Visual walkthrough — Change of basis matrix

We work in the flat plane you can draw on paper — the set of all arrows starting at one fixed dot (the origin). This set is called ${\mathbb{R}}^2$. Everything generalises, but you should watch it happen in 2D first.

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Step 1 — An arrow is real; its numbers are not

WHAT. Draw one arrow $v$ from the origin. It just sits there. Now lay down two rulers — two directions we agree to measure along. Call the first ruler-direction $b_1$ and the second $b_2$. Together $\{b_1,b_2\}$ is a basis: two directions that (a) don't lie on the same line and (b) let us reach every point by combining them. (See Basis and Dimension for why two independent directions is exactly enough in the plane.)

WHY. A column of numbers means nothing until we say "numbers of what, measured how." The basis is the "measured how." Change the rulers and the numbers change — but the arrow does not move a millimetre.

PICTURE. The black arrow $v$ is fixed. The red pair of rulers $b_1,b_2$ is our choice.

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Step 2 — Two friends describe the same arrow

WHAT. Bring in a second basis $C=\{c_1^{\prime },c_2^{\prime }\}$ — a different pair of rulers. (I write the vectors of $C$ with primes, $c_1^{\prime },c_2^{\prime }$, so they never get confused with coordinate numbers $c_1,c_2$.) The same arrow $v$ now has two coordinate columns: $[v{]}_B$ and $[v{]}_C$.

WHY. Our whole goal is a machine that turns one column into the other. To design the machine we first need both columns staring at the same arrow.

PICTURE. One black arrow $v$. Grid of $B$-rulers in thin black; the single red thing is the new ruler pair $C$. Same tip, two grids reading it.

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Step 3 — Write the arrow in the OLD language (all we're given)

WHAT. We are handed $[v{]}_B=(c_1,c_2)$. Unpack what that literally means: $v=\underset{\text{steps along }{b}_1}{\underbrace{c_1}}{b}_1+\underset{\text{steps along }{b}_2}{\underbrace{c_2}}{b}_2.$

WHY. This is the only fact in our hands. We cannot use $C$ yet — we don't yet know how $C$ sees anything. Start from what's given.

PICTURE. The red arrow $v$ built as a staircase: walk $c_1$ copies of $b_1$, then $c_2$ copies of $b_2$, land on the tip.

Term by term: $c_1{b}_1$ is the first leg of the staircase, $c_2{b}_2$ is the second leg, and their vector sum $v$ is the diagonal from origin to tip.

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Step 4 — Teach the machine ONE word at a time

WHAT. Each old ruler $b_j$ is itself just an arrow, so it too has a $C$-description. Ask the $C$-friend: "how do you write $b_1$? how do you write $b_2$?" The answers are columns $[b_1{]}_C=\left[\begin{array}{c}{p}_{11}\\ p_{21}\end{array}\right],[b_2{]}_C=\left[\begin{array}{c}{p}_{12}\\ p_{22}\end{array}\right].$ The symbol $p_{ij}$ reads "row $i$, column $j$": how many steps along the $i$-th $C$-ruler you need to reproduce the $j$-th $B$-ruler.

WHY. This is the heart of translation. A dictionary is not built for whole sentences — it's built word by word. The words are the basis vectors $b_1,b_2$. If I can say each ruler in the new language, I can say any combination of them, because combining is just addition and stretching, which coordinates respect.

PICTURE. The single red ruler $b_1$ shown decomposed against the $C$-grid: its shadow $p_{11}$ along $c_1^{\prime }$ and $p_{21}$ along $c_2^{\prime }$.

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Step 5 — Coordinates add and stretch (the linear law)

WHAT. Take $C$-coordinates of both sides of Step 3's equation. The map "arrow $\mapsto$ its $C$-column", written $[\cdot {]}_C$, obeys two rules: $[u+w{]}_C=[u{]}_C+[w{]}_C,[ku{]}_C=k[u{]}_C.$ Together these two rules make $[\cdot {]}_C$ a linear map (see Linear Transformations). Applying them: $[v{]}_C=[c_1{b}_1+c_2{b}_2{]}_C=c_1[b_1{]}_C+c_2[b_2{]}_C.$

WHY. Because coordinates are unique (only one recipe per arrow per basis), doubling the arrow doubles every step-count, and adding arrows adds step-counts. So we may pull the sum apart term by term — turning "translate the whole arrow" into "translate each word, then recombine with the same weights $c_1,c_2$."

PICTURE. Same staircase as Step 3, but now every rung is read on the $C$-grid: the red total column $[v{]}_C$ is literally $c_1$ of the first dictionary column plus $c_2$ of the second.

Notice: the weights $c_1,c_2$ did not change — they are still the old $B$-coordinates. Only the directions being weighted switched from arrows $b_j$ to their $C$-columns $[b_j{]}_C$.

