Inner Product in Hilbert Space

Question

Let $H$ be a Hilbert space and $\phi_{1}, \dots, \phi_{n} \in H$ are linearly independent vectors. How can we construct the inner product on $H$ such that $\phi_{1}, \dots, \phi_{n}$ become orthogonal and it still remains a Hilbert space?

Answer 1

Let $V$ be the span of $\phi_1,\dots,\phi_n$, and let $V^\perp$ be its orthogonal complement. Introduce new inner product on $V$ so that $\langle \phi_i,\phi_j \rangle_V = \delta_{ij}$ (this formula defines it for basis vectors; extend by linearity).

Every vector $u$ in $H$ is the sum of vectors $u_1\in V$ and $u_2\in V^\perp$. Define $$\langle u,v\rangle = \langle u_1,v_1\rangle_V+\langle u_2,v_2\rangle_H$$

What happens here is that the space $H$ is treated as a direct sum of two Hilbert spaces. Changing the inner product on the finite-dimensional summand does not change anything essential.