A vector v \mathbf{v} v before you pick any basis. A coordinate vector is the list of numbers you write down to describe that same arrow once you've chosen a ruler-system (a basis). Change the basis → the arrow stays put, but the numbers change. Change of basis is the dictionary that translates one list of numbers into another for the same arrow.
Definition Coordinate vectorLet B = { b 1 , … , b n } \mathcal{B} = \{\mathbf{b}_1, \dots, \mathbf{b}_n\} B = { b 1 , … , b n } V V V v ∈ V \mathbf{v}\in V v ∈ V unique expansion
v = c 1 b 1 + c 2 b 2 + ⋯ + c n b n . \mathbf{v} = c_1\mathbf{b}_1 + c_2\mathbf{b}_2 + \cdots + c_n\mathbf{b}_n. v = c 1 b 1 + c 2 b 2 + ⋯ + c n b n . coordinate vector of v \mathbf{v} v B \mathcal{B} B [ v ] B = [ c 1 ⋮ c n ] . [\mathbf{v}]_{\mathcal{B}} = \begin{bmatrix} c_1 \\ \vdots \\ c_n\end{bmatrix}. [ v ] B = c 1 ⋮ c n .
WHY is it unique? Because B \mathcal{B} B linearly independent . If two expansions existed, subtracting them gives a non-trivial combination equal to 0 \mathbf{0} 0 whole reason coordinates are well-defined.
You know [ v ] B [\mathbf{v}]_{\mathcal{B}} [ v ] B [ v ] C [\mathbf{v}]_{\mathcal{C}} [ v ] C C \mathcal{C} C P P P [ v ] C = P C ← B [ v ] B . [\mathbf{v}]_{\mathcal{C}} = \underset{\mathcal{C}\leftarrow\mathcal{B}}{P}\;[\mathbf{v}]_{\mathcal{B}}. [ v ] C = C ← B P [ v ] B .
Start from what [ v ] B [\mathbf{v}]_{\mathcal{B}} [ v ] B means :
v = c 1 b 1 + ⋯ + c n b n . \mathbf{v} = c_1\mathbf{b}_1 + \cdots + c_n\mathbf{b}_n. v = c 1 b 1 + ⋯ + c n b n .
Now take coordinates of both sides relative to C \mathcal{C} C [ ⋅ ] C [\,\cdot\,]_{\mathcal{C}} [ ⋅ ] C linear (proved below), so
[ v ] C = c 1 [ b 1 ] C + ⋯ + c n [ b n ] C . [\mathbf{v}]_{\mathcal{C}} = c_1[\mathbf{b}_1]_{\mathcal{C}} + \cdots + c_n[\mathbf{b}_n]_{\mathcal{C}}. [ v ] C = c 1 [ b 1 ] C + ⋯ + c n [ b n ] C .
The right side is exactly a matrix times [ c 1 ⋮ c n ] \begin{bmatrix}c_1\\\vdots\\c_n\end{bmatrix} c 1 ⋮ c n [ b j ] C [\mathbf{b}_j]_{\mathcal{C}} [ b j ] C
Why coordinate maps are linear: writing u = ∑ a i b i \mathbf{u}=\sum a_i\mathbf{b}_i u = ∑ a i b i w = ∑ d i b i \mathbf{w}=\sum d_i\mathbf{b}_i w = ∑ d i b i u + α w = ∑ ( a i + α d i ) b i \mathbf{u}+\alpha\mathbf{w}=\sum(a_i+\alpha d_i)\mathbf{b}_i u + α w = ∑ ( a i + α d i ) b i [ u + α w ] B = [ u ] B + α [ w ] B [\mathbf{u}+\alpha\mathbf{w}]_{\mathcal{B}} = [\mathbf{u}]_{\mathcal{B}}+\alpha[\mathbf{w}]_{\mathcal{B}} [ u + α w ] B = [ u ] B + α [ w ] B
Swapping roles gives P B ← C \underset{\mathcal{B}\leftarrow\mathcal{C}}{P} B ← C P P B ← C = ( P C ← B ) − 1 . \underset{\mathcal{B}\leftarrow\mathcal{C}}{P} = \Big(\underset{\mathcal{C}\leftarrow\mathcal{B}}{P}\Big)^{-1}. B ← C P = ( C ← B P ) − 1 . Why? Translating to C \mathcal{C} C B \mathcal{B} B I I I
R n \mathbb{R}^n R n E \mathcal{E} E B = [ b 1 ⋯ b n ] B=[\mathbf{b}_1\cdots\mathbf{b}_n] B = [ b 1 ⋯ b n ] P E ← B = B , P B ← E = B − 1 . \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}=B,\qquad \underset{\mathcal{B}\leftarrow\mathcal{E}}{P}=B^{-1}. E ← B P = B , B ← E P = B − 1 . B B B B \mathcal{B} B B − 1 B^{-1} B − 1 P C ← B = C − 1 B . \underset{\mathcal{C}\leftarrow\mathcal{B}}{P}=C^{-1}B. C ← B P = C − 1 B .
