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@jemoka / Jemoka Knowledge Base / raw/concept/kbhzero_times_vector.md
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--- title: "0v=0" source: https://www.jemoka.com/posts/kbhzero_times_vector/ --- \begin{align} 0v &= (0+0)v \\ &= 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.