Discover Linear Algebra: A first course in linear algebra

A directed line segment (or arrow) could be thought of dynamically as describing a change in position, from the initial point to the terminal point. A two-dimensional vector in the plane or a three-dimensional vector in space captures just the change part of “change in position,” leaving the position part (that is, the initial and terminal points) unspecified. For example, in the plane, the instructions “move two units right and three units down” describe a way to change positions, but don’t actually specify from where or to where the change in position is occurring. So a vector corresponds to an infinite number of directed line segments, where each of these directed line segments has a different initial point but all of them require the same “change” to change positions from initial point to terminal point. Continuing our example, every change in position between some initial and terminal points in the plane that requires moving two units right and three units down can be represented by the same vector.

We describe a two-dimensional vector in the plane with a pair of numerical components: the change in $x$ and the change in $y\text{.}$ If $P(x_1,y_1)$ and $Q(x_2,y_2)$ are points in the plane, then the vector associated to the directed line segment $\overrightarrow{PQ}$ has components $\mathbf{v}=(\mathrm{\Delta }x,\mathrm{\Delta }y)=(x_2-x_1,y_2-y_1)\text{.}$ (A three-dimensional vector in space requires a third component as well: the change in $z\text{.}$)

Notice what happens when we use the origin $O(0,0)$ as the initial point and an arbitrary point $R(x,y)$ as the terminal point in a directed line segment: the vector associated to $\overrightarrow{OR}$ is then $(x-0,y-0)=(x,y)\text{.}$ So when the initial point is the origin, the components of the vector are exactly the coordinates of the terminal point. In Discovery 11.1 we saw that this works in reverse as well. That is, if we have a vector $\mathbf{v}=(v_1,v_2)\text{,}$ then we could consider the numbers $v_1,v_2$ as coordinates of a point $R(x,y)$ with $x=v_1$ and $y=v_2\text{,}$ and then the vector associated to $\overrightarrow{OR}$ is just $\mathbf{v}$ again. In this way, every vector corresponds to one unique directed line segment with initial point at the origin, and so that is sort of the “natural position” of the vector as a directed line segment. But we will find that it is often convenient to consider other directed line segments that correspond to a particular vector.

We live in a three-dimensional world (or, at least, it appears that way to us), and our little human brains cannot visualize points or arrows in four- or higher-dimensional spaces. However, we can still describe such imaginary objects using our experience from two- and three-dimensional points and vectors. For example, if we had two points $P$ and $Q$ in an imaginary four-dimensional space, they would each require four coordinates, so we would describe them as $P(w_1,x_1,y_1,z_1)$ and $Q(w_2,x_2,y_2,z_2)\text{.}$ Then the vector corresponding to the directed line segment $\overrightarrow{PQ}$ would have four components and we would compute it as $\mathbf{v}=(\mathrm{\Delta }w,\mathrm{\Delta }x,\mathrm{\Delta }y,\mathrm{\Delta }z)=(w_2-w_1,x_2-x_1,y_2-y_1,z_2-z_1)\text{.}$

A vector describes a change in position. If we chain two changes in position together, by making the initial point of the second vector the same as the terminal point of the first vector, then we could consider the overall change in position.

If these are points and vectors in the plane, then clearly the change in position from $P$ to $R$ will be described by the total net change in $x$ and the total net change in $y\text{,}$ as we discovered in Discovery 11.2. So, in the diagram above, we obtain the components for the vector corresponding to $\overrightarrow{PR}$ by adding corresponding components of $\mathbf{u}$ and $\mathbf{v}\text{.}$ That is, if $\mathbf{u}=(u_1,u_2)$ and $\mathbf{v}=(v_1,v_2)\text{,}$ then the components of the dashed arrow labelled with a question mark are $(u_1+v_1,u_2+v_2)\text{.}$ For obvious reasons, we call this the sum of $\mathbf{u}$ and $\mathbf{v}\text{.}$

In Discovery 11.2 we also considered the result of interchanging the order of a pair of vectors that have been chained together.