Worked example Example 1 — basicR 2 \mathbb{R}^2 R 2 B = { b 1 , b 2 } \mathcal{B}=\{\mathbf{b}_1,\mathbf{b}_2\} B = { b 1 , b 2 } b 1 = [ 1 1 ] , b 2 = [ 1 − 1 ] \mathbf{b}_1=\begin{bmatrix}1\\1\end{bmatrix},\ \mathbf{b}_2=\begin{bmatrix}1\\-1\end{bmatrix} b 1 = [ 1 1 ] , b 2 = [ 1 − 1 ] [ v ] B [\mathbf{v}]_{\mathcal{B}} [ v ] B v = [ 4 2 ] \mathbf{v}=\begin{bmatrix}4\\2\end{bmatrix} v = [ 4 2 ]
We need c 1 b 1 + c 2 b 2 = v c_1\mathbf{b}_1+c_2\mathbf{b}_2=\mathbf{v} c 1 b 1 + c 2 b 2 = v B [ v ] B = v B\,[\mathbf{v}]_{\mathcal{B}}=\mathbf{v} B [ v ] B = v [ v ] B = B − 1 v [\mathbf{v}]_{\mathcal{B}}=B^{-1}\mathbf{v} [ v ] B = B − 1 v Why this step? In the standard basis the coordinates of v \mathbf{v} v B − 1 B^{-1} B − 1 B \mathcal{B} B B = [ 1 1 1 − 1 ] , B − 1 = − 1 2 [ − 1 − 1 − 1 1 ] = 1 2 [ 1 1 1 − 1 ] . B=\begin{bmatrix}1&1\\1&-1\end{bmatrix},\quad B^{-1}=-\tfrac12\begin{bmatrix}-1&-1\\-1&1\end{bmatrix}=\tfrac12\begin{bmatrix}1&1\\1&-1\end{bmatrix}. B = [ 1 1 1 − 1 ] , B − 1 = − 2 1 [ − 1 − 1 − 1 1 ] = 2 1 [ 1 1 1 − 1 ] . [ v ] B = 1 2 [ 1 1 1 − 1 ] [ 4 2 ] = [ 3 1 ] . [\mathbf{v}]_{\mathcal{B}}=\tfrac12\begin{bmatrix}1&1\\1&-1\end{bmatrix}\begin{bmatrix}4\\2\end{bmatrix}=\begin{bmatrix}3\\1\end{bmatrix}. [ v ] B = 2 1 [ 1 1 1 − 1 ] [ 4 2 ] = [ 3 1 ] . 3 b 1 + 1 b 2 = [ 3 3 ] + [ 1 − 1 ] = [ 4 2 ] . 3\mathbf{b}_1+1\mathbf{b}_2=\begin{bmatrix}3\\3\end{bmatrix}+\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}4\\2\end{bmatrix}. 3 b 1 + 1 b 2 = [ 3 3 ] + [ 1 − 1 ] = [ 4 2 ] .
