Consider the vector space spanned by: \begin{align*} & H_0(x) = 1 \\& H_1(x) = 2x \\& H_2(x) = 4x^2 - 2 \\& H_3(x) = 8x^3 - 12x \end{align*}
$$V = span(\{H_0, H_1, H_2, H3\})$$ where $\{H_0, H_1, H_2, H_3\}$ constitute a basis of $V$
Given the equation: $$L_2(x) = \frac{3}{2}x^2 - \frac{1}{2}$$
is $L_2(x) \in V$?
Yes, each element of the basis can be tuned such that $L_2$ falls within the span of the basis.
i.e. a linear combination of the basis elements can form $L_2$
Question:
What would be a similar element that does not fall within the basis? Also what would be a new (slightly modified) basis that would contain that element?
I am trying to understand the connection between $V$ and basis
Thanks!
Assuming that the vectors are functions from $\Bbb R$ to $\Bbb R$ and that the vector space is over the field of reals (so that, e.g., $H_1(x)=2x$ means that $H_1$ is the function $f:\Bbb R\to \Bbb R$ such that $f(x)=2x $ for all $x\in \Bbb R$):
$V=$ Span$( \{H_0,H_1,H_2,H_3\} )= \{\sum_{j=0}^3r_jH_j: r_0,r_1,r_2,r_3\in \Bbb R\}.$
$V$ is a vector space. So if $U\subset V$ then any finite linear combination of members of $U$ is a member of $V.$ So if $U\subset V$ then the set of all finite linear combinations of members of $U,$ which is Span$(U),$ is a subset of $V.$
We have:
$A_0=1=H_0\in V.$
$A_1=x=H_1/2-H_0\in V.$
$A_2=x^2=H_2/4-H_0/2\in V.$
$A_3=x^3=H_3/8-3H_1/4\in V.$
So with $A=\{A_0,A_1,A_2,A_3\}$ we have $A\subset V$ so Span$(A)\subset V.$
Now Span$(A)=\{\sum_{j=0}^3r_jx^j:r_0,r_1,r_2,r_3\in \Bbb R\}$ is the set $P_3$ of all polynomial functions from $\Bbb R$ to $\Bbb R$ of degree $3$ or less. Now observe that any linear combination of $H_0,H_1,H_2,H_3$ is a polynomial of degree $3$ or less. So $V=$Span$(\{H_0,H_1,H_2,H_3\}\subset P_3=$Span$(A).$
We now have Span$(A)\subset V$ and $V\subset $ Span$(A),$ so $V=$Span$(A)=P_3.$
We have $L_2\in P_3$ so $L_2\in V.$
Let $0^*$ denote the "origin" (the additive identity) of $V$. That is, $0^*$ is the function $f:\Bbb R\to \Bbb R$ such that $f(x)=0$ for all $x\in \Bbb R.$
Now $A$ and $\{H_0,H_1,H_2,H_3\}$ are bases for $V$ because (i) Their spans are each equal to $V$, and (ii) Each of them is a linearly independent set. That is, if $r_0,r_1,r_2,r_3\in \Bbb R$ and $\sum_{j=0}^3r_jH_j=0^*$ or $\sum_{j=0}^3r_jA_j=0^* $ then $r_0=r_1=r_2=r_3=0.$
Let $H=\{H_0,H_1,H_2,H_3\}.$ We have seen that $H$ is a basis for $V.$ If $0^*\ne v\in V$ and $v\not \in H,$ let $v'\in H$ such that deg$(v')=$deg$(v).$ Then $\{v\} \cup (H\setminus \{v'\})$ is also a basis for $V.$