Let \(f_1,\ldots,f_n\) be continuous real-valued functions on \([a,b]\). Show that \(\{f_1,\ldots,f_n\}\) is linearly dependent on \([a,b]\) if and only if \[ \det\left(\int_a^b f_i(x)f_j(x)\,dx\right)_{i,j=1}^n=0. \]
Solution
Let \[ G \coloneqq \left( \int_a^b f_i(x)f_j(x)\,dx \right)_{i,j=1}^n. \]
1. If \(f_1,\ldots,f_n\) are linearly dependent, then\[ \det G=0. \]
Choose real numbers \[ c_1,\ldots,c_n, \] not all zero, such that \[ \sum_{j=1}^n c_jf_j(x)=0 \] for every \(x\in[a,b]\). Let \[ c=(c_1,\ldots,c_n)^t. \] For each \(i\), \[ \begin{aligned} (Gc)_i &= \sum_{j=1}^n c_j \int_a^b f_i(x)f_j(x)\,dx\\ &= \int_a^b f_i(x) \left( \sum_{j=1}^n c_jf_j(x) \right) \,dx\\ &=0. \end{aligned} \] Thus \[ Gc=0 \] with \(c\neq0\), so \(G\) is singular and \(\det G=0\).
2. Suppose\[ \det G=0. \]Then there is a nonzero vector\[ c=(c_1,\ldots,c_n)^t\in\RR^n \]such that\[ Gc=0. \]
A square matrix has determinant zero exactly when its kernel is nontrivial.
3. For the vector \(c\) from step 2,\[ \int_a^b \left( \sum_{j=1}^n c_jf_j(x) \right)^2 \,dx =0. \]
Since \(Gc=0\), \[ c^tGc=0. \] Expanding the left side, \[ \begin{aligned} c^tGc &= \sum_{i,j=1}^n c_ic_j \int_a^b f_i(x)f_j(x)\,dx\\ &= \int_a^b \left( \sum_{i=1}^n c_if_i(x) \right) \left( \sum_{j=1}^n c_jf_j(x) \right) \,dx, \end{aligned} \] which is the displayed integral.
4. One has\[ \sum_{j=1}^n c_jf_j(x)=0 \]for every \(x\in[a,b]\).
The function \[ F(x) \coloneqq \left( \sum_{j=1}^n c_jf_j(x) \right)^2 \] is continuous and nonnegative. Step 3 gives \[ \int_a^bF(x)\,dx=0. \] If \(F(x_0)>0\) at some point, continuity would make \(F\) positive on a nonempty interval around \(x_0\), forcing its integral to be positive. Hence \(F\equiv0\), which gives the claim.
5. If \(\det G=0\), then \(f_1,\ldots,f_n\) are linearly dependent.
The coefficients \(c_j\) from step 2 are not all zero, and step 4 shows that their linear combination of the \(f_j\) vanishes identically.
6. Q.E.D.
Step 1 proves one implication and step 5 proves the converse.