Preface:
I am preparing for the GRE Math subject test, and one of the books that I use is this one by Charles Rambo. I have currently been solving through the “Linear Algebra #1” section of the book, and there is one problem that I think has an incorrect answer in the solutions section. Let me state the problem as it was drawn in the book:
Let $V$ be the set of all possible polynomials $p(x)$ with coefficients in $\mathbb{R}$. And let the linear operators $T$ and $S$ be defined over $V$ as follows: $$ T(p(x)) = xp(x) \\ S(p(x)) = \frac{d}{dx}p(x) = p’(x) $$ We denote $ST$ and $TS$ as the expected compositions $S\circ T(p(x))$ and $T \circ S(p(x))$ respectively. Which of the following is true?
- $ST = 0$
- $ST = T$
- $ST = TS$
- $ST - TS$ is the identity map of $V$
- $ST + TS$ is the identity map of $V$
My attempt:
In my opinion, the correct option is (4), since:
- $ST = \frac{d}{dx}\left[xp(x)\right] = x’p(x) + xp’(x) = p(x)+xp’(x)$
- $TS = x \cdot \frac{d}{dx}p(x) = xp’(x)$
- $F(p(x)) = (ST-TS)(p(x))$ is a transformation defined over $V$, then $F(p(x)) = (ST - TS)(p(x)) = p(x)+xp’(x)-xp’(x) = p(x)$, hence $F : V \mapsto V$, in other terms $F$ forms an identity map over $V$.
Book’s solution:
In fact, the only thing written in the book about this problem (page 84 in the pdf) was that the answer is (5), and that one had to use the product rule to obtain the correct answer (no explicit solution given).
And so I wonder whether there is any fallacy in my logic above, or it is the book, which is mistaken.
Any help will be greatly appreciated. Thank you in advance!
