Linear Algebra

Consider for a moment the line \(\ell\) through the points \(P\) and \(Q\text{.}\) When \(P,Q\in\R^{2}\text{,}\) we can describe \(\ell\) with an equation of the form \(y=mx+b\) (provided it isn’t a vertical line), but if \(P,Q\in\R^{3}\text{,}\) it’s much harder to describe \(\ell\) with an equation. We can solve this problem by using vectors.

Let \(\vec d=\overrightarrow{PQ}\) and consider the set of points (or vectors) \(\vec x\) that can be expressed as

for \(t\in \R\text{.}\) Geometrically, this is the set of all points we get by starting at \(P\) and displacing by some multiple of \(\vec d\text{.}\) This is a line!

We simultaneously interpret this line as a set of points (the points that make up the line) and as a set of vectors rooted at the origin (the vectors pointing from the origin to the line). Note that sometimes we draw vectors as directed line segments. Other times, we draw each vector by marking only its ending point because drawing each vector as a line segment would make it hard to see what is going on.

Which picture below do you think best represents \(\ell\text{?}\)

The line \(\ell\) described above can be written in set-builder notation as:

Notice that in set-builder notation, we write “for some \(t\in \R\text{.}\)” Make sure you understand why replacing “for some \(t\in\R\)” with “for all \(t\in \R\)” would be incorrect.

Writing lines with set-builder notation all the time can be overkill, so we will allow ourselves to describe lines in a shorthand called vector form

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\(y=mx+b\) form of a line is also shorthand. The line \(\ell\) described by the equation \(y=mx+b\) is actually the set \(\Set{(x,y)\in\R^2\given y=mx+b}\text{.}\)

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We can also use coordinates when writing a line in vector form. For example,

corresponds to the line passing through \(\mat{p_1\\p_2}\) with \(\mat{d_1\\d_2}\) as a direction vector.

The “\(t\)” that appears in a vector form is called the parameter variable, and for this reason, some textbooks use the term parametric form in place of “vector form”.

Writing a line in vector form requires only a point on the line and a direction for the line

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Notice that a direction vector for a line \(\ell\) is different than a vector in a line \(\ell\text{.}\)

, which makes converting from another form into vector form straightforward.

It’s important to note that when we write a line in vector form, it is a specific shorthand notation. If we augment the notation, we no longer have written a line in “vector form”.

But, why is vector form useful? For starters, every line can be expressed in vector form (you cannot write a vertical line in \(y=mx+b\) form, and in \(\R^{3}\text{,}\) you would need two linear equations to represent a line). But, the most useful thing about expressing a line in vector form is that you can easily generate points on that line.

Suppose \(\ell\) can be represented in vector form as \(\vec x=t\vec d+\vec p\text{.}\) Then, for every \(t\in \R\text{,}\) the vector \(t\vec d +\vec p\in\ell\text{.}\) Not only that, but as \(t\) ranges over \(\R\text{,}\) all points on \(\ell\) are “traced out”. Thus, we can find points on \(\ell\) without having to “solve” any equations.

The downside to using vector form is that it is not unique. There are multiple direction vectors and multiple points for every line. Thus, merely by looking at the vector equation for two lines, it can be hard to tell if they’re equal.

For example,

all represent the same line. In the second equation, the direction vector is parallel but scaled, and in the third equation, a different point on the line was chosen.

Recall that in vector form, the variable \(t\) is called the parameter variable. It is an instance of a dummy variable. In other words, \(t\) is a placeholder—just because “\(t\)” appears in two different vector forms, doesn’t mean it’s the same quantity.

To drive this point home, let’s think about vector form in terms of the sets it specifies. Let \(\vec d_{1},\vec d_{2}\neq\vec 0\) and \(\vec p_{1},\vec p_{2}\) be vectors and define the lines

and

These lines have vector forms \(\vec x=t\vec d_{1}+\vec p_{1}\) and \(\vec x=t\vec d_{2}+\vec p_{2}\text{.}\) However, declaring that \(\ell_{1}=\ell_{2}\) if and only if \(t\vec d_{1}+\vec p_{1}=t\vec d_{2}+\vec p_{2}\) does not make sense. Instead, as per the definition, \(\ell_{1}=\ell_{2}\) if \(\ell_{1}\subseteq\ell_{2}\) and \(\ell_{2}\subseteq\ell_{1}\text{.}\) If \(\vec x\in\ell_{1}\) then \(\vec x=t\vec d_{1}+\vec p_{1}\) for some \(t\in\R\text{.}\) If \(\vec x\in\ell_{2}\) then \(\vec x=t\vec d_{2}+\vec p_{2}\) for some possibly different \(t\in \R\text{.}\) This can get confusing really quickly. The easiest way to avoid confusion is to use different parameter variables when comparing different vector forms.

The geometry of lines in space (\(\R^{3}\) and above) is a bit more complicated than that of lines in the plane. Lines in the plane either intersect or are parallel. In space, we have to be careful about what we mean by “parallel lines,” since lines with entirely different directions can still fail to intersect

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Recall that in Euclidean geometry two lines are defined to be parallel if they coincide or never intersect.

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Any two distinct points define a line. To define a plane, we need three points. But there’s a caveat: the three points cannot be on the same line, otherwise they’d define a line and not a plane. Let \(A,B,C\in\R^{3}\) be three points that are not collinear and let \(\mathcal{P}\) be the plane that passes through \(A\text{,}\) \(B\text{,}\) and \(C\text{.}\)

Just like lines, planes have direction vectors. For \(\mathcal{P}\text{,}\) both \(\vec d_{1}=\overrightarrow{AB}\) and \(\vec d_{2}=\overrightarrow{AC}\) are direction vectors. Of course, \(\vec d_{1}\text{,}\) \(\vec d_{2}\) and their multiples are not the only direction vectors for \(\mathcal{P}\text{.}\) There are infinitely many more, including \(\vec d_{1}+\vec d_{2}\text{,}\) and \(\vec d_{1}-7\vec d_{2}\text{,}\) and so on. However, since a plane is a two-dimensional object, we only need two different direction vectors to describe it.

Like lines, planes have a vector form. Using the direction vectors \(\vec d_{1}=\overrightarrow{AB}\) and \(\vec d_{2}=\overrightarrow{AC}\text{,}\) the plane \(\mathcal{P}\) can be written in vector form as

Using vectors, we can describe more than just lines and planes—we can describe all sorts of geometric objects.

Recall that when we write \(\vec x=t\vec d+\vec p\) to describe the line \(\ell\text{,}\) what we mean is

The line \(\ell\) stretches off infinitely in both directions. But, what if we wanted to describe just a part of \(\ell\text{?}\) We can do this by placing additional restrictions on \(t\text{.}\) For example, consider the ray \(R\) and the line segment \(S\text{:}\)

We can also make polygons by adding restrictions to the vector form of a plane. Let \(\vec a=\mat{2\\1}\) and \(\vec b=\mat{-1\\1}\) and consider the unit square \(U\) and the parallelogram \(P\) defined by

Each set so far is a set of linear combinations, and we have made different shapes by restricting the coefficients of those linear combinations. There are two ways of restricting linear combinations that arise often enough to get their own names.

You can think of non-negative linear combinations as vectors you can arrive at by only displacing “forward”. Convex linear combinations can be thought of as weighted averages of vectors (the average of \(\vec v_{1},\ldots, \vec v_{n}\) would be the convex linear combination with coefficients \(\alpha_{i}=\frac{1}{n}\)). A convex linear combination of two vectors gives a point on the line segment connecting them.