Matrix of a linear transformation

To build it, apply \(T\) to each vector of the domain basis and express the result in coordinates of the codomain basis; each result is a column. In the standard bases, column \(j\) is simply \(T(e_j)\). For example, if \(T(x,y)=(2x+y,\,3x-y)\), then \(T(e_1)=(2,3)\) and \(T(e_2)=(1,-1)\), and the matrix is \(\begin{pmatrix}2&1\\3&-1\end{pmatrix}\).

With the matrix, evaluating is multiplying: the coordinates of \(T(v)\) are obtained by multiplying the matrix by the coordinates of \(v\). If \(T:\mathbb{R}^n\to\mathbb{R}^m\), the matrix is \(m\times n\): as many columns as the dimension of the domain and as many rows as that of the codomain. Also, the matrix of the composition \(T\circ S\) is the product \([T][S]\): \(S\) acts first, so it sits on the right.

The matrix depends on the chosen bases. The same transformation has another matrix in another basis, and a good choice can make it very simple, for instance diagonal. Another example: the derivative on polynomials of degree at most 2, with basis \(\{1,x,x^2\}\), has columns \((0,0,0)\), \((1,0,0)\) and \((0,2,0)\).

Typical mistakes: writing the images as rows instead of columns, and multiplying in the wrong order when composing. This construction is what will make eigenvalues approachable: a transformation is best understood in the basis where its matrix is simplest.