Mathematical Immaturity

2.8 Exercises

31. $\quad$ Let $V$ be the linear space of all real polynomials $p(x).$ Let $R,$ $S,$ $T$ be the functions which map an arbitrary polynomial $p(x) = c_0 + c_1 x + \cdots + c_n x^n$ in $V$ onto the polynomials $r(x),$ $s(x),$ and $t(x),$ respectively, where \begin{align*} r(x) = p(0), \quad s(x) = \sum_{k=1}^n c_k x^{k-1}, \quad t(x) = \sum_{k=0}^n c_k x^{k+1}. \end{align*} (a) $\quad$ Let $p(x) = 2 + 3x - x^2 + x^3$ and determine the image of $p$ under each of the following transformations: $R,$ $S,$ $T,$ $ST,$ $TS,$ $(TS)^2,$ $T^2 S^2,$ $S^2 T^2,$ $TRS,$ $RST.$
(b) $\quad$ Prove that $R,$ $S,$ and $T$ are linear and determine the null space and range of each.
(c) $\quad$ Prove that $T$ is one-to-one and determine its inverse.
(d) $\quad$ If $n \geq 1,$ express $(TS)^n$ and $S^n T^n$ in terms of $I$ and $R.$

Solution.
(a) $\quad$ Let $p(x) = 2 + 3x - x^2 + x^3$ and determine the image of $p$ under each of the following transformations: $R,$ $S,$ $T,$ $ST,$ $TS,$ $(TS)^2,$ $T^2S^2,$ $S^2 T^2,$ $TRS,$ $RST.$ \begin{align*} R[p(x)] &= r(x) \\ &= p(0) \\ &= 2 \\ S[p(x)] &= s(x) \\ &= \sum_{k=1}^n c_k x^{k-1} \\ &= 3 - x + x^2 \\ \\ T[p(x)] &= t(x) \\ &= \sum_{k=0}^n c_k x^{k+1} \\ &= 2x + 3x^2 - x^3 + x^4 \\ \\ ST[p(x)] &= S[T[p(x)]] \\ &= S[2x + 3x^2 - x^3 + x^4] \\ &= 2 + 3x - x^2 + x^3 \\ \\ TS[p(x)] &= T[S[p(x)]] \\ &= T[3 - x + x^2] \\ &= 3x - x^2 + x^3 \\ \\ (TS)^2[p(x)] &= TS[TS[p(x)]] \\ &= T[S(3x - x^2 + x^3)] \\ &= T(3 - x + x^2) \\ &= 3x - x^2 + x^3 \\ \\ T^2S^2[p(x)] &= T^2S[3 - x + x^2] \\ &= T[T(-1 + x)] \\ &= T(-x + x^2) \\ &= -x^2 + x^3 \\ \\ S^2T^2[p(x)] &= S^2T(2x + 3x^2 - x^3 + x^4) \\ &= S[S(2x^2 + 3x^3 - x^4 + x^5)] \\ &= S(2x + 3x^2 - x^3 + x^4) \\ &= 2 + 3x - x^2 + x^3 \\ \\ TRS[p(x)] &= T[R(3 - x + x^2)] \\ &= T(3) \\ &= 3x \\ \\ RST[p(x)] &= R[ST[p(x)]] \\ &= R(2 + 3x - x^2 + x^3) \\ &= 2 \quad \blacksquare \end{align*}

(b) $\quad$ Prove that $R,$ $S,$ and $T$ are linear and determine the null space and range of each.

Proof. $\quad$ Recall the definition of a linear transformation from Section 1:

$\quad$ Definition. $\quad$ If $V$ and $W$ are linear spaces , a function $T: V \to W$ is called a linear transformation of $V$ into $W$ if it has the following two properties:
$\quad$ (a) $\quad T(x + y) = T(x) + T(y) \quad$ for all $x$ and $y$ in $V,$
$\quad$ (b) $\quad T(cx) = cT(x) \quad$ for all $x$ in $V$ and all scalars $c.$

