Vectors As Arrows Before Coordinates

A $2\times2$ matrix can be read as instructions for how the plane moves. Feed in every point of a shape and the matrix sends it somewhere new. The course problem is to predict that new shape without guessing.

If you start with matrices as tables of numbers, it is tempting to think the only path is compute, compute, compute. The faster path is to build a picture first. A matrix acts on vectors, and a vector is best understood as an arrow that represents a displacement. Once the arrow picture is solid, coordinates become a convenient way to describe the same arrow, not the definition of what it is.

By the end of this lesson you should be able to look at a vector expression like $2u-v$ and say what motion it represents in the plane, then translate that motion into coordinates when you need to calculate. That is the bridge to the next lessons, where a matrix will become a machine that transforms every arrow in a consistent way. What does it mean for a transformation to be consistent, and how can a few arrows determine what happens to an entire shape?

Vectors as displacements

A vector is a directed displacement. You can draw it as an arrow from a starting point to an ending point. The key feature is that only the displacement matters. Sliding the arrow without rotating or resizing it does not change the vector.

Vector addition answers a concrete question. If one move takes you from $A$ to $B$, and a second move takes you from $B$ to $C$, what single move takes you from $A$ to $C$? Place the second arrow head to tail after the first. The arrow from the start of the first to the end of the second is the sum.

Scalar multiplication answers a different question. If one move is useful, what about repeating it, or reversing it?

• Multiplying by $2$ keeps the direction and doubles the length, like taking the same step twice.
• Multiplying by $\tfrac12$ keeps the direction and halves the length.
• Multiplying by $-1$ keeps the length but flips the direction. It is the same displacement backward.

A common mistake is to think $u+v$ means attach the tails and then draw the arrow between the heads. That arrow is not $u+v$ in general. It measures how far apart the endpoints are, not the combined displacement. A quick counterexample is $v=u$. The true sum is $2u$, which points the same way as $u$, but the head-to-head arrow would be the zero vector.

Look closely at how the head to tail picture and the parallelogram picture encode the same sum.

Once you trust these geometric rules, coordinates will feel like labels on the picture rather than the picture itself.

Coordinates as a description of an arrow

In the plane, a grid gives you two special direction arrows. One step to the right and one step up. Call them basis vectors because they set the directions you measure along. Using the usual grid, those are

$$\hat{\imath}=\begin{pmatrix}1\\0\end{pmatrix},\qquad \hat{\jmath}=\begin{pmatrix}0\\1\end{pmatrix}.$$

This notation is not the definition of $\hat{\imath}$ and $\hat{\jmath}$. It is the coordinate description of those two arrows in this grid.

Now take an arbitrary arrow $u$. Suppose moving by $u$ is the same as moving $3$ steps in the $\hat{\imath}$ direction and $2$ steps in the $\hat{\jmath}$ direction. Then you can build it as

$$u=3\hat{\imath}+2\hat{\jmath},$$

and the coordinate column

$$u=\begin{pmatrix}3\\2\end{pmatrix}$$

is a compact record of those two components.

The subtle point is that the arrow is the geometric object, and the pair of numbers depends on the grid you chose. Change the basis directions, and the same physical arrow can get different coordinates. That dependence is not a flaw. It is the whole reason linear algebra can switch viewpoints while keeping the geometry fixed.

Use the diagram to connect one arrow to its component counts along the chosen basis directions, and notice what stays the same when the grid changes.