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If $\Lambda$ is spltsplit basic, then by Gabriel's theorem it is isomorphic to $\Bbbk Q$ where $Q$ is a finite acyclic quiver. Up to isomorphism you can assume $e$ is the sum of empty paths running over some subset $X$ of vertices. Then $e\Lambda e$ is isomorphic to the path algebra on the full (i.e. induced) subquiver on the vertex set $X$. Thus it is hereditary.

If $\Lambda$ is splt basic, then by Gabriel's theorem it is isomorphic to $\Bbbk Q$ where $Q$ is a finite acyclic quiver. Up to isomorphism you can assume $e$ is the sum of empty paths running over some subset $X$ of vertices. Then $e\Lambda e$ is isomorphic to the path algebra on the full (i.e. induced) subquiver on the vertex set $X$. Thus it is hereditary.

If $\Lambda$ is split basic, then by Gabriel's theorem it is isomorphic to $\Bbbk Q$ where $Q$ is a finite acyclic quiver. Up to isomorphism you can assume $e$ is the sum of empty paths running over some subset $X$ of vertices. Then $e\Lambda e$ is isomorphic to the path algebra on the full (i.e. induced) subquiver on the vertex set $X$. Thus it is hereditary.

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Benjamin Steinberg
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If $\Lambda$ is splt basic, then by Gabriel's theorem it is isomorphic to $\Bbbk Q$ where $Q$ is a finite acyclic quiver. Up to isomorphism you can assume $e$ is the sum of empty paths running over some subset $X$ of vertices. Then $e\Lambda e$ is isomorphic to the path algebra on the full (i.e. induced) subquiver on the vertex set $X$. Thus it is hereditary.