Let \(E\) be a finite-dimensional complex vector space with Hermitian inner product \(\langle x,y\rangle\).
- Prove that \(E\) has an orthonormal basis.
- Let \(f:E\times E\to\mathbb C\) be linear in the first variable and conjugate-linear in the second. Show that there is a linear map \(A:E\to E\) such that \[ f(x,y)=\langle Ax,y\rangle \] for all \(x,y\in E\).
Solution
Let \[ m\coloneqq\dim_\CC E. \] If \(m=0\), the empty set is an orthonormal basis and part (2) holds for the unique map \(A:0\to0\). Assume henceforth that \(m\ge1\).
1. Every linearly independent family\[ v_1,\ldots,v_r \]in \(E\) can be converted into an orthonormal family\[ e_1,\ldots,e_r \]with\[ \operatorname{span}(e_1,\ldots,e_j) = \operatorname{span}(v_1,\ldots,v_j) \]for every \(j\).
Apply the Gram–Schmidt construction. Set \[ u_1=v_1, \qquad e_1=\frac{u_1}{\|u_1\|}. \] Inductively, after constructing \[ e_1,\ldots,e_{j-1}, \] define \[ u_j \coloneqq v_j-\sum_{i=1}^{j-1} \langle v_j,e_i\rangle e_i. \] For \[ 1\le k<j, \] orthonormality gives \[ \begin{aligned} \langle u_j,e_k\rangle &= \langle v_j,e_k\rangle - \sum_{i=1}^{j-1} \langle v_j,e_i\rangle \langle e_i,e_k\rangle\\ &=0. \end{aligned} \] Moreover \(u_j\ne0\): otherwise \(v_j\) would belong to the span of \(v_1,\ldots,v_{j-1}\), contradicting linear independence. Hence \[ e_j\coloneqq\frac{u_j}{\|u_j\|} \] is defined, has norm \(1\), and is orthogonal to the preceding vectors. The defining formula also gives the asserted equality of spans at each stage.
2. The space \(E\) has an orthonormal basis.
Choose any basis \[ v_1,\ldots,v_m \] of \(E\). Step 1 produces an orthonormal family \[ e_1,\ldots,e_m \] with the same span as the original basis. Hence it is an orthonormal basis of \(E\). This proves part (1).
3. Fix an orthonormal basis\[ e_1,\ldots,e_m \]as in step 2 and define\[ Ax \coloneqq \sum_{i=1}^m f(x,e_i)e_i. \]Then\[ A:E\to E \]is linear.
For \[ x,x'\in E \] and \[ \alpha,\beta\in\CC, \] linearity of \(f\) in its first variable gives \[ \begin{aligned} A(\alpha x+\beta x') &= \sum_{i=1}^m f(\alpha x+\beta x',e_i)e_i\\ &= \sum_{i=1}^m \bigl( \alpha f(x,e_i)+\beta f(x',e_i) \bigr)e_i\\ &= \alpha Ax+\beta Ax'. \end{aligned} \]
4. For every\[ y=\sum_{i=1}^m y_i e_i, \]one has\[ f(x,y) = \sum_{i=1}^m \overline{y_i}\,f(x,e_i). \]
The form \(f\) is conjugate-linear in its second variable. Therefore \[ \begin{aligned} f(x,y) &= f\left( x, \sum_{i=1}^m y_i e_i \right)\\ &= \sum_{i=1}^m \overline{y_i}\,f(x,e_i). \end{aligned} \]
5. For all\[ x,y\in E, \]the map \(A\) from step 3 satisfies\[ \boxed{ f(x,y)=\langle Ax,y\rangle. } \]
4. This proves part (2).
6. Q.E.D.
Step 2 proves existence of an orthonormal basis, and steps <1>3–<1>5 construct the required linear operator and verify its representation formula.