I am fairly new to linear algebra and am trying my best to keep everything straight. That being said, I have a question on the following:
Consider the following set of polynomials: $$P = \{1, x^2, x^4, x^6 \} $$ and the set created by spanning the elements of P: $$U = \text{span}(P)$$ *Show that $U$ is a vector space. The span of $P$ is always a vector space correct? So how do you properly show this? Would simply giving an example of being closed under addition and scalar multiplication be sufficient?
*Consider the linear operator $T$, defined on $U$ such that $$Tu = v = \frac {du}{dx}$$ for all $u \in U$. Identify the vector space $V$ into which the above define operator maps into, such that $$T: U \to V$$
What is this exactly looking for? I can easily take the derivative of P(?) Is that the same as the derivative of the $\text{span}(P)$? But that just gives me another vector... Is that considered an 'operator'?
*Identify a basis for $U$ ($\beta _U$) and a basis for $V$ ($\beta _V$) to then identify a matrix representation $A$ of the linear operator $T$ so that for each $u \in U$ and $v \in V$ such that $v = Tu$ there is:
$$A[u]_{\beta _U} = [v]_{\beta _V}$$
Isn't $P$ a basis for $U$? Or do you have to include $\{x, x^3, x^5\}$ as well since $P$ is at most $x^6$? This doesn't make too much sense to me though, because how would you create a linear combination of any of the terms $\{x, x^3, x^5\}$? Unless they're always zero? In a similar vein, $\{1, x, x^2, x^3, x^4, x^5\}$ wouldn't be a basis for $V$ correct?
You show that something is a vector space by verifying that it satisfies the definition of a vector space. Showing that a set is closed under addition and scalar multiplication is sufficient only when that set is contained in a known vector space.
It's asking for a vector space $V$ that contains the range of $T$. You have to figure out what the range of $T$ is, and verify that it is a vector space.
You can't take the derivative of $P$ or $span(P)$. You can take the derivative of the elements in $P$ or $span(P)$.
Yes, $T$ would be considered a (linear) operator.
Yes, your intuition is correct. If you add $\{x,x^3,x^5\}$ to the basis, then as you have said, the coefficients next to $x$, $x^3$ and $x^5$ in any linear combination of elements of $U$ would always be $0$, so you would be representing $U$ as a subspace of the vector space with $\{1,x,\dots,x^6\}$ as its basis.