A calculus of the absurd
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20.8.4 Orthogonal stuff
I know, it’s an imaginative title for this section. Hopefully you know what perpendicular means (if not, you might want to review a fair amount of mathematics before returning to this section). We would (well, maybe you wouldn’t, but what can one do) like to generalise this notion to
a more abstract setting; we can say that
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Definition 20.8.1 Let \(\textsf {V}\) be an inner product space over a field \(\textsf {F}\), then two vectors are orthogonal if and only if their inner product is zero; that is if we let \(x, y \in \textsf
{V}\)
\(\seteqnumber{0}{20.}{159}\)
\begin{equation}
\langle x, y \rangle = 0 \iff \text { $x$ and $y$ are orthonormal}
\end{equation}
We can denote this as \(x \bot y\) (read “\(x\) and \(y\) are orthogonal”).
This is not the most exciting definition, but it is a very useful one!
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Example 20.8.1 Consider the vector space \(\textsf {V} = \mathbb {R}^2\). The two vectors
\(\seteqnumber{0}{20.}{160}\)
\begin{equation}
\begin{pmatrix} 1 \\ 0 \end {pmatrix}, \begin{pmatrix} 0 \\ 1 \end {pmatrix}
\end{equation}
are orthogonal with respect to the dot product.
We can just apply the definition, recall that
\(\seteqnumber{0}{20.}{161}\)
\begin{align}
\begin{pmatrix} 1 \\ 0 \end {pmatrix} \cdot \begin{pmatrix} 0 \\ 1 \end {pmatrix} &= 1 \times 0 + 0 \times 1 \\ &= 0
\end{align}
And therefore, the two vectors are orthogonal.