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Step 6 — Recognise the matrix hiding in the sum

WHAT. The expression $c_1[b_1{]}_C+c_2[b_2{]}_C$ is exactly what "matrix times column" means: stack the two dictionary columns side by side, feed it $(c_1,c_2)$.

$$[v{]}_C=\underset{P_{C\leftarrow B}}{\underbrace{[[b_1{]}_C|[b_2{]}_C]}}\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]=P_{C\leftarrow B}[v{]}_B.$$

WHY. A matrix–vector product $M\mathbf{x}$ is defined as "$x_1$ times column 1 plus $x_2$ times column 2 …". Step 5 produced precisely that shape. So the machine we wanted is real, and its columns are forced to be the dictionary entries $[b_j{]}_C$ — nothing else fits.

PICTURE. The two red dictionary columns slotting in as the two columns of a box labelled $P_{C\leftarrow B}$; input column $[v{]}_B$ on the right, output $[v{]}_C$ dropping out.

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Step 7 — The standard-basis shortcut (why $C^{-1}B$)

WHAT. In ${\mathbb{R}}^2$ there is a free ruler pair: the standard basis $E=\{(1,0),(0,1)\}$, the axes themselves. For it, $[b_j{]}_E=b_j$ literally (a vector's standard coordinates are just its entries). So the dictionary "$B\to E$" needs no work: $P_{E\leftarrow B}=[b_1\mid b_2]=:B(\text{the matrix whose columns ARE the basis vectors}).$ To go the other way, $E\to B$, we undo that machine — the inverse: $P_{B\leftarrow E}=B^{-1}.$ Now chain two translations, $B\to E\to C$: $P_{C\leftarrow B}=P_{C\leftarrow E}{P}_{E\leftarrow B}=C^{-1}B.$

WHY. Translating everything through the standard axes lets us avoid ever solving "how does $C$ see $b_j$" by hand — matrix inversion does it in one shot. Reading right-to-left: first $B$ carries $B$-coords to standard, then $C^{-1}$ carries standard to $C$-coords.

PICTURE. Three grids in a row — $B$-grid, then standard axes, then $C$-grid — with the arrow's column being relabelled at each hop; the two hops $B$ and $C^{-1}$ are marked in red.

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Step 8 — Edge and degenerate cases (never hit an unshown scenario)

WHAT & WHY & PICTURE for each corner case:

(a) $C=B$ (translate a language into itself). Then $[b_j{]}_C=[b_j{]}_B=e_j$, so the columns are $e_1,e_2$: the machine is the identity $I$. Nothing to do — the arrow already speaks the target language.

(b) $C=E$ (target is standard). Columns become $[b_j{]}_E=b_j$, so $P_{E\leftarrow B}=B$: just stack the basis vectors. This is the "easy direction" — no inverse. Example: $B=\{(2,1),(-1,1)\}$, $[v{]}_B=(4,-1{)}^T$ gives $[v{]}_E=B[v{]}_B=(9,3{)}^T$.

(c) $B=E$ (source is standard). Now $P_{C\leftarrow E}=C^{-1}$: the hard direction, an inverse is required. Example: $w=(5,0)$, $B=\{(2,1),(-1,1)\}$ gives $[w{]}_B=B^{-1}w=(5/3,-5/3{)}^T$. Check $\frac{5}{3}(2,1)-\frac{5}{3}(-1,1)=(5,0)$. ✓

(d) The forbidden case: rulers on one line. If $b_1,b_2$ point the same (or opposite) way, they are not a basis — they can't reach off-line arrows, and the matrix $B$ has no inverse. The whole construction refuses to start. That's why bases must be independent, and why every genuine change of basis matrix is invertible.

PICTURE. Left panel — two independent red rulers spanning the plane (allowed). Right panel — two collinear red rulers collapsing to a line, the shaded "unreachable" region (forbidden).

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The one-picture summary

Everything on this page is one sentence made visual: the columns of $P_{C\leftarrow B}$ are the old rulers $b_j$ redrawn on the new grid; multiplying then rebuilds the arrow with the same weights. The figure compresses Steps 3–6 into a single flow.

Recall Feynman retelling — explain the whole walkthrough to a friend

A treasure sits in a park; it never moves. Friend $B$ gives directions in her steps ($b_1,b_2$); friend $C$ in his ($c_1^{\prime },c_2^{\prime }$). I only have $B$'s directions ($c_1$ of her first step, $c_2$ of her second). To convert, I don't re-find the treasure — I just ask $C$ once: "how do you say each of $B$'s two steps in your own steps?" Those two answers are my two dictionary columns. Then I reuse the same counts $c_1,c_2$ — because doubling the walk doubles every step and adding walks adds steps — and stack the dictionary columns into a box. Feeding my $B$-counts to that box spits out $C$-counts pointing at the identical treasure. If I ever want to go the other way I flip the box over (its inverse). And through the standard axes the boxes are free: $B$ to stack, $C^{-1}$ to peel — so $P_{C\leftarrow B}=C^{-1}B$.

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Connections

• Basis and Dimension — why two independent rulers exactly cover the plane.
• Coordinate Vectors — the columns we translate.
• Invertible Matrices — case (d): a real change of basis is always invertible.
• Similar Matrices — $A^{\prime }=P^{-1}AP$ reuses this machine on a map.
• Eigenvectors and Diagonalization — change to the eigenbasis.
• Linear Transformations — the linearity of $[\cdot {]}_C$ used in Step 5.