Answer to Question #121754 in Functional Analysis for Henry

Question

Please, this is urgent , solve it as soon as possible.

Suppose V is finite-dimensional and phi is a linear functional on V. Then
there is a unique vector u in V such that
phi(v)=< v, u> for every v in V.

Answer 1

The vector $u$  is unique. Let $u_1$  and $u_2$  be such that $\phi(v) = \langle v, u_1\rangle$  and $\phi(v) = \langle v, u_2\rangle$  for every $v\in V$. Then $\langle v, u_1-u_2\rangle = \langle v, u_1\rangle - \langle v, u_2\rangle = \phi(v) - \phi(v) = 0$  for every $v\in V$ . Substituting $u_1-u_2$  for $v$ , we get $\langle u_1-u_2, u_1-u_2\rangle = 0$ , hence $u_1-u_2 = 0$ .

The vector $u$  exists. Since $V$ is finite-dimensional, the Gram-Schmidt procedure gives an orthonormal basis $\{e_i\}_i$  of $V$. Choose $u = \sum_i \overline{\phi(e_i)} e_i$ .

Let $v\in V$. Since $\{e_i\}_i$  is a basis, there are scalars $\{a_i\}_i$  such that $v = \sum_i a_i e_i$ . Hence

$$\phi(v) = \phi\left(\sum_i a_i e_i\right) = \sum_i a_i \phi(e_i).$$

On the other hand,

$$\langle v, u\rangle = \left\langle \sum_i a_i e_i, \sum_i \overline{\phi(e_i)} e_i\right\rangle \\= \sum_i \sum_j a_i \phi(e_j) \langle e_i, e_j\rangle = \sum_i a_i \phi(e_i)$$

where the last equality follows from the orthonormality of $\{e_i\}_i$ . Therefore, $\phi(v) = \langle v, u\rangle$ for every $v\in V$.

The expression $\overline{a}$  denotes the complex conjugate of a complex scalar $a$  and is used for complex vector spaces. For real vector spaces, the proof is the same as above except that complex conjugation should be eliminated.