The same map looks different in different bases, and a well-chosen basis can make it look simple. Changing basis is a matter of translating coordinates in, doing the map, and translating back, and the three steps multiply into one matrix.

A problem posed in miles can be solved in kilometres: convert the data, solve, and convert the answer back. When the second system suits the problem better, the detour is the short way.

The map of $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ in a basis fitted to it. It stretches $\mathbf{b}_1=(1,1)$ along its own line to $A\mathbf{b}_1=(3,3)=3\mathbf{b}_1$, and leaves $\mathbf{b}_2=(-1,1)$ where it is. In the basis $\mathbf{b}_1,\mathbf{b}_2$ its matrix is therefore $\begin{pmatrix}3&0\\0&1\end{pmatrix}$.

Let $B=(\mathbf{b}_1,\ldots,\mathbf{b}_n)$ be a basis of ${\mathbb{R}}^n$ and the matrix with these vectors as columns. Then , so turns -coordinates into standard ones, and , which exists because the columns are a basis, turns them back. If is the map of , its matrix in the basis is . Matrices related in this way, for some invertible , are called similar: they are the same map written in two bases.

• $[\mathbf{x}]_B=P^{-1}\mathbf{x}$: to find coordinates, solve $P\mathbf{c}=\mathbf{x}$.

Start from the coordinates $[\mathbf{x}]_B$. Multiplying by $P$ gives the vector $\mathbf{x}$ itself, multiplying by $A$ applies the map, and multiplying by $$ turns the result back into -coordinates. The product therefore takes to , which is what does.

$P$ has the basis vectors as its columns, so it turns coordinates into the vector they describe.

Apply the map, which in standard coordinates is $A$.

Translate the result back into $B$-coordinates with $P^{-1}$.

This matrix turns $[\mathbf{x}]_B$ into $[T\mathbf{x}]_B$ for every $\mathbf{x}$, which is what the matrix of in does.

Put the basis vectors in as the columns of $P$. Multiplying by $P$ turns $B$-coordinates into the vector; multiplying by $P^{-1}$ turns a vector into its $B$-coordinates.

$A$ and $P^{-1}AP$ describe one map in two bases. Anything that belongs to the map, rather than to the basis, is the same for both.

• $P$, with the basis vectors as columns, turns $B$-coordinates into vectors, and $P^{-1}$ turns them back.
• The matrix of the map of $A$ in the basis $B$ is .

Great work! You now know how to:

• turn coordinates into vectors with P and back with its inverse;
• write the matrix of a map in a new basis as $P^{-1}AP$;
• recognise similar matrices and what they share;
• see how a basis fitted to a map makes its matrix diagonal.