Answer:
The matrix representation of the linear transformation is
[tex] \begin{pmatrix} 0 & 1 & 0 & 0 & \cdots & 0\\ 0 & 0 & 2 & 0 & \cdots & 0\\ 0 & 0 & 0 & 3 & \cdots & 0\\ \cdot & \cdot & \cdot & \cdot & \cdots & \cdot\\ 0 & 0 & 0 & 0 & \cdots & n-1\\ 0 & 0 & 0 & 0 & \cdots & 0\end{pmatrix}[/tex],
which is a matrix of dimension [tex]n\times n[/tex].
Step-by-step explanation:
First, as we are working with vector spaces, in particular with [tex]P_{n-1}[/tex] the space of real polynomials of degree at most n-1, we need to set a base. As usual we will use the canonical base: [tex]\{1,t,t^2,\ldots,t^{n-1}\}[/tex].
The differential operator [tex]D[/tex] is such that [tex]D:P_{n-1}\rightarrow P_{n-1}[/tex]. So, in the image we will use the canonical base too.
Recall that, if we have a polynomial [tex]p(t) = c_0 + c_1t + \cdots +c_{n-1}t^{n-1}[/tex], its derivative is [tex]p'(t) = c_1 + 2c_2t +\cdots +(n-1)c_{n-1}t^{n-2}[/tex], i.e. [tex]p' = Dp[/tex]. In order to construct the matrix representation of a linear transformation we must evaluate the transformation at the elements of the base. In this case we must find
[tex]\{D1, Dt, Dt^2,\ldots, Dt^{n-1}\}[/tex].
It is no difficult to notice that
[tex]\{D1, Dt, Dt^2,\ldots, Dt^{n-1}\} = \{0, 1, 2t, \ldtos, (n-1)t^{n-2}\}.[/tex]
The next step is to write the image of each element of the base as a linear combination of the base of [tex]P_{n-2}[/tex]. Then,
[tex] 0 = 0*1+0*t+\cdots + 0*t^{n-2}[/tex]
[tex] 1 = 1*1+0*t+\cdots + 0*t^{n-2}[/tex]
[tex] 2t = 0*1+2*t+\cdots + 0*t^{n-2}[/tex]
[tex] 3t^2 = 0*1+0*t+3t^2+\cdots + 0*t^{n-2}[/tex]
and so on, until
[tex] 0 = 0*1+0*t+\cdots + (n-1)*t^{n-2}[/tex].
The coefficients of this linear combinations are written as columns of matrix, which is the representation we are looking for:
[tex] \begin{pmatrix} 0 & 1 & 0 & 0 & \cdots & 0\\ 0 & 0 & 2 & 0 & \cdots & 0\\ 0 & 0 & 0 & 3 & \cdots & 0\\ \cdot & \cdot & \cdot & \cdot & \cdots & \cdot\\ 0 & 0 & 0 & 0 & \cdots & n-1\\ 0 & 0 & 0 & 0 & \cdots & 0\end{pmatrix}[/tex].
In this case we have a matrix of dimension [tex]n\times n[/tex], because we are considering a linear transformation from [tex]P_{n-1}[/tex] to itself, rather to [tex]P_{n-2}[/tex].
