# Problem: Linear Independence and Basis in Fun(\(\mathbb{R}, \mathbb{R}\)) Given the set \( S = \{ e^{cx} \mid c \in \mathbb{R} \} \) in the vector space \( \mathrm{Fun}(\mathbb{R}, \mathbb{R}) \), determine: - Whether \( S \) is linearly independent. - Whether \( S \) is a basis for \( \mathrm{Fun}(\mathbb{R}, \mathbb{R}) \). --- ## 1. Complete Solution with Calculations ### 1.1. **Linear Independence** #### **Definition** A set \(\{f_i\}\) in a vector space \(V\) is **linearly independent** if for any finite set of distinct \(f_1, \dots, f_n\) and scalars \(a_1, \dots, a_n\), the equation \[ a_1 f_1(x) + a_2 f_2(x) + \dots + a_n f_n(x) = 0 \text{ for all } x \] implies that \(a_1 = a_2 = \dots = a_n = 0\). #### **Apply to \(S = \{e^{c x}\}\)** Take any finite subset \( \{e^{c_1 x}, \dots, e^{c_n x}\} \) with \( c_1, \dots, c_n \) distinct real numbers, and consider: \[ a_1 e^{c_1 x} + a_2 e^{c_2 x} + \dots + a_n e^{c_n x} = 0 \qquad \forall x \in \mathbb{R} \] Divide both sides by \( e^{c_1 x} \) (since it is never zero): \[ a_1 + a_2 e^{(c_2-c_1)x} + a_3 e^{(c_3-c_1)x} + \dots + a_n e^{(c_n - c_1)x} = 0 \qquad \forall x \] Set \( d_i = c_i - c_1 \), so \( d_1 = 0, d_2 \neq 0, \dots, d_n \neq 0 \), and the equation becomes: \[ a_1 + a_2 e^{d_2 x} + a_3 e^{d_3 x} + \dots + a_n e^{d_n x} = 0 \] The set \( \{1, e^{d_2 x}, \dots, e^{d_n x}\} \) is linearly independent on \( \mathbb{R} \). This can be shown using the theory of exponential polynomials (Vandermonde argument) or by differentiating at \( x = 0 \) and forming a system of equations whose determinant (a Vandermonde determinant) is nonzero if all \( d_i \) are distinct. Thus, all \( a_i = 0 \). Therefore, **the set \( S \) is linearly independent**. --- ### 1.2. **Is \( S \) a Basis for \( \mathrm{Fun}(\mathbb{R}, \mathbb{R}) \)?** #### **Definition** A set is a **basis** for \( V \) if it is linearly independent and **spans** \( V \). #### **Spanning \( \mathrm{Fun}(\mathbb{R}, \mathbb{R}) \)** The space \( \mathrm{Fun}(\mathbb{R}, \mathbb{R}) \) consists of **all** functions from \( \mathbb{R} \) to \( \mathbb{R} \). To span \( \mathrm{Fun}(\mathbb{R}, \mathbb{R}) \), any function \( f: \mathbb{R} \to \mathbb{R} \) must be expressible as a finite linear combination of elements of \( S \): \[ f(x) = \sum_{i=1}^n a_i e^{c_i x} \] for some \( a_i \in \mathbb{R} \), \( c_i \in \mathbb{R} \). However, not all functions can be written in this way. For example: - The function \( f(x) = x \) cannot be written as a finite linear combination of exponentials \( e^{c x} \). - The function \( f(x) = \sin x \) cannot be written as such a combination (unless complex exponents are permitted, and even then, only certain functions can be represented). - Indicator functions, step functions, and many others cannot be written as finite sums of exponential functions. Therefore, **\( S \) does not span \( \mathrm{Fun}(\mathbb{R}, \mathbb{R}) \)**. --- ## 2. Explanation of Each Part ### 2.1. **Linear Independence** - Any finite linear combination of distinct exponentials \( e^{c x} \) is zero for all \( x \) only if all coefficients are zero. - This follows from the properties of the exponential function and the linear independence of exponentials with distinct exponents. ### 2.2. **Spanning** - To span \( \mathrm{Fun}(\mathbb{R}, \mathbb{R}) \), every function must be a finite (not infinite) linear combination of elements from \( S \). - Most functions, such as polynomials, trigonometric, or piecewise functions, cannot be written as finite sums of exponentials with real exponents. --- ## 3. Final Conclusion - **The set \( S = \{ e^{c x} \mid c \in \mathbb{R} \} \) is linearly independent in \( \mathrm{Fun}(\mathbb{R}, \mathbb{R}) \).** - **The set \( S \) is not a basis for \( \mathrm{Fun}(\mathbb{R}, \mathbb{R}) \) because it does not span the space.**