Let $u_1,u_2,\dots,u_n$ be a basis for $\mathbb{C}^n$ show that it is an orthonormal basis for some inner product.

Question

Let $u_1,u_2,\dots,u_n$ be a basis for $\mathbb{C}^n$ show that it is an orthonormal basis with respect to some inner product.

How would i go about doing this?

Answer 1

Every complex vector $z \in \mathbb C^n$ can be written uniquely in the form $$z = \lambda_1 u_1 + \cdots + \lambda_n u_n$$ where each $\lambda_n \in \mathbb C$. You can define the inner product $$\langle \lambda_1 u_1 + \cdots + \lambda_n u_n, \mu_1 u_1 + \cdots + \mu_n u_n \rangle = \lambda_1 \bar{\mu_1} + \cdots + \lambda_n \bar{\mu_n}$$ and check it has the required properties.

Answer 2

There's a correspondence between inner-products and positive-definite matrices: $$ \langle x, y \rangle_A = x^* A y, $$ where $x, y \in \mathbb C^n$ and $x^*$ denotes conjugate-transpose. The usual inner-product on $\mathbb C^n$ has $A = I_n$, the identity matrix.

Let $e_1, \dots, e_n$ be the standard (orthonormal!) basis of $\mathbb C^n$; since $u_1, \dots, u_n$ and $e_1, \dots, e_n$ are bases, there exists a (change-of-basis) matrix $B$ such that $B u_i = e_i$ for all $i \in \{1, \dots, n\}$. Define the inner-product matrix by $B^* B$. (Is it positive-definite?) Then $$ \langle u_i, u_j \rangle_B = u_i^* B^* B u_j = (B u_i)^* (B u_j) = e_i^* e_j = \delta_{i,j}. $$ Hence $u_1, \dots, u_n$ is orthonormal with respect to the inner-product $\langle \cdot, \cdot \rangle_{B^* B}$.