Show that $T$ is a linear map and finds its coordinates $[T]_B*$ with respect to the dual basis

Question

Let $B = \{x, x^3, x^5\}$ so that $B$ is a basis of $V_1 = \{a_1x + a_2x^3 + a_3x^5 \mid a_1, a_2, a_3 \in \mathbb{R}\}$.

Define a map $T: V_1 \rightarrow \mathbb{R}$ by

$T(f) = -f(3)$ for all $f \in V_1$

Show that $T$ is a linear map and finds its coordinates $[T]_B*$ with respect to the dual basis

$B^* = \{x^*, (x^3)^*, (x^5)^*\}$

I showed that $T$ was a linear map by showing that it preserved the operations of addition and scalar multiplication.

However I'm unsure about the second part of the question.

I've found coordinates with respect to another basis in previous courses - however my problems involved matrices, which aren't used in this particular course. Could somebody please help?

Answer 1

In general, to find the coordinates of $T$ with respect to the dual basis, you first write it as a general sum of dual vectors

$$T=c_1(x)^*+c_3(x^3)^*+c_5(x^5)^*,$$

and then to isolate the coefficients, consider $c_i=T(x^i)$.

Answer 2

I'm assuming when you say $x*$ you mean the function that pulls out the coefficient of $x$. I'm going to call this $a_1$, and the other two, I will call $a_3$ and $a_5$ respectively.

Now if $f=f_1x+f_3x^3+f_5x^5$, then $T(f)=-f(3)=-f_1\cdot3-f_3\cdot 27-f_5\cdot 243=-3a_1(f)-27a_3(f)-243a_5(f)$. Therefore $T=-3a_1-27a_3-243a_5$.

Answer 3

The index $k$ runs over $\{1,3,5\}$ in the following. We first have to identify the dual basis vectors $e_k$. These vectors compute the coefficients of arbitrary elements $f\in V$ with respect to the basis $B:=(x,x^3,x^5)$ of $V$. It follows that $$e_k(f)={f^{(k)}(0)\over k!}\ .$$ Since $$T(f)=-f(3)=-\sum_k{f^{(k)}(0)\over k!}3^k=\sum_k(-3^k)e_k(f)$$ it follows that the coordinates of $T$ with respect to the basis $(e_1,e_3,e_5)$ of $V^*$ are $(-3,-27,-243)$.