Worked example Example 2 — between two non-standard basesB \mathcal{B} B C = { c 1 , c 2 } \mathcal{C}=\{\mathbf{c}_1,\mathbf{c}_2\} C = { c 1 , c 2 } c 1 = [ 1 0 ] , c 2 = [ 1 2 ] \mathbf{c}_1=\begin{bmatrix}1\\0\end{bmatrix},\ \mathbf{c}_2=\begin{bmatrix}1\\2\end{bmatrix} c 1 = [ 1 0 ] , c 2 = [ 1 2 ] P C ← B \underset{\mathcal{C}\leftarrow\mathcal{B}}{P} C ← B P
Use P = C − 1 B P=C^{-1}B P = C − 1 B Why? Columns of P P P [ b j ] C = C − 1 b j [\mathbf{b}_j]_{\mathcal{C}}=C^{-1}\mathbf{b}_j [ b j ] C = C − 1 b j C − 1 B C^{-1}B C − 1 B C = [ 1 1 0 2 ] , C − 1 = [ 1 − 1 2 0 1 2 ] . C=\begin{bmatrix}1&1\\0&2\end{bmatrix},\quad C^{-1}=\begin{bmatrix}1&-\tfrac12\\0&\tfrac12\end{bmatrix}. C = [ 1 0 1 2 ] , C − 1 = [ 1 0 − 2 1 2 1 ] . P = C − 1 B = [ 1 − 1 2 0 1 2 ] [ 1 1 1 − 1 ] = [ 1 2 3 2 1 2 − 1 2 ] . P=C^{-1}B=\begin{bmatrix}1&-\tfrac12\\0&\tfrac12\end{bmatrix}\begin{bmatrix}1&1\\1&-1\end{bmatrix}=\begin{bmatrix}\tfrac12&\tfrac32\\\tfrac12&-\tfrac12\end{bmatrix}. P = C − 1 B = [ 1 0 − 2 1 2 1 ] [ 1 1 1 − 1 ] = [ 2 1 2 1 2 3 − 2 1 ] . = [ b 1 ] C =[\mathbf{b}_1]_{\mathcal{C}} = [ b 1 ] C 1 2 c 1 + 1 2 c 2 = 1 2 [ 1 0 ] + 1 2 [ 1 2 ] = [ 1 1 ] = b 1 \tfrac12\mathbf{c}_1+\tfrac12\mathbf{c}_2=\tfrac12\begin{bmatrix}1\\0\end{bmatrix}+\tfrac12\begin{bmatrix}1\\2\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}=\mathbf{b}_1 2 1 c 1 + 2 1 c 2 = 2 1 [ 1 0 ] + 2 1 [ 1 2 ] = [ 1 1 ] = b 1
Worked example Example 3 — Forecast-then-Verify (polynomials)V = P 2 V=P_2 V = P 2 B = { 1 , t , t 2 } \mathcal{B}=\{1,t,t^2\} B = { 1 , t , t 2 } C = { 1 , 1 + t , 1 + t + t 2 } \mathcal{C}=\{1,\,1+t,\,1+t+t^2\} C = { 1 , 1 + t , 1 + t + t 2 } p ( t ) = 2 + 3 t + t 2 p(t)=2+3t+t^2 p ( t ) = 2 + 3 t + t 2 Forecast: C \mathcal{C} C a ( 1 ) + b ( 1 + t ) + c ( 1 + t + t 2 ) = 2 + 3 t + t 2 a(1)+b(1+t)+c(1+t+t^2)=2+3t+t^2 a ( 1 ) + b ( 1 + t ) + c ( 1 + t + t 2 ) = 2 + 3 t + t 2
t 2 t^2 t 2 c = 1 c=1 c = 1 Why? Only c 3 \mathbf{c}_3 c 3 t 2 t^2 t 2 t t t b + c = 3 ⇒ b = 2 b+c=3\Rightarrow b=2 b + c = 3 ⇒ b = 2 const: a + b + c = 2 ⇒ a = − 1 a+b+c=2\Rightarrow a=-1 a + b + c = 2 ⇒ a = − 1 [ p ] C = [ − 1 2 1 ] [p]_{\mathcal{C}}=\begin{bmatrix}-1\\2\\1\end{bmatrix} [ p ] C = − 1 2 1 negative — coordinates needn't be positive.
