Change of basis matrix

1. Setup — What are we even translating?

WHAT is a change of basis matrix? Given two bases $B$ (old) and $C$ (new), it's the matrix $P_{C\leftarrow B}$ such that $[v]_C = P_{C\leftarrow B}\,[v]_B.$ It takes $B$ -coordinates in and spits $C$ -coordinates out.

2. Derivation from first principles

We want a matrix that, fed $[v]_B$ , returns $[v]_C$ . Let's build it.

Step 1 — Write $v$ in the old basis. By definition, $v = c_1 b_1 + \cdots + c_n b_n.$ Why this step? This is the only thing we know: $[v]_B = (c_1,\dots,c_n)$ .

Step 2 — Express each old basis vector in the new basis. Each $b_j$ is itself a vector, so it has $C$ -coordinates: $[b_j]_C = \begin{bmatrix} p_{1j} \\ \vdots \\ p_{nj}\end{bmatrix}, \quad\text{i.e. } b_j = \sum_i p_{ij}\, c_i.$ Why this step? Translation is "term by term": if I know how to say each building block $b_j$ in the new language, I can say any combination.

Step 3 — Take $C$ -coordinates of the whole sum. Coordinates are linear, so $[v]_C = \Big[\sum_j c_j b_j\Big]_C = \sum_j c_j\,[b_j]_C.$ Why this step? The coordinate map $v \mapsto [v]_C$ is a linear map (uniqueness of expansion guarantees this), so it distributes over sums and scalars.

Step 4 — Recognize the matrix–vector product. Stack $[b_j]_C$ as columns: $[v]_C = \underbrace{\Big[\,[b_1]_C \;\big|\; [b_2]_C \;\big|\; \cdots \;\big|\; [b_n]_C\,\Big]}_{P_{C\leftarrow B}} \begin{bmatrix} c_1 \\ \vdots \\ c_n\end{bmatrix} = P_{C\leftarrow B}\,[v]_B. \qquad\blacksquare$

3. The Standard-Basis Shortcut

In $\mathbb{R}^n$ with standard basis $E=\{e_1,\dots,e_n\}$ :

$P_{E\leftarrow B} = \big[\,b_1 \mid b_2 \mid \cdots \mid b_n\,\big]$ — just put the basis vectors as columns! (Because $[b_j]_E = b_j$ literally.) Call this $B$ (abuse of notation: the matrix of basis vectors).
Going the other way: $P_{B\leftarrow E} = B^{-1}$ , since translating then translating back is the identity: $P_{B\leftarrow E}\,P_{E\leftarrow B} = I.$

4. Worked Examples

Worked example Example 1 — Two bases of

$\mathbb{R}^2$ Let $B=\{(1,0),(1,1)\}$ and $C=\{(1,1),(1,-1)\}$ . Find $P_{C\leftarrow B}$ and convert $v$ with $[v]_B=(3,2)^T$ .

Step 1: Build $B$ , $C$ matrices. $B=\begin{bmatrix}1&1\\0&1\end{bmatrix},\quad C=\begin{bmatrix}1&1\\1&-1\end{bmatrix}.$ Why? Columns are the basis vectors in standard coords.

Step 2: $P_{C\leftarrow B}=C^{-1}B$ . $C^{-1}=\frac{1}{-2}\begin{bmatrix}-1&-1\\-1&1\end{bmatrix}=\begin{bmatrix}\tfrac12&\tfrac12\\\tfrac12&-\tfrac12\end{bmatrix}.$ Why $C^{-1}B$ ? Go $B$ -coords $\to$ standard ( $\times B$ ), then standard $\to C$ -coords ( $\times C^{-1}$ ). $P_{C\leftarrow B}=\begin{bmatrix}\tfrac12&\tfrac12\\\tfrac12&-\tfrac12\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}\tfrac12&1\\\tfrac12&0\end{bmatrix}.$

Step 3: Apply to $[v]_B=(3,2)^T$ . $[v]_C=\begin{bmatrix}\tfrac12&1\\\tfrac12&0\end{bmatrix}\begin{bmatrix}3\\2\end{bmatrix}=\begin{bmatrix}3.5\\1.5\end{bmatrix}.$ Check (Forecast-then-Verify): Actual vector $v = 3(1,0)+2(1,1)=(5,2)$ . Then $3.5(1,1)+1.5(1,-1)=(5,2)$ . ✓

5. Common Mistakes (Steel-manned)

6. Active Recall

Recall What is the

$j$ -th column of $P_{C\leftarrow B}$ ? The old basis vector $b_j$ written in $C$ -coordinates: $[b_j]_C$ .

