Throughout, E={e1,e2} is the standard basis of R2, meaning e1=(1,0) and e2=(0,1) — the plain "one step right, one step up" directions. When we write a matrix like B=[b1∣b2] we mean: stack the basis vectors b1,b2 as the columns, each written in standard coordinates. That single convention drives everything.
How to read this figure (we will keep pointing back to it): the orange box on the left holds a coordinate list in B-language, the plum box on the right holds the same information in C-language, and the teal box at the bottom is the standard-E language everyone shares. The small circle in the centre is the one physical arrow that never moves. The three labelled arrows are exactly the translator matrices ×B (orange, B→E), ×C−1 (teal, E→C), and their composite PC←B=C−1B (the top edge). Every exercise below is a walk along one or more of these edges — we will name which edge each time.
WHAT the subscript says:PC←B eats B-coordinates (they sit on the right, matching the B on the right of the arrow) and returns C-coordinates. On the figure, this is the top orange-to-plum edge — we start in the orange box and land in the plum box.
[v]C=PC←B[v]B=[2103][11]=[24].
So the output is [v]C=(2,4)T — the new (C) coordinates. The vector itself never moved (it is the fixed centre circle in the figure); only its description did.
Recall Solution L1.2
WHY a single column tells us about b1: matrix–vector multiplication reads off columns. Feeding the matrix the column (1,0)T selects column 1, feeding (0,1)T selects column 2 — that is just how the product is defined. So to expose what column 1means, feed the input that isolates it.
The right input is [b1]B=(1,0)T: this is the first old basis vector described in its own basis B (one step along b1, zero along b2). Since PC←B turns any B-coordinate list into its C-coordinate list, the output must be b1 spoken in C:
PC←B[10]=[21]=[b1]C.
So column 1 is [b1]C — the old basis vector b1 written in the new basis C. Column j of PC←B is always [bj]C.
WHY no inverse here: the columns of B are already written in standard coordinates, so "decode B→ standard" is just multiply by B. This is the orange edge (×B, B→E) of the figure.
[v]E=B[v]B=[31−12][21]=[54].Verify by rebuilding the arrow:2(3,1)+1(−1,2)=(6−1,2+2)=(5,4). ✓
Recall Solution L2.2
WHY an inverse now: we want to go standard →B, which is the orange edge run backwards in the figure, so we need B−1.
For B=[31−12], detB=3⋅2−(−1)⋅1=7, so
B−1=71[2−113].[w]B=B−1w=71[2−113][54]=71[147]=[21].
Consistent with L2.1 — as it must be, since we just reversed it. ✓
Recall Solution L2.3
WHY the formula C−1B: go B-coords → standard (orange edge, multiply by B), then standard →C-coords (teal edge, multiply by C−1). Chaining the two edges gives the top edge PC←B=C−1B.
B=[1011],C=[2001],C−1=[21001].PC←B=C−1B=[21001][1011]=[210211].
This is a full trip around the figure: orange box → teal box (×B) → plum box (×C−1).
B=[1011],C=[111−1],detC=−2,C−1=[212121−21].PC←B=C−1B=[212121−21][1011]=[212110].[v]C=PC←B[v]B=[212110][42]=[42].Verify (through the teal box): the actual arrow is v=4(1,0)+2(1,1)=(6,2). Rebuild from C: 4(1,1)+2(1,−1)=(6,2). ✓ Same arrow, two descriptions.
Recall Solution L3.2
WHY the inverse: to translate C→B we run the top edge of the figure backwards — flip the translator card: PB←C=(PC←B)−1.
detPC←B=21⋅0−1⋅21=−21, so
PB←C=(PC←B)−1=−211[0−21−121]=[012−1].[u]B=PB←C[u]C=[012−1][12]=[4−1].Verify: arrow from C: u=1(1,1)+2(1,−1)=(3,−1). Rebuild from B: 4(1,0)−1(1,1)=(3,−1). ✓
WHY this formula: to apply T while working in B-coordinates, decode B→ standard (×B, the orange edge), apply A, then re-encode standard →B (×B−1, orange edge backwards). So A′=B−1AB — the Similar Matrices relation.
B=[1011],B−1=[10−11].AB=[2013][1011]=[2033].A′=B−1(AB)=[10−11][2033]=[2003].
The map looks diagonal in this basis — because (1,0) and (1,1) happen to be eigenvectors of A (eigenvalues 2 and 3).
Recall Solution L4.2
WHY change of basis: diagonalizing = choosing the eigenbasis as your new basis; then T just stretches along each axis.
Eigenvalues: det(A−λI)=(2−λ)(3−λ)=0⇒λ=2,3.
λ=3: (A−3I)x=0⇒[−1010]x=0⇒x1=x2, eigenvector (1,1).
P=[1011],P−1=[10−11],P−1AP=[2003]=[λ100λ2].
Same answer as L4.1 — because "diagonalize" is "change to the eigenbasis."
Strategy: route everything through the standard basis — orange edge then teal edge. PC←B=C−1B.
B=[2111],C=[1012],detC=2,C−1=[10−2121].PC←B=C−1B=[10−2121][2111]=[23212121].[v]C=PC←B[v]B=[23212121][13]=[32].Verify: arrow v=1(2,1)+3(1,1)=(5,4). Rebuild from C: 3(1,0)+2(1,2)=(5,4). ✓
Recall Solution L5.2
[v]C=PC←B[00]=[00].WHY it must be zero in every basis: the zero vector is the one arrow of length 0; it has no direction to describe, so every basis assigns it the coordinate list (0,…,0). Any change of basis matrix maps 0↦0 (a linear map always fixes the origin — the centre circle of the figure sitting exactly at the origin). This is the boundary case that confirms your matrix is behaving.
Recall Solution L5.3
C=[100−1]. Since C−1=C here ((−1)−1=−1),
[w]C=C−1w=[100−1][35]=[3−5].What it means: the second C-axis points downward, so to reach a point that is 5 units up, you must take −5 steps along the downward axis. Geometrically C−1reflects the vertical component — a mirror flip across the horizontal axis. Sign handling matters: pick the basis direction backwards and every second coordinate flips sign.
Recall Feynman check — say the whole ladder in one breath
Give it the coordinates in one basis, multiply by C−1B to get the coordinates in another basis; flip the card (B−1C) whenever you want to go back the other way; and to see a linear map in a friendlier basis, sandwich its matrix as P−1AP where P holds the new basis vectors — and if that friendlier basis happens to be the eigenbasis, the sandwich comes out diagonal. Throughout, the arrow at the centre of the figure never budges; only our language for it changes.
