Linear Algebra

Recall that for vectors \(\vec a\) and \(\vec b\text{,}\) the relationship \(\vec a\cdot \vec b=0\) can hold for two reasons: (i) either \(\vec a=\vec 0\text{,}\) \(\vec b=\vec 0\text{,}\) or both or (ii) \(\vec a\) and \(\vec b\) meet at \(90^{\circ}\text{.}\) Thus, the dot product can be used to tell if two vectors are perpendicular. There is some strangeness with the zero vector here, but it turns out this strangeness simplifies our lives mathematically.

The definition of orthogonal encapsulates both the idea of two vectors forming a right angle and the idea of one of them being \(\vec 0\text{.}\)

Before we continue, let’s pin down exactly what we mean by the direction of a vector. There are many ways we could define this term, but we’ll go with the following.

The vector \(2\xhat\) points in the direction of \(\xhat\) since \(\frac{1}{2}(2\xhat)=\xhat\text{.}\) Since \(\frac{1}{2}>0\text{,}\) \(2\xhat\) also points in the positive direction of \(\xhat\text{.}\) In contrast, \(-\xhat\) points in the direction \(\xhat\) but not the positive direction of \(\xhat\text{.}\)

When it comes to the relationship between two vectors, there are two extremes: they point in the same direction, or they are orthogonal. The dot product can be used to tell you which of these cases you’re in, and more than that, it can tell you to what extent one vector points in the direction of another (even if they don’t point in the same direction).

Let \(\vec n=\mat{1\\2}\text{.}\) If a vector \(\vec v=\mat{v_1\\v_2}\) is orthogonal to \(\vec n\text{,}\) then

and so \(v_{1}=-2v_{2}\text{.}\) In other words, \(\vec v\) is orthogonal to \(\vec n\) exactly when \(\vec v\in\Span\Set{\mat{-2\\1}}\text{.}\) What have we learned? The set of all vectors orthogonal to \(\vec n\) forms a line \(\ell=\Span\Set{\mat{-2\\1}}\text{.}\) In this case, we call \(\vec n\) a normal vector for \(\ell\text{.}\)

In \(\R^{2}\text{,}\) normal vectors provide yet another way to describe lines, including lines which don’t pass through the origin.

Let \(\vec n=\mat{1\\2}\) as before, and fix \(\vec p=\mat{1\\1}\text{.}\) If we draw the set of all vectors orthogonal to \(\vec n\) but root all the vectors at \(\vec p\), again we get a line, but this time the line passes through \(\vec p\text{.}\)

In fact, the line we get is \(\ell_{2}=\Span\Set{\mat{-2\\1}}+\Set{\mat{1\\1}}=\ell+\Set{\vec p}\text{,}\) which is just \(\ell\) (the parallel line through the origin) translated by \(\vec p\text{.}\)

Let’s relate this to dot products and normal vectors. By definition, for every \(\vec v\in \ell\text{,}\) we have \(\vec n\cdot \vec v=0\text{.}\) Since \(\ell_{2}\) is a translate of \(\ell\) by \(\vec p\text{,}\) we deduce the relationship that for every \(\vec v\in \ell_{2}\text{,}\)

When a line is expressed as above, we say it is expressed in normal form.

Though the definition doesn’t explicitly state it, if a line \(\ell\in \R^{2}\) is expressed in normal form as \(\vec n\cdot (\vec x-\vec p)=0\text{,}\) then \(\vec n\) is necessarily a normal vector for \(\ell\). (Think about what the solution set to \(\vec n\cdot (\vec x-\vec p)=0\) would be if \(\vec n\) happened to be \(\vec 0\text{!}\))

What about in \(\R^{3}\text{?}\) Fix a non-zero vector \(\vec n\in\R^{3}\) and let \(\mathcal{Q}\subseteq\R^{3}\) be the set of vectors orthogonal to \(\vec n\text{.}\) \(\mathcal{Q}\) is a plane through the origin, and again, we call \(\vec n\) a normal vector of the plane \(\mathcal{Q}\text{.}\)

In a similar way to the line, \(\mathcal{Q}\) is the set of solutions to \(\vec n\cdot \vec x=0\text{.}\) And, for any \(\vec p\in\R^{3}\text{,}\) the translated plane \(\mathcal{Q}+\Set{\vec p}\) is the solution set to

Thus, we see planes in \(\R^{3}\) have a normal form just like lines in \(\R^{2}\) do.

In \(\R^{2}\text{,}\) only lines have a normal form, and in \(\R^{3}\) only planes have a normal form. In general, we call objects in \(\R^{n}\) which have a normal form hyperplanes.

Hyperplanes always have dimension one less than the space they’re contained in. So, hyperplanes in \(\R^{2}\) are (one-dimensional) lines, hyperplanes in \(\R^{3}\) are regular (two-dimensional) planes, and hyperplanes in \(\R^{4}\) are (three-dimensional) volumes.

Suppose \(\vec n,\vec p\in \R^{3}\) and \(\vec n\neq \vec 0\text{.}\) Then, solutions to

define a plane \(\mathcal{P}\text{.}\) But, \(\vec n\cdot (\vec x-\vec p)=0\) if and only if

Since \(\vec n\) and \(\vec p\) are fixed, \(\alpha\) is a constant. Expanding using coordinates, we see

and so \(\mathcal{P}\) is the set of solutions to

Equation (4.3.1) is sometimes called scalar form of a plane. For us, it will not be important to distinguish between scalar and normal form, but what is important is that we can use the row reduction algorithm to write the complete solution to (4.3.1), and this complete solution will necessarily be written in vector form.