Recall If

$B,C$ are matrices of basis vectors (standard coords), what is $P_{C\leftarrow B}$ ? $P_{C\leftarrow B}=C^{-1}B$ .

Recall How do you reverse a change of basis?

Take the inverse: $P_{B\leftarrow C}=(P_{C\leftarrow B})^{-1}$ .

Recall (Feynman, explain to a 12-year-old)

Imagine a treasure sits in a park. One friend gives directions using "north/east steps," another using "toward-the-oak / toward-the-pond steps." The treasure never moves! The change of basis matrix is a little translator card: you tell it the steps in one friend's language, it tells you the same spot in the other friend's language. To swap which friend is talking, you flip the card over (that's the inverse).

7. Connections

Basis and Dimension — coordinates require a basis to exist.
Coordinate Vectors — the columns we manipulate.
Invertible Matrices — a change of basis matrix is always invertible (bases are independent).
Similar Matrices — $A' = P^{-1}AP$ uses change of basis to rewrite a linear map in a new basis.
Eigenvectors and Diagonalization — diagonalizing = changing to the eigenbasis.
Linear Transformations — active (vectors move) vs passive (description changes) views.

What does

P_{C\leftarrow B}

do to a coordinate vector?

Takes

B

-coordinates as input and returns

C

-coordinates:

[v]_C = P_{C\leftarrow B}[v]_B

What is column

j

of the change of basis matrix

P_{C\leftarrow B}

[b_j]_C

, the old basis vector

b_j

expressed in the new basis

C

B,C

hold basis vectors as columns (in standard coords),

P_{C\leftarrow B}=

C^{-1}B

How to translate standard coords into basis

B

coords?

Multiply by

B^{-1}

[v]_B=B^{-1}v

How to translate basis

B

coords back to standard?

Multiply by

B

[v]_E=B[v]_B

(columns of

B

are basis vectors in standard coords).

How do you reverse

P_{C\leftarrow B}

Use its inverse

P_{B\leftarrow C}=(P_{C\leftarrow B})^{-1}

Does a change of basis move the vector?

No — the vector is fixed; only its coordinate description changes (passive view).

Why is every change of basis matrix invertible?

Its columns are coordinates of a basis (independent, spanning), so it's square with full rank.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, ek vector ek fixed arrow hai — wo apni jagah pe baitha hai, hilta nahi. Lekin usko numbers me likhne ke liye humein ek basis (reference directions ka set) choose karna padta hai. Jab basis badalte ho, vector wahi rehta hai, sirf uske coordinates (numbers) badalte hain. Yahi cheez change of basis matrix handle karti hai — wo ek dictionary hai jo ek basis ki language se doosri basis ki language me coordinates translate karta hai.

Formula yaad rakho aise: $P_{C\leftarrow B}$ ka matlab hai "input $B$ -coords, output $C$ -coords." Iski har column kya hai? Old basis vector $b_j$ ko new basis $C$ ke coordinates me likha hua. Standard basis ke saath shortcut: agar $B$ aur $C$ matrices hain jinke columns basis vectors hain, to $P_{C\leftarrow B}=C^{-1}B$ . Mnemonic "CBC" — C-inverse Bites B.

Sabse common galti: log sochte hain columns naye basis ke vectors hain — nahi! Columns purane $b_j$ hote hain, lekin $C$ -coordinates me. Doosri galti: $P$ aur $P^{-1}$ ka confusion. Arrow ko subscript ki tarah padho: $C \leftarrow B$ matlab $B$ andar jaata hai, $C$ bahar aata hai. Ulta karna ho to inverse lo. Aur teesri galti — yeh maan lena ki vector "move" ho gaya; nahi yaar, vector fixed hai, sirf description badli (passive view).

Yeh topic isliye important hai kyunki aage diagonalization aur similar matrices ( $A'=P^{-1}AP$ ) sab change of basis pe khade hain. Eigenvectors me hum basically aise basis me jaate hain jahan matrix simple (diagonal) dikhe. To yeh foundation solid karlo!