How many controls does a vector need?

Imagine four controls, each responsible for one number in a record. You want the record (2,−1,0,3)(2,-1,0,3). How should you set the controls, and what should the zero in the third slot do?

You already know how to add vectors and multiply them by numbers. Here we will use those operations to build a vector one component at a time.

Keep a record of your investigation

Start with the reference record (0,0,0,0)(0,0,0,0). Before each adjustment, write the four entries you expect. Change one coordinate control at a time, keeping the other three fixed, and record the new record. Compare consecutive records: which slot changed, and by how much? Use those differences to explain what one control contributes.

Build the record without a four-dimensional drawing

Before setting Coordinate 1 to 2, predict all four entries in its contribution. Make the change and record the result. Then predict and test Coordinate 2 at −1, followed by Coordinate 4 at 3, recording each change separately.

Leave the third control at zero. Does its slot disappear, or does it remain part of the record?

Build a four-component record

Target: (2, −1, 0, 3). Each control changes just one slot.

Contribution of each coordinate control
ControlContribution to the record
0e1(0, 0, 0, 0)
0e2(0, 0, 0, 0)
0e3(0, 0, 0, 0)
0e4(0, 0, 0, 0)
Current record: (0, 0, 0, 0). Compare each slot with the target.

Explain which entries change when you move just one control.

Compare your explanation

The four contributions are (2,0,0,0)(2,0,0,0), (0,−1,0,0)(0,-1,0,0), (0,0,0,0)(0,0,0,0), and (0,0,0,3)(0,0,0,3). Their sum is the target. The third coordinate is still present; it simply contributes zero.

We do not need to draw four perpendicular arrows to calculate with four components. A table makes the separate slots visible.

The building blocks for coordinates

In the plane, define

e1=[10],e2=[01].\mathbf e_1=\begin{bmatrix}1\\0\end{bmatrix},\qquad \mathbf e_2=\begin{bmatrix}0\\1\end{bmatrix}.

Multiplying e1\mathbf e_1 changes only the first coordinate; multiplying e2\mathbf e_2 changes only the second. For example,

3e1−2e2=[30]+[0−2]=[3−2].3\mathbf e_1-2\mathbf e_2= \begin{bmatrix}3\\0\end{bmatrix}+ \begin{bmatrix}0\\-2\end{bmatrix}= \begin{bmatrix}3\\-2\end{bmatrix}.

These are standard unit vectors. Each has a 1 in one coordinate and zeros elsewhere. The subscript tells you which coordinate gets the 1; it is a label, not a multiplier.

They form the standard basis: every vector can be assembled using them, and its component values specify the coefficients uniquely. We will later investigate other choices of basis.

What does Rⁿ mean?

The notation Rn\mathbb R^n means the set of ordered lists of nn real numbers. In R4\mathbb R^4, there are four standard coordinate vectors. For instance,

e3=[0010].\mathbf e_3=\begin{bmatrix}0\\0\\1\\0\end{bmatrix}.

Our target becomes

[2−103]=2e1−e2+0e3+3e4.\begin{bmatrix}2\\-1\\0\\3\end{bmatrix} =2\mathbf e_1-\mathbf e_2+0\mathbf e_3+3\mathbf e_4.

We can omit the zero term and write 2e1−e2+3e42\mathbf e_1-\mathbf e_2+3\mathbf e_4. We cannot omit the zero coordinate from the resulting tuple: (2,−1,3)(2,-1,3) is a different, three-component object.

The dimensions of the unit vectors come from context. The e1\mathbf e_1 used in R2\mathbb R^2 has two entries; the one used in R4\mathbb R^4 has four.

Why the coefficients cannot be ambiguous

Suppose ae1+be2=(3,−2)a\mathbf e_1+b\mathbf e_2=(3,-2). The left side is simply (a,b)(a,b). Equality of ordered pairs requires a=3a=3 and b=−2b=-2. The same reasoning works one slot at a time in any Rn\mathbb R^n.

For an ML data record, a slot might represent a particular feature. The order matters: swapping two entries changes which feature receives each value. Coordinate arithmetic is meaningful only when records use the same feature order.

Decode a new expression

Without the explorer, convert −2e1+5e3-2\mathbf e_1+5\mathbf e_3 in R4\mathbb R^4 into component form. Then write (0,4,−1,2)(0,4,-1,2) as a sum of standard unit vectors.

Check the slots

The first vector is (−2,0,5,0)(-2,0,5,0). The second is 4e2−e3+2e44\mathbf e_2-\mathbf e_3+2\mathbf e_4. Missing terms contribute zero; they do not shift the remaining components left.

Design your own comparison

Choose your own four-component target using values from −4 to 4, including a zero and a negative entry. Write its standard-unit-vector expression before building it. Then choose one nonzero control to set to zero while holding the other three fixed. Predict the resulting record and compare it with the original target.

Could changing a different control repair the missing coordinate? Test one such attempt and explain the result from the coordinate slots.

Write a short conclusion: what you predicted, what you observed, and how the evidence changed or supported your explanation.

Check the reasoning in your investigation

Setting coordinate jj to zero removes its contribution in slot jj only. A different standard coordinate control cannot repair that slot. The record still has four entries, including zeros; omitting a term from the expression does not remove its coordinate.

Question 1 of 30 checked · 0 correct
In R⁴, what is 2e₂−3e₄?