Linear Algebra Notes

Recall that in algebra, a line is given by a Cartesian equation, e.g. of the form $y=mx+b\text{,}$ where $x,y$ are variables, and $m,b$ are constants (slope-intercept form). More precisely, a line is an (infinite) set of points in the plane. In particular, the line $y=mx+b$ could be represented as the set,

That is, the set of all points of the form $(x,mx+b)\text{,}$ where $x$ ranges over all real numbers. The number $x$ here is called a parameter, because it is a quantity which specifies each point on the line. Here, it happens to represent the $x$-coordinate of the line, but its label does not change the line. That is,

both represent the same set of points, and the same line. We typically represent parameters by $t\text{.}$ Representing points in the plane as position vectors, this line can be written as the set of all vectors,

We can rewrite this vector in terms of the part which depends on the parameter $t$ and that part which doesn't,

Roughly, the $y$-intercept $b$ specifies a "starting point" for the line, and the slope $m$ specifies the "direction" the line is in (when moving 1 unit right, we move $m$ units up).

Using the language of vectors, we can form a more vector focused form of a line.

Suppose we want a line to pass through the point ${\mathbf{r}}_0=(x_0,y_0)\text{.}$ This vector will act as a “starting point” for the line. Then, we specify another vector $\mathbf{v}=(a,b)\text{,}$ which represents the direction from the starting point ${\mathbf{r}}_0$ which we want the line to travel, called a direction vector. Note that a line goes in two opposite directions, so $-\mathbf{v}$ would also work. Then, consider adding some multiple of $\mathbf{v}$ to ${\mathbf{r}}_0\text{.}$ Say, of the form,

where $t\in \mathbb{R}\text{.}$ The resuling vectors, for varying $t\text{,}$ trace out a line in the plane. As $t$ runs from $-\mathrm{\infty }$ to $\mathrm{\infty }\text{,}$ the vectors sweep out the entire line. Then, we say that the set,

is the vector parametric form of the line.

Note that the direction vector $\mathbf{v}$ is not unique, because any other vector parallel to it would also specify the same line. Also, the point on the line is not unique, because we could have used any other point on the line.

Another way to think about this, is that $\mathbf{r}$ is a point on the line if and only if $\mathbf{r}-{\mathbf{r}}_0$ is parallel to $\mathbf{v}\text{,}$ i.e. $\mathbf{r}-{\mathbf{r}}_0=t\mathbf{v}\text{,}$ which is an equivalent equation.

More explicitly, we can compare components,

Then, by equating components, this results in the equations,

This is called the scalar parametric form of a line. You may have encountered this form if you have studied parametric equations in pre-calculus or calculus.

Another method of specifying a line is to specify two points ${\mathbf{r}}_0,{\mathbf{r}}_1$ on the line. This is because a direction vector for the line is the difference between the points, $\mathbf{v}={\mathbf{r}}_1-{\mathbf{r}}_0\text{,}$ and one of the points (say ${\mathbf{r}}_0$) can be used as the starting point. Thus, the vector parametric form is,

Notice that ${\mathbf{r}}_0$ is on the line, corresponding to $t=0\text{,}$ and the point ${\mathbf{r}}_1$ is on the line, corresponding to $t=1\text{.}$

Recall that a line can also be specified using two points $(x_1,y_1),(x_2,y_2)\text{.}$ Then, a direction vector is the vector that is the difference of the points, as vectors,

Then, using the point ${\mathbf{r}}_0=(x_1,y_1)$ as a refernce point, the vector parametric equation of the line is again $\mathbf{r}={\mathbf{r}}_0+t\mathbf{v}\text{.}$ Or, in scalar parametric form,

In ${\mathbb{R}}^3\text{,}$ the equation of a line builds off the case for ${\mathbb{R}}^2\text{.}$ Notice that in Cartesian coordinates, it is unclear how to extend the plane equation $y=mx+b\text{,}$ to an equation in ${\mathbb{R}}^3$ involving all of $x,y,z\text{.}$

Again, a line can be completely determined by a point (say ${\mathbf{r}}_0=(x_0,y_0,z_0)\in {\mathbb{R}}^3$), and a vector (say $\mathbf{v}=(a,b,c)\in {\mathbb{R}}^3$).

Then, if $\mathbf{r}=(x,y,z)$ is a point on the line, then $\mathbf{r}-{\mathbf{r}}_0$ is parallel to $\mathbf{v}\text{.}$ That is, $\mathbf{r}-{\mathbf{r}}_0=t\mathbf{v}$ for some $t\in \mathbb{R}\text{.}$ Then, the line can be written as,

Then, similarly, we can compare components, to obtain the scalar parametric form of a line,

Again, the direction vector $\mathbf{v}$ and point ${\mathbf{r}}_0$ are both not unique.

Similarly, we can use two points ${\mathbf{r}}_0,{\mathbf{r}}_1$ to specify the line, given by,

Using some algebraic manipulations, we can eliminate the parameter $t\text{,}$ to write the equation of a line in Cartesian form. Recall that,

Then, solving for $t$ in each equation,

For every $t\text{,}$ the three quantities on the right-hand side of these equations must be equal. That is,

This is the Cartesian equation of a line in ${\mathbb{R}}^3\text{.}$ Really, it is two equations, say,

Let $\mathbf{u},\mathbf{v}$ be vectors. Then, consider the line segment connecting the two heads of the vectors in standard position. The midpoint of this line segment is the point $\frac{1}{2}(\mathbf{u}+\mathbf{v})\text{.}$ That is, the linear combination,

More generally, by varying the weights of this linear combination, but requiring that the weights sum to 1, we can reach every point along this line segment. Further, if we allow the weights to vary over all real numbers (but still require them to sum to 1), then we can access all points on the line extending in the direction of this segment. Equivalently, the line along $\mathbf{b}-\mathbf{a}\text{.}$

Then, if we are given two points $\mathbf{u},\mathbf{v}$ in the plane, we can form the line between them by considering the set of all linear combinations of the vectors, where the weights sum to 1. We specify this by denoting one weight by $t\text{,}$ and the other weight by $1-t\text{.}$ Then,

Linear combinations where the weights sum to 1 are called affine combinations.