P P P new basis vectors c j \mathbf{c}_j c j Why it feels right: you're going to C \mathcal{C} C C \mathcal{C} C The fix: the columns are the old vectors b j \mathbf{b}_j b j new coordinates: [ b j ] C [\mathbf{b}_j]_{\mathcal{C}} [ b j ] C consumes B \mathcal{B} B B \mathcal{B} B
B − 1 C B^{-1}C B − 1 C C − 1 B C^{-1}B C − 1 B Why it feels right: both letters appear; order seems arbitrary. The fix: check units. P C ← B = C − 1 B \underset{\mathcal{C}\leftarrow\mathcal{B}}{P}=C^{-1}B C ← B P = C − 1 B B B B B \mathcal{B} B C − 1 C^{-1} C − 1 C \mathcal{C} C
Common mistake Thinking the vector itself changes.Why it feels right: the numbers visibly change. The fix: only the description changes. v \mathbf{v} v [ v ] C = P [ v ] B [\mathbf{v}]_{\mathcal{C}}=P[\mathbf{v}]_{\mathcal{B}} [ v ] C = P [ v ] B
Recall Quick self-test (cover the answers)
Q: What are the columns of P C ← B \underset{\mathcal{C}\leftarrow\mathcal{B}}{P} C ← B P [ b j ] C [\mathbf{b}_j]_{\mathcal{C}} [ b j ] C
Q: In R n \mathbb{R}^n R n B , C B,C B , C P C ← B \underset{\mathcal{C}\leftarrow\mathcal{B}}{P} C ← B P C − 1 B C^{-1}B C − 1 B
Q: Relation between P C ← B \underset{\mathcal{C}\leftarrow\mathcal{B}}{P} C ← B P P B ← C \underset{\mathcal{B}\leftarrow\mathcal{C}}{P} B ← C P
Q: Why is [ v ] B [\mathbf{v}]_{\mathcal{B}} [ v ] B
What is the coordinate vector [ v ] B [\mathbf{v}]_{\mathcal{B}} [ v ] B The unique column of scalars
c i c_i c i such that
v = ∑ c i b i \mathbf{v}=\sum c_i\mathbf{b}_i v = ∑ c i b i .
Why are coordinates unique? Because a basis is linearly independent, so the expansion is unique.
What are the columns of the change-of-basis matrix P C ← B P_{\mathcal{C}\leftarrow\mathcal{B}} P C ← B The old basis vectors expressed in the new basis:
[ b j ] C [\mathbf{b}_j]_{\mathcal{C}} [ b j ] C .
For R n \mathbb{R}^n R n B , C B,C B , C P C ← B = ? P_{\mathcal{C}\leftarrow\mathcal{B}}=? P C ← B = ? How do P C ← B P_{\mathcal{C}\leftarrow\mathcal{B}} P C ← B P B ← C P_{\mathcal{B}\leftarrow\mathcal{C}} P B ← C They are inverses of each other.
Does changing basis change the vector? No — only its coordinate description changes; the vector is invariant.
Why is the coordinate map linear? Adding/scaling vectors adds/scales their
B \mathcal{B} B -coefficients termwise.
Recall Feynman: explain to a 12-year-oldImagine the same toy on a treasure map. One friend measures from the door, another from the window. The toy hasn't moved, but the two friends write down different "steps left, steps up" numbers. Change of basis is the little rulebook that turns one friend's numbers into the other's — just translating directions, never moving the toy.
Mnemonic Remember the order"New columns of Old, then C-inverse-B."
Columns = old vectors b j \mathbf{b}_j b j new clothes. And C − 1 B C^{-1}B C − 1 B C − 1 C^{-1} C − 1 C \mathcal{C} C B B B B \mathcal{B} B
Basis and dimension — coordinates exist only once a basis is fixed.Linear independence — guarantees coordinate uniqueness.Invertible matrices — B , C B,C B , C Similar matrices and diagonalization — A ′ = P − 1 A P A'=P^{-1}AP A ′ = P − 1 A P operators .Linear transformations and their matrices — coordinate maps make abstract maps concrete.
old vectors in new coords
Columns are b_j in C coords
Standard basis shortcut P equals B
Intuition Hinglish mein samjhoDekho, ek vector v \mathbf{v} v pehle . Jab hum koi basis B \mathcal{B} B coordinate vector [ v ] B [\mathbf{v}]_{\mathcal{B}} [ v ] B linearly independent hota hai — do alag descriptions possible hi nahi.
Ab change of basis ka matlab: arrow wahi ka wahi rehta hai, sirf numbers badalte hain jab basis badalta hai. Translation matrix P C ← B \underset{\mathcal{C}\leftarrow\mathcal{B}}{P} C ← B P b j \mathbf{b}_j b j [ b j ] C [\mathbf{b}_j]_{\mathcal{C}} [ b j ] C khaata hai).
R n \mathbb{R}^n R n B B B C C C P C ← B = C − 1 B \underset{\mathcal{C}\leftarrow\mathcal{B}}{P}=C^{-1}B C ← B P = C − 1 B B B B B \mathcal{B} B C − 1 C^{-1} C − 1 C \mathcal{C} C I I I
Yeh topic kyun matter karta hai? Kyunki diagonalization (A ′ = P − 1 A P A'=P^{-1}AP A ′ = P − 1 A P
