Zero Vector is Orthogonal to All Vectors
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Theorem
Let $\mathbf V$ be a vector space of $n$ dimensions.
Let $\bszero \in \mathbf V$ be the zero vector.
Let $\mathbf a \in \mathbf V$ be an arbitrary vector in $\mathbf V$
Then $\bszero$ is orthogonal to $\mathbf a$.
Proof
By definition, $\mathbf a$ and $\mathbf b$ are orthogonal if and only if their dot product is zero:
- $\mathbf a \cdot \mathbf b = 0$
We have:
\(\ds \mathbf a\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n a_k \mathbf e_k\) | ||||||||||||
\(\ds \mathbf b\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n b_k \mathbf e_k\) |
where $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is the standard ordered basis of $\mathbf V$.
By definition of dot product:
- $\ds \mathbf a \cdot \mathbf b = \sum_{k \mathop = 1}^n a_k b_k$
By definition of zero vector:
\(\ds \bszero\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n 0 \mathbf e_k\) |
Hence:
\(\ds \bszero \cdot \mathbf a\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n 0 \times a_k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
Hence the result.
$\blacksquare$
Sources
- For a video presentation of the contents of this page, visit the Khan Academy.