Let \(V\) be the vector space over the complex numbers of all functions from \(\RR\) into \(C\), i.e., the space of all complex-valued functions on the real line. Let \(f_1(x) = 1\), \(f_2(x) = e^{ix}\), \(f_3(x) = e^{-ix}\).
Prove that \(f_1\), \(f_2\), and \(f_3\) are linearly independent.
Let \(g_1(x) = 1\), \(g_2(x) = \cos x\), \(g_3(x) = \sin x\). Find an invertible \(3 \times 3\) matrix \(P\) such that \[ g_j = \sum_{i=1}^{3} P_{ij} f_i. \]
Solution
1. The functions \(f_1,f_2,f_3\) are linearly independent.
Suppose \[ a+b e^{ix}+c e^{-ix}=0 \] for all real \(x\). Evaluating at \(x=0\) and \(x=\pi\) gives \[ a+b+c=0, \qquad a-b-c=0. \] Hence \(a=0\) and \(b+c=0\). Evaluating at \(x=\pi/2\) then gives \[ ib-ic=0, \] so \(b=c\). Together with \(b+c=0\), this yields \(2b=0\), hence \(b=c=0\). Thus the three functions are independent.
2. The required matrix is\[ P= \begin{bmatrix} 1&0&0\\ 0&\frac12&-\frac{i}{2}\\ 0&\frac12&\frac{i}{2} \end{bmatrix}. \]
Euler’s formulas give \[ 1=f_1, \qquad \cos x=\frac12 f_2+\frac12 f_3, \] \[ \sin x=-\frac{i}{2}f_2+\frac{i}{2}f_3. \] These are exactly the column relations \(g_j=\sum_iP_{ij}f_i\). Moreover \[ \det P=\frac{i}{2}\ne0, \] so \(P\) is invertible.