Visual walkthrough — Coordinate vectors — change of basis

---

Step 1 — One arrow, no numbers yet

WHAT. Draw a single arrow $\mathbf{v}$ on blank paper. It points somewhere. That is all it is: a length and a direction. No numbers are attached to it yet.

WHY. The whole subject rests on one idea people forget: the arrow exists before you measure it. Numbers are something we add later by choosing rulers. If we skip this, every later formula looks like magic.

PICTURE. Look at the amber arrow below. There is deliberately no grid behind it. It is just a thing in space.

---

Step 2 — A basis is a pair of rulers

WHAT. Lay down two arrows ${\mathbf{b}}_1$ and ${\mathbf{b}}_2$. They must point in genuinely different directions (not parallel). This pair is a basis, which we name $\mathcal{B}=\{{\mathbf{b}}_1,{\mathbf{b}}_2\}$.

WHY these two conditions. We need "different directions" so that every arrow in the plane can be reached by mixing them, and so that the recipe to reach it is unique. Two arrows that point the same way could only ever reach one line — useless as a coordinate system. This "genuinely different" requirement is exactly Linear independence, and having enough of them to fill the space is Basis and dimension.

PICTURE. Below, the cyan arrows ${\mathbf{b}}_1,{\mathbf{b}}_2$ are our new rulers. The faint white grid they generate is the "graph paper" of basis $\mathcal{B}$ — a slanted graph paper.

---

Step 2b — The one basis we secretly already use: the standard basis

WHAT. There is one special pair of rulers everyone uses without saying so: the standard basis $\mathcal{E}=\{{\mathbf{e}}_1,{\mathbf{e}}_2\}$ with ${\mathbf{e}}_1=\left[\begin{array}{c}1\\ 0\end{array}\right]\text{(one step right)},{\mathbf{e}}_2=\left[\begin{array}{c}0\\ 1\end{array}\right]\text{(one step up)}.$

WHY it matters here. When we write a vector as $\mathbf{v}=\left[\begin{array}{c}4\\ 2\end{array}\right]$, those two entries are already coordinates — they mean "$4$ steps right, $2$ steps up", i.e. $\mathbf{v}=4{\mathbf{e}}_1+2{\mathbf{e}}_2$. So the plain column $\left[\begin{array}{c}4\\ 2\end{array}\right]$ is nothing but $[\mathbf{v}{]}_{\mathcal{E}}$, the coordinate vector in the standard basis. "Standard graph paper" from here on means the square grid drawn by ${\mathbf{e}}_1,{\mathbf{e}}_2$.

PICTURE. The square grid with ${\mathbf{e}}_1$ pointing right, ${\mathbf{e}}_2$ pointing up, and $\mathbf{v}=(4,2)$ read straight off the axes.

---

Step 3 — Coordinates = "how much of each ruler"

WHAT. Slide along ${\mathbf{b}}_1$ some amount $c_1$, then along ${\mathbf{b}}_2$ some amount $c_2$, until you land exactly on the tip of $\mathbf{v}$. The pair $(c_1,c_2)$ is the coordinate vector $[\mathbf{v}{]}_{\mathcal{B}}$.

$\mathbf{v}=\underset{\text{how much }{\mathbf{b}}_1}{\underbrace{c_1}}\underset{\text{ruler 1}}{\underbrace{{\mathbf{b}}_1}}+\underset{\text{how much }{\mathbf{b}}_2}{\underbrace{c_2}}\underset{\text{ruler 2}}{\underbrace{{\mathbf{b}}_2}},[\mathbf{v}{]}_{\mathcal{B}}=\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right].$

WHY it is a column of numbers. Because once the rulers are fixed, the two amounts are the only freedom left. And because the rulers are independent, there is exactly one way to land on $\mathbf{v}$ — so the column is well-defined, never ambiguous.

PICTURE. The dashed amber path shows the two slides: first $c_1{\mathbf{b}}_1$, then $c_2{\mathbf{b}}_2$, arriving at $\mathbf{v}$. The parallelogram closes — that closure is the equation above.

---

Step 4 — The matrix $B$ rebuilds the arrow from its numbers

WHAT. Stack the two ruler-arrows as the columns of a matrix and call it $B$: $B=\left[{\mathbf{b}}_1{\mathbf{b}}_2\right].$ Then the slide-recipe of Step 3 is exactly a matrix times the coordinate column: $\mathbf{v}=B[\mathbf{v}{]}_{\mathcal{B}}=[{\mathbf{b}}_1{\mathbf{b}}_2]\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]=\underset{\text{col 1}\times c_1}{\underbrace{c_1{\mathbf{b}}_1}}+\underset{\text{col 2}\times c_2}{\underbrace{c_2{\mathbf{b}}_2}}.$

Here the left side $\mathbf{v}$ is really $[\mathbf{v}{]}_{\mathcal{E}}$ (Step 2b): $B$ takes $\mathcal{B}$-numbers in and hands back the arrow's standard numbers.

