$R$ is a ring and $H$ is the intersection of all non-zero left ideals of $R.$ Then prove if $H$ not equal to $0$ then $H$ is a two sided ideal in $R$ $R$ is a ring and $H$ is the intersections of all non-zero left ideals of $R$.Then prove if $H$ not equal to zero then $H$ is a two sided ideal in $R$ and that we have $H^2=0$ or else $R$ is a division ring
Since $H$ is the set of all non zero left ideals then we can find a $x$ which belongs to $H$ and hence $H$ is itself a left ideal .
Since a left ideal not necessary equals right ideal.Then how to prove it please someone explain.
 A: If $H$ is nonzero, then $H$ is obviously a minimal left ideal. For any $r\in R$, $Hr$ is another left ideal, in fact, a homomorphic image of $H$. Because $H$ is simple, $Hr$ is either $\{0\}$ or simple. If $Hr$ is nonzero, then because we know $H\subseteq Hr$ and $Hr$ is simple, we have $H=Hr$. So in either case, $Hr\subseteq H$, and $H$ is a right ideal.$^\ast$ It's obviously a left ideal (being an intersection of left ideals.) So it is a two-sided ideal.
If $H=R$, then we obviously have a division ring. Otherwise, $R$ has nontrivial maximal left ideals. In that case, $H$ is contained in the intersection of all maximal left ideals, a.k.a. the Jacobson radical. 
But the Jacobson radical annihilates simple left modules including $H$. Therefore $\{0\}\subseteq H^2\subseteq J(R)H=\{0\}$ implies $H^2=\{0\}$.

$^\ast$ Alternatively: $H$ is then contained in the left socle $Soc(_RR)$, which is always a two-sided ideal. (Help if you need it.)  But if $Soc(_RR)$ was a direct sum of more than one minimal left ideal, $H$ could not be contained in the other minimal ideals. So $Soc(_RR)=H$ is a two-sided ideal.