In the diagram above, the vector for $\overrightarrow{P{Q}^{\prime }}$ is the same as that for $\overrightarrow{QR}\text{,}$ because they represent the same change in position, just with a different initial point. Accordingly, we have labelled both vectors with $\mathbf{v}\text{.}$ And the same applies to $\mathbf{u}$ with respect to $\overrightarrow{PQ}$ and $\overrightarrow{Q^{\prime }R}\text{.}$

The diagram illustrates that if we start at $P$ and chain together the change-of-position instructions contained in vectors $\mathbf{u}$ and $\mathbf{v}\text{,}$ the order that we do so does not matter — the overall change in position will be from $P$ to $R\text{.}$ Thus, the order of vector addition doesn’t matter. Algebraically, we could have predicted that this would be the case because it doesn’t matter what order you add components: the identities $u_1+v_1=v_1+u_1$ and $u_2+v_2=v_2+v_2$ are both valid. But it’s useful conceptually to have the above geometric picture of vector addition because, whether you believe this about yourself or not, humans are spatial thinkers. And the geometric version of the vector identity $\mathbf{v}+\mathbf{u}=\mathbf{u}+\mathbf{v}$ makes a pretty picture of a parallelogram, so we call it the parallelogram rule.

There is one special change in position that is unlike any other — the one where the initial and terminal points are the same, so that there is actually no change in position. In two dimensions, this means there is no change in either $x$ or $y\text{,}$ so the components are $(0,0)\text{.}$ Similarly, in any number of dimensions we have the zero vector $\mathbf{0}=(0,0,\dots ,0)\text{.}$

As we explored in Discovery 11.3, if we chain together a vector $\mathbf{v}\text{,}$ representing some change in position, with the zero vector, which represents no change, then the net result is just the change of $\mathbf{v}\text{.}$ That is, $\mathbf{v}+\mathbf{0}=\mathbf{v}\text{,}$ and also $\mathbf{0}+\mathbf{v}=\mathbf{v}\text{.}$

If we move from $P$ to $Q\text{,}$ and from there move from $Q$ back to $P\text{,}$ the net result is no change in position, which is represented by the zero vector. This means if we add the vector corresponding to $\overrightarrow{PQ}$ to the one corresponding to $\overrightarrow{QP}\text{,}$ the result is $\mathbf{0}\text{.}$ So if we label the vector for $\overrightarrow{PQ}$ as $\mathbf{v}\text{,}$ it seems reasonable to label the vector for $\overrightarrow{QP}$ as $-\mathbf{v}\text{,}$ the negative of $\mathbf{v}\text{,}$ so that we have $\mathbf{v}+(-\mathbf{v})=\mathbf{0}\text{.}$

Remembering that the “natural” position for a vector is with its tail at the origin, it’s useful to visualize negatives in the following manner.

That is, the negative of a vector will change positions by the same distance but in the opposite direction.

To subtract vectors, we add to a vector the negative of another.

Here, the diagonal vector labelled $\mathbf{u}-\mathbf{v}$ is obtained by adding $\mathbf{u}$ and $-\mathbf{v}\text{.}$ As we explored in Discovery 11.5, we get an interesting pattern if we draw in another copy of the vector labelled $\mathbf{u}-\mathbf{v}$ with its initial point at $R\text{.}$

Triangle $\mathrm{△}R{P}^{\prime }P$ creates the vector addition pattern $\mathbf{u}+(-\mathbf{v})=\mathbf{u}-\mathbf{v}\text{.}$ But notice that $\mathrm{△}ORP$ creates a vector addition pattern starting at $O$ and ending up at $P\text{,}$ by $\mathbf{v}+(\mathbf{u}-\mathbf{v})=\mathbf{u}\text{.}$ So we can think of a difference of two vectors as a vector that runs between the heads of the two vectors in the difference when they share the same initial point. Algebraically, we can think of the $\mathbf{v}$ and $-\mathbf{v}$ cancelling in the expression $\mathbf{v}+(\mathbf{u}-\mathbf{v})\text{,}$ leaving just $\mathbf{u}\text{.}$

Of course, there are two vectors that run between the heads of $\mathbf{u}$ and $\mathbf{v}\text{,}$ namely $\mathbf{u}-\mathbf{v}$ and its negative.

