A vector does not change when we rename the measuring rods around it, but its list of coordinates can change completely. The vector $x = 2e_1 + e_2$ in a plane is not a different vector if we measure it with the slanted basis $(e_1+e_2,e_2)$; only the coordinate record changes. Change of basis is the algebra that separates the vector from the coordinates used to describe it.

The problem becomes urgent as soon as a [linear map](/page/Linear%20Map) is present. A map may have a complicated matrix in one basis and a diagonal matrix in another; a quadratic form may contain a cross term in one basis and split into squares after a suitable coordinate choice; a [Lie algebra](/page/Lie%20Algebra) representation may become understandable only after the right basis exposes invariant pieces. Each setting has its own compatibility conditions, but the shared lesson is the same: the mathematical object is basis-independent, while computation happens through coordinates.

[example: The Same Vector in Two Bases] Let $V=\mathbb{R}^2$ with the standard basis $\mathcal E=(e_1,e_2)$, and let $\mathcal B=(b_1,b_2)$ where $b_1=e_1+e_2$ and $b_2=e_2$. For $x=2e_1+e_2$, we compute its $\mathcal B$-coordinates by finding $a_1,a_2\in\mathbb{R}$ such that \begin{align*} x=a_1b_1+a_2b_2. \end{align*} Substituting the definitions of $b_1$ and $b_2$ gives \begin{align*} a_1b_1+a_2b_2=a_1(e_1+e_2)+a_2e_2. \end{align*} Distributing $a_1$ and collecting the $e_2$ terms gives \begin{align*} a_1(e_1+e_2)+a_2e_2=a_1e_1+a_1e_2+a_2e_2=a_1e_1+(a_1+a_2)e_2. \end{align*} Since $x=2e_1+e_2$, equality with $a_1e_1+(a_1+a_2)e_2$ forces equality of the coefficients of the ordered basis $\mathcal E$: \begin{align*} a_1=2,\qquad a_1+a_2=1. \end{align*} Substituting $a_1=2$ into the second equation gives \begin{align*} 2+a_2=1. \end{align*} Subtracting $2$ from both sides gives \begin{align*} a_2=-1. \end{align*} Therefore \begin{align*} x=2b_1-b_2. \end{align*} In standard coordinates, the same vector is $[x]_{\mathcal E}=(2,1)^\top$, while in the basis $\mathcal B$ it is $[x]_{\mathcal B}=(2,-1)^\top$. The vector has not moved; only the coordinate column has changed because the basis vectors used to reconstruct it have changed. [/example]

This example already shows the central tension. Coordinates are indispensable for calculation, but they are not the object itself. Change of basis gives a controlled way to translate between coordinate languages without changing the underlying linear algebra.

## Definition

The page topic is not merely a matrix trick. A change of basis is the act of keeping the [vector space](/page/Vector%20Space) fixed while replacing the coordinate system used to describe its elements. The definition must therefore mention both sides: the old ordered basis whose coordinates we start with, and the new ordered basis whose coordinates we want.

[definition: Change of Basis] Let $V$ be a finite-dimensional vector space over a field $k$. A change of basis on $V$ is a passage from one ordered basis $\mathcal B=(b_1,\ldots,b_n)$ of $V$ to another ordered basis $\mathcal C=(c_1,\ldots,c_n)$ of $V$. [/definition]

This definition names the mathematical situation, but the word "ordered" is doing real work. A basis alone is not enough for coordinates, because coordinates are an ordered list. Swapping two basis vectors swaps two entries in every coordinate column. To make the phrase "the first coordinate" meaningful, we first record the basis vectors in a definite order.

[definition: Ordered Basis] Let $V$ be a vector space over a field $k$. An ordered basis of $V$ is a finite tuple $\mathcal B = (v_1,\ldots,v_n)$ such that the set $\{v_1,\ldots,v_n\}$ is a basis of $V$. [/definition]

An ordered basis gives names to the slots in a coordinate column, but it does not yet say how a vector fills those slots. The next concept answers the reconstruction question: which scalars multiply the basis vectors to produce a given vector? This is the data that will later be converted from one basis to another.

[definition: Coordinate Vector] Let $V$ be an $n$-dimensional vector space over a field $k$, and let $\mathcal B=(v_1,\ldots,v_n)$ be an ordered basis of $V$. For $x \in V$, the coordinate vector of $x$ with respect to $\mathcal B$ is the column vector $[x]_{\mathcal B} = (a_1,\ldots,a_n)^\top \in k^n$ defined by \begin{align*} x &= \sum_{i=1}^n a_i v_i. \end{align*} [/definition]

The coordinate vector depends on $\mathcal B$, but the assignment $x\mapsto [x]_{\mathcal B}$ is systematic. It sends addition and scalar multiplication in $V$ to addition and scalar multiplication in $k^n$. The next definition packages the whole coordinate system as a single linear isomorphism, which is the cleanest way to compare two bases.

[definition: Coordinate Isomorphism] Let $V$ be an $n$-dimensional vector space over a field $k$, and let $\mathcal B=(v_1,\ldots,v_n)$ be an ordered basis of $V$. The coordinate isomorphism determined by $\mathcal B$ is the map \begin{align*} \Phi_{\mathcal B}: V &\to k^n \end{align*} defined by \begin{align*} \Phi_{\mathcal B}(x) &= [x]_{\mathcal B}. \end{align*} [/definition]

Now suppose two ordered bases describe the same vector space. A vector can be decoded from its $\mathcal B$-coordinates and then encoded again using $\mathcal C$-coordinates. The resulting operation is the abstract change of basis before any matrix entries are written down.

[definition: Change of Basis Map] Let $V$ be an $n$-dimensional vector space over a field $k$, and let $\mathcal B$ and $\mathcal C$ be ordered bases of $V$. The change of basis map from $\mathcal B$-coordinates to $\mathcal C$-coordinates is the linear map \begin{align*} \Phi_{\mathcal C}\circ \Phi_{\mathcal B}^{-1}: k^n &\to k^n \end{align*} defined by \begin{align*} (\Phi_{\mathcal C}\circ \Phi_{\mathcal B}^{-1})(u) &= \Phi_{\mathcal C}(\Phi_{\mathcal B}^{-1}(u)). \end{align*} [/definition]

Computations need the matrix of this map. The matrix is not a new object acting on $V$ itself; it acts on coordinate columns. Naming it separately keeps the vector space, the coordinate spaces, and the conversion operation from being conflated.

[definition: Change of Basis Matrix] Let $V$ be an $n$-dimensional vector space over a field $k$, and let $\mathcal B$ and $\mathcal C$ be ordered bases of $V$. The change of basis matrix from $\mathcal B$ to $\mathcal C$ is the matrix $P_{\mathcal C \leftarrow \mathcal B} \in k^{n\times n}$ representing the map $\Phi_{\mathcal C}\circ \Phi_{\mathcal B}^{-1}$ with respect to the standard basis of $k^n$. [/definition]

The notation $P_{\mathcal C \leftarrow \mathcal B}$ should be read from right to left: it starts with $\mathcal B$-coordinates and returns $\mathcal C$-coordinates. Thus $[x]_{\mathcal C} = P_{\mathcal C \leftarrow \mathcal B}[x]_{\mathcal B}$. To compute the matrix in practice, we need a column rule that turns the definition into a finite list of coordinate calculations.

[quotetheorem:8323]

This column rule is the practical recipe: write each old basis vector in the new basis, and assemble the resulting coordinate columns. The rest of the chapter repeatedly uses this single idea in increasingly structured settings.