Actually, I would think that the best way to show this is not a linear transformation would be to show that is not always equal to . And do that by taking .

However, since you are asked to do this by showing that there is no matrix representing this linear transformation, take , , and as basis for P2 and 1 as basis for P0. Then . A standard way to construct a matrix for a linear transformation, given a basis for domain and range spaces, is to take that transformation of each of the basis vectors of the domain, in turn, and write the result as a linear combination of the basis vecotrs in the range. The coefficients are the columns of the matrix. Here, , L(x)= and . That is,ifthis were a linear transformation, its matrix in the standard bases would be the row matrix [1 0 0].

But while [tex]L(-x^2)= |-1|= 1.