WHY a matrix and not just "add them up". Because a matrix is the machine "multiply each column by the matching input number and add". That is precisely the slide-recipe. Writing it as $B[\mathbf{v}{]}_{\mathcal{B}}$ turns a picture into an object we can later invert. (A matrix whose columns are an independent basis is always invertible — see Invertible matrices.)

PICTURE. Here $\mathbf{v}=\left[\begin{array}{c}4\\ 2\end{array}\right]=[\mathbf{v}{]}_{\mathcal{E}}$ on the standard square grid, and $B$'s two columns are the cyan arrows. The amber arrow is the same $\mathbf{v}$ rebuilt as $3{\mathbf{b}}_1+1{\mathbf{b}}_2$.

---

Step 5 — A second basis $\mathcal{C}$ over the same arrow

WHAT. Now bring in a different pair of rulers $\mathcal{C}=\{{\mathbf{c}}_1,{\mathbf{c}}_2\}$ and stack them as $C=[{\mathbf{c}}_1{\mathbf{c}}_2]$. The arrow $\mathbf{v}$ has not moved. But its number-list is different, call it $[\mathbf{v}{]}_{\mathcal{C}}$.

$[\mathbf{v}{]}_{\mathcal{E}}=C[\mathbf{v}{]}_{\mathcal{C}}.$

WHY. Same arrow, different rulers ⇒ different amounts of each ruler needed. Both equations $[\mathbf{v}{]}_{\mathcal{E}}=B[\mathbf{v}{]}_{\mathcal{B}}$ and $[\mathbf{v}{]}_{\mathcal{E}}=C[\mathbf{v}{]}_{\mathcal{C}}$ describe the one fixed arrow's standard numbers — so their right-hand sides must be equal. That equality is the seed of everything.

PICTURE. One amber arrow, two overlaid grids — cyan for $\mathcal{B}$, white-dotted for $\mathcal{C}$. Same tip, two sets of gridlines counting to it.

---

Step 6 — Set the two descriptions equal and solve

WHAT. Both matrices produce the same standard column, so: $\underset{\text{arrow via }\mathcal{C}}{\underbrace{C[\mathbf{v}{]}_{\mathcal{C}}}}=\underset{\text{arrow via }\mathcal{B}}{\underbrace{B[\mathbf{v}{]}_{\mathcal{B}}}}.$ We want $[\mathbf{v}{]}_{\mathcal{C}}$ alone on the left. So multiply both sides on the left by $C^{-1}$: $\underset{=I}{\underbrace{C^{-1}C}}[\mathbf{v}{]}_{\mathcal{C}}=C^{-1}B[\mathbf{v}{]}_{\mathcal{B}}⟹[\mathbf{v}{]}_{\mathcal{C}}=(C^{-1}B)[\mathbf{v}{]}_{\mathcal{B}}.$

WHY multiply by $C^{-1}$ and not something else. We are trying to undo the "$\mathcal{C}$-numbers in, arrow out" machine $C$. The unique tool that undoes a matrix is its inverse (Step 4 box) — that is the reason $\mathcal{C}$ must be a genuine basis: only then does $C^{-1}$ exist (Invertible matrices).

PICTURE. A pipeline: $\mathcal{B}$-numbers enter, $B$ turns them into the standard column, then $C^{-1}$ reads that column off in $\mathcal{C}$-numbers. The middle "standard numbers" is the meeting point where the two halves join.

---

Step 7 — Numbers on the picture (a full run)

WHAT. Use $\mathcal{B}$ with ${\mathbf{b}}_1=\left[\begin{array}{c}1\\ 1\end{array}\right],{\mathbf{b}}_2=\left[\begin{array}{c}1\\ -1\end{array}\right]$ and $\mathcal{C}$ with ${\mathbf{c}}_1=\left[\begin{array}{c}1\\ 0\end{array}\right],{\mathbf{c}}_2=\left[\begin{array}{c}1\\ 2\end{array}\right]$. Then $C=\left[\begin{array}{cc}1& 1\\ 0& 2\end{array}\right],C^{-1}=\left[\begin{array}{cc}1& -\frac{1}{2}\\ 0& \frac{1}{2}\end{array}\right],P=C^{-1}B=\left[\begin{array}{cc}\frac{1}{2}& \frac{3}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right].$

