---
title: "0v=0"
source: https://www.jemoka.com/posts/kbhzero_times_vector/
---

\begin{align} 0v &amp;= (0+0)v \\ &amp;= 0v+0v \end{align}
Given scalar multiplication is closed, \(0v \in V\), which means \(\exists -0v:0v+(-0v)=0\). Applying that to both sides:
\begin{equation} 0 = 0v\ \blacksquare \end{equation}
The opposite proof of \(\lambda 0=0\) but vectors work the same exact way.