Now $\mathrm{△}P{P}^{\prime }R$ creates a vector addition pattern starting at $P$ and ending up at $R\text{,}$ so that $\mathbf{v}+(-\mathbf{u})=\mathbf{v}-\mathbf{u}\text{.}$ But also $\mathrm{△}OPR$ creates a vector addition pattern starting at $O$ and ending up at $R\text{,}$ so that $\mathbf{u}+(\mathbf{v}-\mathbf{u})=\mathbf{v}\text{.}$ And finally, the fact that $\mathbf{u}-\mathbf{v}$ and $\mathbf{v}-\mathbf{u}$ both run between $P$ and $R\text{,}$ but in opposite directions, verifies geometrically that $-(\mathbf{u}-\mathbf{v})=\mathbf{v}-\mathbf{u}\text{,}$ as we would expect algebraically.

Geometrically, when we scalar multiply a vector we “stretch” or scale its length by the scale factor. (If this scale factor is negative, then we also flip the vector around in the opposite direction.) Here are some examples.

Notice how each of these vectors either points in the same direction as or in the opposite direction to $\mathbf{v}\text{.}$ In particular, they are all parallel to one another. This happens precisely when the vectors are scalar multiples of one another.

Thinking of the vectors in the diagram above as vectors in the plane, if we scale $\mathbf{v}$ by a factor of $2\text{,}$ then our knowledge of similar triangles tells us that the change in both $x$ and $y$ must be double.

We can connect scalar multiplication to addition, as we explored in Discovery 11.6. If we add a vector to itself, then the sum vector will be twice as long as the original.

So we have $\mathbf{v}+\mathbf{v}=2\mathbf{v}\text{.}$

We will provide more rules of vector algebra as Proposition 11.5.1 in Subsection 11.5.1. In Discovery 11.12, we decided that the algebra of vectors is the same as the algebra of column matrices (which we have already been referring to as column vectors), so we should be able to anticipate a number of the vector algebra rules that will appear in that proposition.

In Discovery 11.8, we encountered two very special vectors in the plane, ${\mathbf{e}}_1=(1,0)$ and ${\mathbf{e}}_2=(0,1)\text{.}$ These two vectors could be considered the fundamental changes in position in the plane — ${\mathbf{e}}_1$ represents a change by one unit right, and ${\mathbf{e}}_2$ represents a change by one unit up.

Any change in position can be built out of these two fundamental changes in position. Using the example in Discovery 11.8, the vector $\mathbf{v}=(5,2)$ represents a change in position by $5$ units right and $2$ units up. We can achieve the “$5$ units right” part with $5{\mathbf{e}}_1=(5,0)$ and the “$2$ units up part” with $2{\mathbf{e}}_2=(0,2)\text{.}$ To get the total change in position represented by $\mathbf{v}\text{,}$ we can combine these two building blocks in the linear combination $\mathbf{v}=5{\mathbf{e}}_1+2{\mathbf{e}}_2\text{.}$

As you can imagine, every vector in the plane can be decomposed into a linear combination of ${\mathbf{e}}_1$ and ${\mathbf{e}}_2$ in this manner: for $\mathbf{v}=(v_1,v_2)\text{,}$ we have $\mathbf{v}=v_1{\mathbf{e}}_1+v_2{\mathbf{e}}_2\text{.}$ For this reason, the two vectors ${\mathbf{e}}_1,{\mathbf{e}}_2$ together are called the standard basis vectors in ${\mathbb{R}}^2\text{,}$ as they form a basis from which every other vector can be constructed. To use an analogy with chemistry, these two vectors are the basic atoms of ${\mathbb{R}}^2\text{,}$ and every other vector in ${\mathbb{R}}^2$ is a molecule built out of a specific combination of these atoms. Since there are only two fundamental directions in ${\mathbb{R}}^2$ (right/left and up/down), it is not surprising that we need only two basis vectors to represent all possible directions. This is the reason we call vectors in ${\mathbb{R}}^2$ two-dimensional vectors.