Finding the representation matrix of the linear transformation $S: \mathbb{R}^3\longrightarrow P_2: (a,b,c) \mapsto (a+b+c)+(3b-c)x$

Question

I know that the basis of $\mathbb{R}^3$ is $\{(1,0,0), (0,1,0),(0,0,1)\}$.

Now, let $P_2(\mathbb{R}$ be the set of polynomials with degree less or equal to 2 with coefficients in $\mathbb{R}$ (i.e. $P_2(\mathbb{R}):=\{p(x)\in \mathbb{R}[x] | \text{degree}(p(x))\leq 2\}$.

I also know that the basis of $P_2(\mathbb{R})$ is $\{(1,x)\}$ and with more simple examples I have no problem with this.

Now, let $$S: \mathbb{R}^3\longrightarrow P_2(\mathbb{R}): (a,b,c) \mapsto (a+b+c)+(3b-c)x$$ How do I find the matrix of "S"?

Answer 1

Hint:

The matrix of a linear map has as column vectors the coordinates, in the basis of $P_2$, of the images of the vectors of the basis used for $\mathbf R^3$.

(Incidentally, there is no such thing as THE basis of a vector space: $\mathbf R^3$ has an infinite number of bases. The basis you mention is usually considered as the canonical basis of $\mathbf R^3$).

Answer 2

Do you understand that a matrix represents a given matrix only in a given choice of basis for both domain and range spaces? What bases do you want to use? A natural choice would be {(1,0,0), (0,1,0), (0,0,1)} for the domain space and {1, x} for the range space. Apply the transformation to each of the domain space basis vectors in turn and write the result in terms of the range space basis. Each gives a column of the matrix representation.

This transformation maps (1, 0, 0) into 1+ 0x so first column of the matrix is (1, 0). The transformation maps (0, 1, 0) into 1+ 3x so the second column of the matrix is (1, 3). The transformation maps (0, 0, 1) into 1- x so the third column of the matrix is (1, -1). The matrix representation of the linear transformation using these choices of basis is $\begin{bmatrix}1 & 1 & 1 \\ 0 & 3 & -1\end{bmatrix}$