Take $[\mathbf{v}{]}_{\mathcal{B}}=\left[\begin{array}{c}3\\ 1\end{array}\right]$ (this is $\mathbf{v}=\left[\begin{array}{c}4\\ 2\end{array}\right]=[\mathbf{v}{]}_{\mathcal{E}}$, from Step 4). Push it through $P$: $[\mathbf{v}{]}_{\mathcal{C}}=\left[\begin{array}{cc}\frac{1}{2}& \frac{3}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right]\left[\begin{array}{c}3\\ 1\end{array}\right]=\left[\begin{array}{c}\frac{3}{2}+\frac{3}{2}\\ \frac{3}{2}-\frac{1}{2}\end{array}\right]=\left[\begin{array}{c}3\\ 1\end{array}\right].$

Check on the arrow: $3{\mathbf{c}}_1+1{\mathbf{c}}_2=3\left[\begin{array}{c}1\\ 0\end{array}\right]+\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}4\\ 2\end{array}\right]=\mathbf{v}$. ✓

WHY it landed right. We never touched the arrow — only re-read its address. The check re-assembles $\mathbf{v}$ from the new numbers and new rulers and gets the original standard column $(4,2)$ back.

PICTURE. The amber arrow with both address-labels shown: "$(3,1)$ in $\mathcal{B}$" and "$(3,1)$ in $\mathcal{C}$" — here they coincide numerically by coincidence of these particular bases; the gridlines counting to the tip differ.

---

Step 8 — Degenerate cases (where it breaks, and why)

WHAT. What if the two "rulers" are parallel, e.g. ${\mathbf{c}}_1=\left[\begin{array}{c}1\\ 0\end{array}\right],{\mathbf{c}}_2=\left[\begin{array}{c}2\\ 0\end{array}\right]$?

WHY it fails — two symptoms of the same illness. With ${\mathbf{c}}_1=(1,0),{\mathbf{c}}_2=(2,0)$ we get $\det C=1\cdot 0-0\cdot 2=0$, so $C^{-1}$ does not exist and $P=C^{-1}B$ has no meaning. Geometrically this shows up in two ways at once:

• No solution (off the line). These rulers only reach the horizontal line. Any arrow with a nonzero vertical part — like $(1.5,1.5)$ — is unreachable: it has no $\mathcal{C}$-address at all.
• Too many solutions (on the line). An arrow that does lie on the line, like $(4,0)$, has infinitely many addresses: $4=4{\mathbf{c}}_1+0{\mathbf{c}}_2$, but also $4=2{\mathbf{c}}_1+1{\mathbf{c}}_2$, or $0{\mathbf{c}}_1+2{\mathbf{c}}_2$, … Because the rulers overlap, you can trade one for the other freely. Uniqueness — the whole reason coordinates mean anything — is destroyed.

Both symptoms are why a basis must be independent: independence is precisely the condition "$\det C\ne 0$", which guarantees exactly one address for every arrow.

Zero coordinates and negatives are fine, though. If $[\mathbf{v}{]}_{\mathcal{C}}$ has a zero entry, the arrow just uses none of that ruler. If it has a negative entry, you slide backwards along that ruler — still perfectly legal (recall Example 3's $-1$).

PICTURE. Left panel: two parallel cyan arrows, an off-line target marked "✗ no address", and an on-line arrow $(4,0)$ shown with two different amber decompositions ("∞ many addresses"). Right panel: a valid basis with a negative coordinate, the backwards slide drawn as a dashed reversed amber arrow.

---

The one-picture summary

Everything compresses to one triangle of maps: $\mathcal{B}$-numbers on the left, $\mathcal{C}$-numbers on the right, the standard column $[\mathbf{v}{]}_{\mathcal{E}}$ at the top apex, with $B$, $C$, and their inverses on the edges. Walking $\mathcal{B}\to$ apex $\to \mathcal{C}$ is $C^{-1}B$.

Recall Feynman retelling — the whole walkthrough in plain words

A treasure sits in a field (Step 1: the arrow, before anyone measures). The town's official map already uses "steps east / steps north" — that's the standard basis, and $(4,2)$ is the treasure's official address (Step 2b). Anna prefers her own two step-directions; her "recipe machine" $B$ turns her step-counts into the official address (Step 3–4). Ben uses yet other step-directions with his own machine $C$ (Step 5). To turn Anna's counts into Ben's: run Anna's machine forward to the official address, then run Ben's machine backwards ($C^{-1}$, which exists only because his directions are genuinely different) to read off how he would pace it (Step 6). That combo is $C^{-1}B$; the numbers checked out on a real example (Step 7). And if Ben's two directions were secretly the same line, off-line treasures get no address and on-line treasures get infinitely many — his machine can't be run backwards at all (Step 8). The treasure never moved; only the paperwork changed.