Thus, we wish to show that $R,$ $S,$ and $T$ satisfy these two properties. Let $p$ and $q$ be two polynomials in $V$ given by: \begin{align*} p(x) &= \sum_{k=0}^n c_kx^k, \quad q(x) = \sum_{k=0}^n d_kx^k \end{align*} and let $a$ and $b$ be scalar constants. We then have: \begin{align*} R[ap(x) + bq(x)] &= R\left[\sum_{k=0}^n (ac_k + bd_k)x^k\right] \\ &= ac_0 + bd_0 \\ &= ap(0) + bq(0) \\ &= aR[p(x)] + bR[q(x)] \\ \\ S[ap(x) + bq(x)] &= S\left[\sum_{k=0}^n (ac_k + bd_k)x^k\right] \\ &= \sum_{k=1}^n (ac_k + bd_k)x^{k-1} \\ &= a\sum_{k=1}^n c_kx^{k-1} + b\sum_{k=1}^n d_kx^{k-1} \\ &= aS[p(x)] + bS[q(x)] \\ \\ T[ap(x) + bq(x)] &= T\left[\sum_{k=0}^n (ac_k + bd_k)x^k\right] \\ &= \sum_{k=0}^n (ac_k + bd_k)x^{k+1} \\ &= a\sum_{k=0}^n c_kx^{k+1} + b\sum_{k=0}^n d_kx^{k+1} \\ &= aT[p(x)] + bT[q(x)] \end{align*} As we can see, $R,$ $S,$ and $T$ are linear transformations. Their null spaces and ranges are given as follows:
$\quad$ $N(R)$ is the set of all polynomials in $V$ with $c_0 = 0.$ $R(V)$ is the set of all constant polynomials.
$\quad$ $N(S)$ is the set of all polynomials in $V$ with coefficients $c_1, \dots, c_n$ equal to zero, and arbitrary $c_0.$ In other words, $N(S)$ is the set of all constant polynomials. $S(V) = V.$
$\quad$ $N(T)$ contains only the zero polynomial. $T(V)$ is the set of all polynomials $p$ such that $p(0) = 0. \quad \blacksquare$

(c) $\quad$ Prove that $T$ is one-to-one and determine its inverse.

$\quad$ Proof. $\quad$ By definition, if $T$ is one-to-one on $V,$ then for any two polynomials $p$ and $q$ in $V,$ $p \neq q$ implies $T(p) \neq T(q).$ Now, let $p$ and $q$ be elements of $V$ given by \begin{align*} p(x) &= \sum_{k=0}^n c_kx^k, \quad q(x) = \sum_{k=0}^n d_kx^k \end{align*} where for some $i,$ in $1, \dots, n,$ $c_i \neq d_i.$ In other words, $p \neq q.$ Taking their respective transformations, we get \begin{align*} Tp(x) &= \sum_{k=0}^n c_kx^{k+1}, \quad Tq(x) = \sum_{k=0}^n d_kx^{k+1} \end{align*} But since $c_i \neq d_i,$ $Tp(x) \neq Tq(x).$ Hence, $p \neq q$ implies $Tp \neq Tq,$ making $T$ one-to-one on $V.$

$\quad$ The inverse of $T$ is the transformation $T^{-1}p$ that decreases the power of $x$ by $1$ for all $x^k$ where $k \geq 1.$ Namely, the inverse of $T$ is $S. \quad \blacksquare$

(d) $\quad$ If $n \geq 1,$ express $(TS)^n$ and $S^n T^n$ in terms of $I$ and $R.$

$\quad$ Let $p(x) = \sum_{k=0}^n c_kx^k$ be an arbitrary polynomial in $V.$ From part (c), we know that $S = T^{-1},$ giving us $ST = I_V.$ Applying $S$ as a left-inverse to $T,$ we get $ST = I_V,$ and $(TS)^n$ becomes \begin{align*} (TS)^n[p(x)] &= T(ST)^{n-1}S[p(x)] \\ &= T(I^{n-1})S[p(x)] \\ &= TS\left[\sum_{k=0}^n c_kx^k\right] \\ &= T\left[\sum_{k=1}^n c_kx^{k-1}\right] \\ &= \sum_{k=1}^n c_kx^k \\ &= p(x) - p(0) \end{align*} But by definition, we have $p(x) - p(0) = I[p(x)] - R[p(x)].$ Then, by linearity, we can rewrite $I[p(x)] - R[p(x)] = (I - R)[p(x)]$ to give us: \begin{align*} (TS)^n &= I - R \end{align*} $\quad$ To express $S^nT^n$ in terms of $I$ and $R,$ we recall that $ST = I_V.$ Hence, we can use the associative property of composition and the result of Exercise 21 to give us \begin{align*} S^nT^n &= S^{n-1}(ST)T^{n-1} \\ &= S^{n-1}(I)T^{n-1} \\ &= S^{n-2}(ST)T^{n-2} \\ &\dots \\ &= ST \\ &= I \quad \blacksquare \end{align*}