Any Independent Pair Builds a Grid

By the end of this section you will have dropped an assumption you may not have realized you were making: that a basis has to look like the neat, perpendicular î, ĵ grid. Sections 1 and 2 built a grid out of two perpendicular rulers. That was a convenient choice, not a requirement. This section shows that any two vectors that point genuinely different ways — square to each other or not — build a perfectly good grid of their own.

What You Will Be Able to Do

The Big Picture: Perpendicular Was Never the Rule

Everything in Section 2 worked because $\hat{\imath}$ and $\hat{\jmath}$ point in genuinely different directions — not because they happen to be perpendicular. Swap them for a skewed pair, say $\vec b_1 = (1, 0)$ and $\vec b_2 = (1, 1)$, and the construction works exactly the same way:

$\vec b_1$ and $\vec b_2$ are not at right angles — but $\vec b_2$ still points somewhere $\vec b_1$ alone could never reach, and that is the only property that matters. Scale and combine the two of them and, once again, every point in the plane is reachable: this skewed pair is a valid basis, and its span is still the whole plane.

What changes is how you read a point. Saying a point sits at “(2, 1) in this basis” means exactly $2\vec b_1 + 1\vec b_2$ — the weights on this grid's own rulers, not on the standard $\hat{\imath}, \hat{\jmath}$ axes. The deeper idea waiting in that sentence — that $\vec b_1$ and $\vec b_2$ quietly play the role the number 1 played on the standard grid — is worth sitting with before the next sections make it precise.

Reading a Point in a Skewed Basis

Let's put actual numbers behind “coordinates against a new basis,” using the skewed pair from above: $\vec b_1 = (1, 0)$ and $\vec b_2 = (1, 1)$. Saying a point sits at “(2, 1) in B-coordinates” is a claim about weights, not a position you read straight off a normal grid. It means exactly:

Work it out one component at a time:

So the B-coordinate pair $(2, 1)$ and the standard coordinate pair $(3, 1)$ name the exact same point in the plane — two different recipes for reaching one place. Against the standard basis $\{\hat{\imath}, \hat{\jmath}\}$, that point is “3 rights and 1 up.” Against $B$, the identical point is “2 lots of $\vec b_1$ and 1 lot of $\vec b_2$.” Change the ruler, and the same physical point gets a different name.

Because $\vec b_1$ and $\vec b_2$ are independent, this recipe is the only way to reach each of those points — no other $(x, y)$ pair produces the same standard coordinates. A basis does not just make a point reachable; it makes the recipe for reaching it unique.

The Hidden 1s

Before returning to vectors, look at two throwaway arithmetic facts: $1 \cdot 8 = 8$ and $2 \cdot 1 = 2$. Trivial — but notice what is happening. The number 8 is secretly $8 \cdot 1$; the multiplier 1 is doing real work, you just never write it, because multiplying by 1 changes nothing and so it becomes invisible.

The standard grid runs on exactly this invisible bookkeeping. The point $3$ on a number line is really $3 \cdot \hat{\imath}$; the point $(3, 1)$ in the plane is really $3\hat{\imath} + 1\hat{\jmath}$. $\hat{\imath}$ and $\hat{\jmath}$ are the standard grid's own “1”s — each coordinate is a coefficient on a unit that already equals one unit of its own direction. That is why writing $(3, 1)$ alone is enough: the hats are implied, invisible, because they are the identity's own basis.

Now look back at $(2, 1)$ in $B$-coordinates from the last section — the exact same invisible multiplication is happening in $2\vec b_1 + 1\vec b_2$. $\vec b_1$ and $\vec b_2$ are not the number 1, and they are not $\hat{\imath}, \hat{\jmath}$ — but inside this grid they play exactly the role 1 played on the number line: they are $B$'s own units, $B$'s own “1”s. Nothing about the arithmetic changed — you still multiply and add — only which vectors serve as the invisible multiplier changed.

See It Live: Two Grids, One Plane

Here is a second working example, this time drawn: the basis $\vec u_1 = (1, 1)$, $\vec u_2 = (-1, 1)$ — the columns of $A = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$. The cyan grid you see is the untouched identity — the standard $\hat{\imath}, \hat{\jmath}$ grid, unmoved. The amber grid drawn over it is the constructed grid built from $\vec u_1$ and $\vec u_2$ — turned 45° and stretched by $\sqrt{2}$ from the standard grid.

This particular pair happens to be perpendicular too — but that is a coincidence, not a requirement. Section 3’s whole point was that right angles were never the rule; “genuinely different directions” was always the only condition. Try editing the matrix entries above: as long as the two columns keep pointing different ways, the amber grid stays a full, working plane-covering grid, however skewed.

Now build one of your own. The viewer below opens directly on Stage 4 — Your Grid: drag the red handle ($\vec b_1$) and the green handle ($\vec b_2$) anywhere you like. The ghost cyan identity grid always stays behind as a reference; the amber constructed grid updates live, along with rank, determinant, and area chips computed straight from your two vectors. Try dragging $\vec b_2$ slowly toward $\vec b_1$'s line and watch what happens to the grid as the two directions stop being genuinely different — that collapse is exactly what the next section names.

Common Misconceptions

“Any two different-looking vectors will do”

Why it is tempting: $(1, 2)$ and $(2, 4)$ are written differently, so it is easy to assume they point differently too. Correction: what matters is direction, not the numbers on the page. $(2, 4) = 2 \cdot (1, 2)$ is a scaled copy of $(1, 2)$ — same line, same direction —so this pair cannot build a grid at all. “Looks different” and “is independent” are not the same test. Example: $\{(1, 2), (2, 4)\}$ fails; $\{(1, 2), (2, 4.001)\}$ — nearly identical numbers — succeeds, because that tiny nudge is enough to point somewhere the first vector alone cannot reach.

“Coordinates in a skewed basis behave like normal x-y offsets”

Why it is tempting: a coordinate pair still looks like $(x, y)$, so it is natural to read it the way you always have. Correction: B-coordinates are combination weights, not perpendicular measurements. Moving 1 unit in the $\vec b_2$ direction above physically moves you $\sqrt{2}$ units across the standard grid, not 1 — lengths and distances in a skewed basis only match the standard grid when the basis happens to be orthonormal. Example: B-coordinates $(0, 1)$ against $\vec b_2 = (1, 1)$ land at the standard point $(1, 1)$, a distance of $\sqrt{2}$ from the origin, not 1.

Practice Problems

Beginner

B1. Using $\vec b_1 = (1, 0)$, $\vec b_2 = (1, 1)$, compute the standard coordinates of the point whose B-coordinates are $(3, -1)$.
Hint: $V = 3\vec b_1 - 1\vec b_2$; work out each term, then add.

B2. Is $\{(2, -1), (4, -2)\}$ a valid basis for $\mathbb{R}^2$? Explain in one sentence.
Hint: is the second vector a scalar multiple of the first?

Intermediate

I1. Using the same $\vec b_1, \vec b_2$ as B1, find the B-coordinates $(x, y)$ of the standard point $(5, 2)$.
Hint: $\vec b_2$'s y-component is 1, so $y$ is fixed by the point's y-coordinate directly; then solve for $x$ from the x-coordinate.

Conceptual

C1. In one sentence, explain why the “hidden 1s” idea means every basis, however skewed, is invisible in the same way the standard grid usually is.
Hint: what role do a basis's own vectors play inside its own coordinate recipe?