J$ be a $3\times 3$ matrix with all entries $1$ Then $J$ is $J$ be a $3\times 3$ matrix with all entries $1$ Then $J$ is


*

*Diagonalizable

*Positive semidefinite

*$0,3$ are only eigenvalues of $J$

*Is positive definite
$J$ has minimal polynomial $x(x-3)=0$ so 1, 2,3 are true , am I right?

You are correct. This is a multiple duplicate but I can't find any. 
The matrix has rank $1$, which makes $0$ an eigenvalue of multiplicity $2$. Then the trace tells us what the other eigenvalue is, whence diagonalizability, since it adds the missing dimension. Since $0$ is an eigenvalue, it can't be positive definite. But it is indeed positive semidefinite by diagonalization in an orthonormal basis ($J$ is symmetric, eigenvalues are nonnegative). The minimal polynomial is obvious by diagonalization. 
Alternative: we can observe that $J^2=3J$ so $J$ is annihilated by $X^2-3X=X(X-3)$, which is therefore divided by the minimal polynomial. Since $J$ is nether $I_3$ nor $3I_3$, the latter must be the minimal polynomial. It splits and has simple roots. Whence 1 and 3. 
