Zeros in vector space

Dependencies:

  1. Vector Space

Let $V$ be a vector space over $F$. Let $a \in F$ and $\mathbf{v} \in V$.

  1. $0\mathbf{v} = \mathbf{0}$.
  2. $a\mathbf{0} = \mathbf{0}$.
  3. $a\mathbf{v} = \mathbf{0} \iff a = 0 \vee \mathbf{v} = \mathbf{0}$.

Proof

\[ 0\mathbf{v} = (0 + 0)\mathbf{v} = 0\mathbf{v} + 0\mathbf{v} \Rightarrow 0\mathbf{v} = \mathbf{0} \] \[ a\mathbf{0} = a(\mathbf{0} + \mathbf{0}) = a\mathbf{0} + a\mathbf{0} \Rightarrow a\mathbf{0} = \mathbf{0} \]

By the above 2 statements, $a = 0 \vee \mathbf{v} = \mathbf{0} \Rightarrow a\mathbf{v} = \mathbf{0}$.

Let $a \neq 0$. \begin{align} & a\mathbf{v} = \mathbf{0} \\ &\Rightarrow a^{-1}(a\mathbf{v}) = a^{-1}\mathbf{0} \tag{$a^{-1}$ exists because $a \neq 0$} \\ &\Rightarrow (a^{-1}a)\mathbf{v} = \mathbf{0} \tag{using scalar associativity and result 2 from above} \\ &\Rightarrow \mathbf{v} = \mathbf{0} \end{align}

Therefore, $a\mathbf{v} = \mathbf{0} \Rightarrow a = 0 \vee \mathbf{v} = \mathbf{0}$.

Dependency for:

  1. Eigenvectors of distinct eigenvalues are linearly independent
  2. Negation in vector space

Info:

Transitive dependencies:

  1. Group
  2. Ring
  3. Field
  4. Vector Space