Spanning and linearly independence

A basis of a vector space V is subset $\{v_1,\cdots,v_n\} \subset V$ satisfying the two properties:
(i) $\{v_1,\cdots,v_n\}$ spans V;
(ii) $\{v_1,\cdots,v_n\}$ is linearly independent.

(i) To show spanning: Take an arbitrary vector $v \in V$, and show that $v$ can be written as a linear combination of $v_1, \cdots,v_n$, namely, $v=a_1v_1+\cdots+a_nv_n$.

(ii) To show linearly independence: Prove that $a_1v_1+\cdots+a_nv_n=0$ implies $a_1=a_2=\cdots=a_n=0$.

Let $au_1(x)+bu_2(x)=0$. Differentiating $\Rightarrow au_1'(x)+bu_2'(x)=0$.
If $W(u_1,u_2)=\begin{vmatrix} u_1 & u_2 \\ u_1' & u_2' \end{vmatrix}=u_1u_2'-u_2u_1' \neq 0$, then $a=b=0$ and $u_1$ and $u_2$ are linearly independent.
Note: $W(u_1, u_2)$ is known as the Wronskian determinant of functions $u_1$ and $u_2$.

Examples:
Spanning

$1$ and $i$ are linearly independent in $\mathbb{C}$ regarded as a vector space over $\mathbb{R}$ but they are linearly dependent if $\mathbb{C}$ is regarded as a complex vector space.

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$u_1(x)=\sin x$ and $u_2(x)=\cos x$ are linearly independent.

Proof: Let $a\sin x +b\cos x=0$.
Differentiating $\Rightarrow a\cos x-b\sin x=0$.
So $a=b \sin x (\cos x)^{-1} \Rightarrow b(\sin^2 x (\cos x)^{-1}+\cos x)=0$
$\Rightarrow b (\cos x)^{-1}=0 \Rightarrow b=0$ Thus $a=b=0$.

Alternatively, $W(u_1,u_2)=u_1u_2'-u_2u_1'=-\sin^2 x-\cos^2 x=-1 \Rightarrow$ linear independence.

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Show that the functions $x, xe^x, e^{-x}$ are linearly independent in the vector space $C^\infty(\mathbb{R})$.

Proof: Suppose $ax+bxe^x+ce^{-x}=0$ for all $x \in \mathbb{R}$, where $a,b,c$ are constants.

Method I – Differentiation:
$\begin{align}\: & ax+bxe^x+ce^{-x}=0 \; &(1)\\ \text{Differentiate both sides:}\: & a+be^x+bxe^x-ce^{-x}=0 \; &(2)\\ \text{Differentiate again:}\: & 2be^x+bxe^x+ce^{-x}=0 \; &(3)\\ \text{Differentiate yet again:}\: & 3be^x+bxe^x-ce^{-x}=0 \; &(4)\\ \text{And again:}\: & 4be^x+bxe^x+ce^{-x}=0 \; &(5) \end{align}$

$(5)-(3) \Rightarrow 2be^x=0 \Rightarrow b=0$
Substitute $b=0$ in $(3) \Rightarrow c=0$.
Substitue $b=c=0$ in $(2) \Rightarrow a=0$.

Method II – Limit:
For any $x \neq 0$ divide both sides by $xe^x$. Then, $ae^{-x}+b+cx^{-1}e^{-2x}=0$
L.H.S approaches $b$ as $x \to +\infty \Rightarrow b=0$.
Now $ax+ce^{-x}=0$ for all $x \in \mathbb{R}$.
For any $x \neq 0$ divide both sides by x. Then $a+cx^{-1}e^{-x}=0$.
L.H.S. approaches $a$ as $x \to +\infty \Rightarrow a=0$.
Now $ce^{-x}=0 \Rightarrow c=0$.

Linear Independence in function spaces

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Applications in ordinary differential equations [later]

Reference:
Linear independence