Mathematical Immaturity

2.12 Exercises

20. $\quad$ Refer to Exercise 19. Let $W$ be the image of $V$ under $TD.$ Find bases for $V$ and for $W$ relative to which the matrix of $TD$ is in diagonal form.

Solution. $\quad$ Referring to part (c) of Exercise 19, we found that the matrix of $TD$ relative to basis $(1, x, x^2, x^3)$ is given by: \begin{align*} (t_{ik}) &= \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 6 \\ 0 & 0 & 0 & 0 \end{bmatrix} \end{align*} This matrix corresponds to the following transformations of the ordered basis elements $(1, x, x^2, x^3):$ \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*} $\quad$ Note that $TD(1) = TD(x) = 0,$ where $0$ is the zero element of $W.$ Moreover, because $1$ and $x$ are independent elements of $V,$ this means that they are elements of a basis for the null space of $TD,$ $N(TD),$ which means $\dim N(TD) \geq 2.$ Then, from the transformations of $x$ and $x^2,$ we find that the independent elements $x$ and $x^2$ are part of a basis for $W,$ giving us $\dim W \geq 2.$ But from the nullity plus rank theorem, we know that $\dim N(TD) + \dim W = \dim V = 4,$ which means that $(x, x^2)$ is in fact a basis for $W.$ Thus, we have bases $(1, x, x^2, x^3)$ for $V$ and $(x, x^2)$ for $W.$

$\quad$ For the matrix of $TD$ relative to these bases to be diagonal, we must order the basis elements in such a way that the component $t_{ik} = w_i$ (where $w_i$ is a nonzero scalar) when $i = k$ and $0$ when $i \neq k.$ One such way to order them is $(x^3, x^2, x, 1)$ for $V$ and $(x^2, x)$ for $W,$ giving us the following matrix of $TD$ relative to these bases: \begin{align*} \begin{bmatrix} 6 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \end{bmatrix} \quad \blacksquare \end{align*}