Change of Basis

x = P[x]_B, [x]_B = P^-1x, and A' = P^-1AP is the same map in a new basis.

The idea

A change of basis rewrites the coordinates of vectors and matrices when we describe the same space with a different basis. Coordinates are not the vector itself: they record how to build it from the chosen basis vectors, so a different basis gives the same vector different numbers.

Fix a basis $B = \{b_{1}, \dots, b_{n}\}$. Every vector $x$ has exactly one expression $x = c_{1}b_{1} + \cdots + c_{n}b_{n}$, and the list of weights $(c_{1}, \dots, c_{n})$ is the coordinate vector of $x$ in $B$, written $[x]_{B}$. In the figure, the plane vector $w$ is reached by travelling $a\,v_{1}$ and then $b\,v_{2}$ along the basis $\{v_{1}, v_{2}\}$, so its coordinate vector in that basis is $(a, b)$. Place the basis vectors as the columns of a matrix $P$, the change-of-basis matrix, and the expression becomes $x = P\,[x]_{B}, \qquad [x]_{B} = P^{-1}x,$ so $P$ converts $B$-coordinates to standard coordinates and $P^{-1}$ converts back.

A matrix changes basis the same way. If $A$ represents a linear map in standard coordinates, its matrix in the basis $B$ is $A' = P^{-1}AP$: convert $B$-coordinates to standard ones, apply $A$, and convert the result back. Matrices related in this way are similar. Similar matrices represent one map in two bases, so every quantity that belongs to the map itself — trace, determinant, characteristic polynomial, eigenvalues — is the same for both.

Ways to work on it

• Walkthrough. Express a vector in a given basis by solving a 2 × 2 system.
• Practice. Convert basis coordinates to standard coordinates via x = P[x]_B.
• Hardest. Find a vector's coordinates in a new basis, or identify what similar matrices share.

Not sure where to start? Take the ten-question placement test.