- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.12 Exercises
19. $\quad$ Choose the basis $(1, x, x^2, x^3)$ in the linear space $V$ of all real polynomials of degree $\leq 3.$ Let $D$ denote the differentiation operator and let $T: V \rightarrow V$ be the linear transformation which maps $p(x)$ onto $x p'(x).$ Relative to the given basis, determine the matrix of each of the following transformations: (a) $T;$ (b) $DT;$ (c) $TD;$ (d) $TD - DT;$ (e) $T^2;$ (f) $T^2 D^2 - D^2 T^2.$
Solution. $\quad$ For each example, we will first find the transformation of the four basis elements $1, x, x^2, x^3,$ stated in terms of those same elements. Then, we will use the scalar coefficients of each element to determine the matrix of the transformation.
(a) $\quad$ $T.$ \begin{align*} T(1) &= xp'(x) \\ &= 0 \\ &= 0(1) + 0(x) + 0(x^2) + 0(x^3) \\ \\ T(x) &= xp'(x) \\ &= x \\ &= 0(1) + 1(x) + 0(x^2) + 0(x^3) \\ \\ T(x^2) &= xp'(x) \\ &= 2x^2 \\ &= 0(1) + 0(x) + 2(x^2) + 0(x^3) \\ \\ T(x^3) &= xp'(x) \\ &= 3x^3 \\ &= 0(1) + 0(x) + 0(x^2) + 3(x^3) \end{align*} From this, we can determine the matrix $(t_{ik})$ of $T:$ \begin{align*} (t_{ik}) &= \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix} \quad \blacksquare \end{align*}
(b) $\quad$ $DT.$ \begin{align*} DT(1) &= D[xp'(x)] \\ &= D[x(0)] \\ &= 0 \\ &= 0(1) + 0(x) + 0(x^2) + 0(x^3) \\ \\ DT(x) &= D[xp'(x)] \\ &= D(x) \\ &= 1 \\ &= 1(1) + 0(x) + 0(x^2) + 0(x^3) \\ \\ DT(x^2) &= D[xp'(x)] \\ &= D(2x^2) \\ &= 4x \\ &= 0(1) + 4(x) + 0(x^2) + 0(x^3) \\ \\ DT(x^3) &= D[xp'(x)] \\ &= D(3x^3) \\ &= 9x^2 \\ &= 0(1) + 0(x) + 9(x^2) + 0(x^3) \end{align*} From this, we can determine the matrix $(t_{ik})$ of $DT:$ \begin{align*} (t_{ik}) &= \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 9 \\ 0 & 0 & 0 & 0 \end{bmatrix} \quad \blacksquare \end{align*}
(c) $\quad$ $TD.$ $\quad$ We will use $D[p(x)]$ in place of the equivalent $p'(x)$ for notational clarity. \begin{align*} TD(1) &= T(0) \\ &= xD(0) \\ &= 0 \\ &= 0(1) + 0(x) + 0(x^2) + 0(x^3) \\ \\ TD(x) &= T(1) \\ &= xD(1) \\ &= 0 \\ &= 0(1) + 0(x) + 0(x^2) + 0(x^3) \\ \\ TD(x^2) &= T(2x) \\ &= xD(2x) \\ &= 2x \\ &= 0(1) + 2(x) + 0(x^2) + 0(x^3) \\ \\ TD(x^3) &= T(3x^2) \\ &= xD(3x^2) \\ &= 6x^2 \\ &= 0(1) + 0(x) + 6(x^2) + 0(x^3) \end{align*} From this, we can determine the matrix $(t_{ik})$ of $TD:$ \begin{align*} (t_{ik}) &= \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 6 \\ 0 & 0 & 0 & 0 \end{bmatrix} \quad \blacksquare \end{align*}
(d) $\quad$ $TD - DT$ $\quad$ We can use the results of (b) and (c) to give: \begin{align*} (TD - DT)(1) &= 0(1) + 0(x) + 0(x^2) + 0(x^3) \\ (TD - DT)(x) &= -1(1) + 0(x) + 0(x^2) + 0(x^3) \\ (TD - DT)(x^2) &= 0(1) - 2(x) + 0(x^2) + 0(x^3) \\ (TD - DT)(x^3) &= 0(1) + 0(x) - 3(x^2) + 0(x^3) \end{align*} From this, we can determine the matrix $(t_{ik})$ of $TD:$ \begin{align*} (t_{ik}) &= \begin{bmatrix} 0 & -1 & 0 & 0 \\ 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & -3 \\ 0 & 0 & 0 & 0 \end{bmatrix} \quad \blacksquare \end{align*} [Note: This array can be found by subtracting the components of the matrix $DT$ from the matrix $TD.$]
(e) $\quad$ $T^2.$ $\quad$ We will once again use $D(p)$ to denote the derivative of $p.$ \begin{align*} T^2(1) &= T[xD(1)] \\ &= xD(0) \\ &= 0 \\ &= 0(1) + 0(x) + 0(x^2) + 0(x^3) \\ T^2(x) &= T[xD(x)] \\ &= xD(x) \\ &= x \\ &= 0(1) + 1(x) + 0(x^2) + 0(x^3) \\ T^2(x^2) &= T[xD(x^2)] \\ &= xD(2x^2) \\ &= 4x^2 \\ &= 0(1) - 0(x) + 4(x^2) + 0(x^3) \\ T^2(x^3) &= T[xD(x^3)] \\ &= xD(3x^3) \\ &= 9x^3 \\ &= 0(1) + 0(x) + 0(x^2) + 9(x^3) \end{align*} From this, we can determine the matrix $(t_{ik})$ of $T^2:$ \begin{align*} (t_{ik}) &= \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 9 \end{bmatrix} \quad \blacksquare \end{align*}
(f) $\quad$ $T^2D^2 - D^2T^2.$ $\quad$ We will use the results of (e) in computing $D^2T^2(p).$ As before, we will use $D(p)$ to denote $p'.$ \begin{align*} (T^2D^2 - D^2T^2)(1) &= T^2(0) - D^2(0) \\ &= 0 \\ &= 0(1) + 0(x) + 0(x^2) + 0(x^3) \\ (T^2D^2 - D^2T^2)(x) &= T^2(0) - D^2(x) \\ &= 0 \\ &= 0(1) + 0(x) + 0(x^2) + 0(x^3) \\ (T^2D^2 - D^2T^2)(x^2) &= T^2(2) - D^2(4x^2) \\ &= T[xD(2)] - 8 \\ &= -8 \\ &= -8(1) - 0(x) + 0(x^2) + 0(x^3) \\ (T^2D^2 - D^2T^2)(x^3) &= T^2(6x) - D^2(9x^3) \\ &= T[xD(6x)] - D(27x^2) \\ &= xD(6x) - 54x \\ &= -48x \\ &= 0(1) - 48 (x) + 0(x^2) + 0(x^3) \end{align*} From this, we can determine the matrix $(t_{ik})$ of $T^2:$ \begin{align*} (t_{ik}) &= \begin{bmatrix} 0 & 0 & -8 & 0 \\ 0 & 0 & 0 & -48 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \quad \blacksquare \end{align*}