Change of Basis¶

Definition¶

Let $\mathcal{F}=(\boldsymbol{f}_1,\ldots,\boldsymbol{f}_n)$ be an ordered basis of $\mathbb{R}^n$ , the new basis, and let $\boldsymbol{e}_1,\ldots,\boldsymbol{e}_n$ be the standard basis. The transformation matrix of $\mathcal{F}$ is the $n\times n$ matrix whose columns are the new basis vectors,

F=\begin{pmatrix}\boldsymbol{f}_1&\cdots&\boldsymbol{f}_n\end{pmatrix}.

$F$ transforms the standard basis into the new basis, $F\boldsymbol{e}_i=\boldsymbol{f}_i$ $f_{i}$ , and $F^{-1}$ transforms it back. It is invertible because its columns form a basis.

The coordinate vector of $\boldsymbol{x}$ with respect to $\mathcal{F}$ is the vector $\boldsymbol{y}=[\boldsymbol{x}]_{\mathcal{F}}$ of the unique scalars with $\boldsymbol{x}=y_1\boldsymbol{f}_1+\cdots+y_n\boldsymbol{f}_n$ $y_{n} f_{n}$ . In matrix form,

\boldsymbol{x}=F\boldsymbol{y},\qquad \boldsymbol{y}=F^{-1}\boldsymbol{x}.

Interpretation¶

The vector $\boldsymbol{x}$ does not move; only its description changes. The columns of $F$ say where the new basis vectors are, so multiplying new coordinates by $F$ rebuilds the vector in standard coordinates, and $F^{-1}$ finds the new coordinates of a vector given in standard coordinates.

The vector x = (3,1)^T drawn over the grid of the new basis f_1 = (1,1)^T, f_2 = (1,-1)^T, reached both as 3e_1 + e_2 and as 2f_1 + f_2, so its coordinates in the new basis are (2,1)^T.

The figure shows the first example. The green arrow $\boldsymbol{x}=(3,1)^{\top}$ is one fixed vector. Along the standard axes it is $3\boldsymbol{e}_1+\boldsymbol{e}_2$ $e_{2}$ ; along the tilted grid of $\mathcal{F}$ it is $2\boldsymbol{f}_1+\boldsymbol{f}_2$ $f_{2}$ . Changing basis only changes which grid the vector is read on, so its coordinates change from $(3,1)^{\top}$ to $[\boldsymbol{x}]_{\mathcal{F}}=F^{-1}\boldsymbol{x}=(2,1)^{\top}$ $(2, 1)^{⊤}$ .

Drag $\boldsymbol{x}$ , $\boldsymbol{f}_1$ or $\boldsymbol{f}_2$ : the arrow stays the same while its description changes, from $(3,1)^{\top}$ in the standard basis to $[\boldsymbol{x}]_{\mathcal{F}}=F^{-1}\boldsymbol{x}=(2,1)^{\top}$ along the grid of $\mathcal{F}$ . Switch on the operator $A$ to follow $A\boldsymbol{x}$ as well.

The direction is the usual source of errors. $F$ takes the standard basis vectors to the new ones, but on coordinates it works the other way round: it turns new coordinates $\boldsymbol{y}$ into standard coordinates $\boldsymbol{x}=F\boldsymbol{y}$ $F y$ . Going from standard to new coordinates needs the inverse $F^{-1}$ . Other textbooks call such a matrix a change-of-basis or transition matrix and may attach it to either direction, so check which direction is meant.

Changing basis also changes the matrix representation of a linear map. For an operator on $\mathbb{R}^n$ the result is a similarity transformation; choosing an eigenbasis makes the matrix diagonal.

Key Results¶

Every invertible $n\times n$ matrix is the transformation matrix of the basis formed by its columns.
If $A$ is the matrix of a linear operator in standard coordinates, its matrix in the new basis is $[A]_{\mathcal{F}}=F^{-1}AF$ $F^{- 1} A F$ : translate new coordinates to standard ones with $F$ , apply $A$ , translate back with $F^{-1}$ .
Between two new bases $\mathcal{F}$ and $\mathcal{G}$ with transformation matrices $F$ and $G$ : $[\boldsymbol{x}]_{\mathcal{G}}=G^{-1}F[\boldsymbol{x}]_{\mathcal{F}}$ $G^{- 1} F [x]_{F}$ .
If $\mathcal{F}$ is orthonormal, then $F^{-1}=F^{\top}$ and $y_i=\boldsymbol{f}_i^{\top}\boldsymbol{x}$ $f_{i}^{⊤} x$ .

Examples¶

Coordinates. In $\mathbb{R}^2$ let $\boldsymbol{f}_1=(1,1)^{\top}$ and $\boldsymbol{f}_2=(1,-1)^{\top}$ $(1, - 1)^{⊤}$ , so

F=\begin{pmatrix}1&1\\1&-1\end{pmatrix},\qquad F^{-1}=\tfrac{1}{2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}.

For $\boldsymbol{x}=(3,1)^{\top}$ $(3, 1)^{⊤}$ ,

[\boldsymbol{x}]_{\mathcal{F}}=F^{-1}\boldsymbol{x}=\tfrac{1}{2}(4,2)^{\top}=(2,1)^{\top},

and indeed $2(1,1)^{\top}+1(1,-1)^{\top}=(3,1)^{\top}$ $(3, 1)^{⊤}$ .

An operator. Let $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ in standard coordinates. Then $AF=\begin{pmatrix}3&1\\3&-1\end{pmatrix}$ and

[A]_{\mathcal{F}}=F^{-1}AF =\tfrac{1}{2}\begin{pmatrix}6&0\\0&2\end{pmatrix}=\begin{pmatrix}3&0\\0&1\end{pmatrix}.

In the new basis the map stretches the first coordinate by $3$ and leaves the second alone: $\mathcal{F}$ is an eigenbasis, and this is a diagonalization of $A$ .

Prerequisites¶

Sources¶

Math V: Advanced Linear Algebra · Week 6

Aliases: basis transformation, change of coordinates, transformation matrix, basis change Keywords: coordinate vector, change-of-basis matrix, change-of-coordinates matrix, x = Fy, F^{-1}x, F^{-1}AF Symbols: real numbers

Last updated: 22 October 2026 19:04