Why can we use matrices to represent elements of polynomial vector space

Question

I think I understand the concept of the polynomial vector space over a field F (set of polynomial functions whose coefficients are part of F), but when describing a transformation from one polynomial vector space to another —- why are we allowed to use matrices to describe that? Matrices aren’t elements of the polynomial vector space, but it is its own vector space right? For example, when describing the transformation matrix for the linear transformation that is the derivative of a polynomial, we have a matrix of coefficients and then multiply that by whatever input column coefficient vector we have that represents a polynomial, but that column vector itself is not part of the polynomial vector space so how do we know it describes a polynomial in the first place?

Answer 1

Consider the vector space $P_{n - 1}(k)$ of polynomials of degree $\leq n - 1$ with coefficients in the field $k.$ Observe that the standard basis of $P_{n - 1}(k)$ is $\mathscr B = \{1, x, x^2, \dots, x^{n - 1}\}.$ (Use the derivative to prove that these are linearly independent.) Consider a linear operator $T : P_{n - 1}(k) \to P_{n - 1}(k).$ Given any polynomial $p(x)$ in $P_{n - 1}(k),$ we can write $T(p(x))$ in terms of the basis $\mathscr B.$ Particularly, $$\begin{align*} T(1) &= a_{11} + a_{21} x + \cdots + a_{n1} x^{n - 1}, \\ \\ T(x) &= a_{12} + a_{22} x + \cdots + a_{n2} x^{n-1}, \\ &\phantom{\,\,} \vdots \\ T(x^{n-1}) &= a_{1n} + a_{2n} x + \cdots + a_{nn} x^{n - 1}, \end{align*}$$ from which we obtain an $n \times n$ matrix $A = [a_{ij}]_{1 \leq i, j \leq n}$ with coefficients in $k.$ (Explicitly, $A$ is the matrix whose $j$th column is $T(x^{j - 1}).$) We refer to $A$ as the matrix of $T$ with respect to $\mathscr B.$

Example. Consider the vector space $P_2(\mathbb R)$ of polynomials of degree $\leq 2$ with coefficients in $\mathbb R$ (the real numbers). Recall that the derivative satisfies $\frac d {dx} [C \cdot p(x)] = C \cdot p'(x)$ for all real numbers $C$ and $\frac d {dx} [p(x) + q(x)] = p'(x) + q'(x)$ for all polynomials $p(x)$ and $q(x),$ hence differentiation is a linear transformation. By the Power Rule, we have that $\deg p'(x) = \deg p(x) - 1,$ hence the map $D : P_2(\mathbb R) \to P_2(\mathbb R)$ defined by $D(p(x)) = p'(x)$ is a linear operator. Considering that $$\begin{align*} D(1) &= 0, \\ \\ D(x) &= 1, \text{ and} \\ \\ D(x^2) &= 2x, \end{align*}$$ we find that the matrix of $D$ with respect to $\mathscr B = \{1, x, x^2\}$ is given by $A = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{pmatrix}.$