Prove that the set of even and odd functions is linearly independent I was trying to prove this: "If we have the set of even functions: $\{f_1,\ldots,f_n\}$ (where $f_i(-x)=f_i(x)$) and odd functions $\{t_1,\ldots,t_m\}$ where $t_k(-x) )=-t_k(x)$. Both sets linearly independent. Prove that $\{f_1,\ldots,t_m\}$is linearly independent."
My idea was to write the last set as a linear combination and set it equal to $0$: $$ a_1 f_1+ a_2f_2+\cdots+ a_n f_ n +b_1 t_1+\cdots b_m t_m= 0$$
Then I can pass the terms containing $t_i$ to the other side of the equality and work with the property that $t_i$ is an odd function. however I am unable to continue after that point. I would appreciate some help
 A: We first note the following fact:
Claim 1: Any linear combination of a set of even functions is even, and any linear combination of a set of odd functions is odd.
We now use this to show that the set of even functions and the set of odd functions as given are linearly independent. Let $f_1,\ldots, f_n$ be the even functions and $g_1,\ldots, g_m$ be the odd functions. Then suppose that there are constants $a_1,\ldots, a_n, b_1,\ldots, b_m$ not all $0$, such that the equation
$$\sum^{n}_{j=1} a_jf_j(x) \ + \ \sum^m_{i=1}b_ig_i(x) \ = \ 0 \quad \forall x \in \mathbb{R}$$
holds.
So now, $F$ be the sum of the even functions i.e.,
$$F=\sum^n_{j=1}a_jf_j,$$
and let $G$ be the sum of the odd functions i.e.,.
$$G=\sum^m_{i=1}b_ig_i.$$
Then on the one hand,
$$F+G$$ $$= \sum^{n}_{j=1} a_jf_j(x) \ + \ \sum^m_{i=1}b_ig_i(x)$$ $$= 0 \quad \forall x.$$ On the other hand, by linear independence of $f_1,\ldots, f_n$, the function $F$ is not everywhere $0$. So let $x'$ be such that $F(x')=c$; $c \not = 0$. By Claim 1, $F$ is itself even, so the equation $F(-x')=c$ must hold. But the fact that $F(x)+G(x)$ is $0$ for all $x$ implies the equations
$$G(x')=-F(x)=-c \not = 0$$ and $$G(-x') = -F(-x)$$ $$= -F(x)=-c \not = 0.$$ So then this implies that $G$ cannot be odd, as the above implies $G(x')=G(-x') \not = -G(-x')$. But then $G$ is odd by Claim 1, so this is impossible. So we arrive at a contradiction. Thus there cannot be such constants $a_1,\ldots, a_n,b_1,\ldots, b_m$ as specified above after all, which implies that the entire set $\{f_1,f_2,\ldots, f_n,g_1,\ldots, g_m\}$ are linearly independent if each of $\{f_1,\ldots, f_n\}$ and $\{g_1,\ldots, g_m\}$ are linearly independent.
