Linear Independence of Systems of Functions

Recall from the Orthogonal and Orthonormal Systems of Functions page that if $I$ is an interval in $\mathbb{R}$ then a collection of functions defined on $I$, $\{ \varphi_0, \varphi_1, \varphi_2, ... \}$ is said to be an orthogonal system of functions on $I$ if for all $i \neq j$ we have that $(\varphi_i, \varphi_j) = 0$.

Furthermore, we said that $\{ \varphi_0, \varphi_1, \varphi_2, ... \}$ is an orthonormal system of functions on $I$ if for all $i \neq j$ we have that $(\varphi_i, \varphi_j) = 0$ and for all $i = j$ we have that $(\varphi_i \varphi_j) = 1$.

We will now look at another important term regarding systems of functions.

Definition: Let $\{ \varphi_0, \varphi_1, \varphi_2, ..., \varphi_m \}$ be a finite system of functions on the interval $I$. Then this system is said to be Linearly Independent on $I$ if the equation $\displaystyle{\sum_{k=0}^{m} c_k \varphi_k(x) = c_0\varphi_0(x) + c_1\varphi_1(x) + ... + c_m\varphi_m(x) = 0}$ for all $x \in I$ implies that $c_0, c_1, ..., c_m = 0$. If instead we consider an infinite system of functions $\{ \varphi_0, \varphi_1, \varphi_2, ... \}$ on $I$, then this system is said to be linearly independent on $I$ if every finite subset of functions is linearly independent in the sense as defined above.

What's nice about linear independence is that every orthonormal system of functions on an interval $I$ is actually a linearly independent system of functions as well!

Theorem 1: Let $\{\varphi_0, \varphi_1, \varphi_2, ... \}$ be an orthonormal system of functions on the interval $I$. Then $\{\varphi_0, \varphi_1, \varphi_2, ... \}$ is a linearly independent system of functions on $I$.

• Proof: Let $\{\varphi_{i_0}, \varphi_{i_1}, \varphi_{i_2}, ..., \varphi_{i_m} \} \subset \{ \varphi_0, \varphi_1, \varphi_2, ... \}$ and suppose that for constants $c_0, c_1, c_2, ..., c_m$ we have that for all $x \in I$:

(1)

\begin{align} \quad \sum_{k=0}^{m} c_k \varphi_{i_k} (x) &= 0 \\ \quad c_0 \varphi_{i_0}(x) + c_1 \varphi_{i_1} (x) + ... + c_m \varphi_{i_m}(x) &= 0 \quad (*) \end{align}

• Since $\{\varphi_0, \varphi_1, \varphi_2, ... \}$ is an orthonormal system, we have that for all $n \in \{ 0, 1, 2, ... \}$ that $(\varphi_n, \varphi_n) = 1$. Take the equation $(*)$ above and take the inner product of it with $\varphi_{i_k}$ where $k \in \{ 0, 1, ..., m \}$. Then for each $k \in \{ 0, 1, ..., m \}$:

(2)

\begin{align} \quad (\varphi_{i_k}, c_0 \varphi_{i_0}+ c_1 \varphi_{i_1}+ ... + c_m \varphi_{i_m}) &= ( \varphi_{i_k}, 0) \\ \underbrace{(\varphi_{i_k}, c_0\varphi_{i_0})}_{=0} + \underbrace{(\varphi_{i_k}, c_1\varphi_{i_1})}_{=0} + ... + \underbrace{(\varphi_{i_k}, c_k\varphi_{i_k})}_{=\overline{c_k}} + ... + \underbrace{(\varphi_{i_k}, c_m \varphi_{i_m})}_{=0} &= 0 \\ \overline{c_k} &= 0 \end{align}

• Since $\overline{c_k} = 0$ this means that $c_k = 0$ for each $k$. In other words, $c_0, c_1, ..., c_m = 0$. So, every finite subset of $\{ \varphi_0, \varphi_1, \varphi_2, ... \}$ is linearly independent on $I$ and so by definition, $\{ \varphi_0, \varphi_1, \varphi_2, ... \}$ is linearly independent on $I$. $\blacksquare$