A calculus of the absurd

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

  • 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 \(x, y \in \textsf {V}\)

    \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!

  • Example 20.8.1 Consider the vector space \(\textsf {V} = \mathbb {R}^2\). The two vectors

    \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

\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.