Given a basis $\mathcal{B}$, can I assume that $\mathcal{B}$ is orthonormal?

Question

Let $E$ be a vector space over $\mathbb{C}$ such that $\text{dim}(E)=n \in \mathbb{N}$. Let $\mathcal{B}:=\{e_1,e_2,\cdots, e_n\}$ be a basis of $E$. I know that if $ E $ is vector space with inner product $\langle \cdot, \cdot \rangle : E \times E \longrightarrow \mathbb{C}$, then I can always assume, due to the Gram-Schmidt Process, that $ \mathcal{B}$ is an orthonormal basis, that is, $$||e_i||=1,\; \forall \; i \in \{1,2,\cdots, n\},$$ where $||x||=\sqrt{\langle x, x \rangle }$, for all $x \in E$.

Question. If $E$ it is not a space with an inner product, only a normed space (with norm $||\cdot||$), so I can also assume that $||e_i||=1$ for all $i \in \{1,2, \cdots , n\}$?

Answer 1

A basis $\{e_1,\dots,e_n\}$ of an inner product space $E$ is orthonormal iff $\langle e_i,e_j\rangle=\delta_{ij}$, not only $\|e_i\|=1$.

Next, as specifically your question says that "is it possible to assume $\|e_i\|=1$ for a basis $\{e_1,\dots,e_n\}$?". So it is true. You can assume. Because if not, then you can take elements of the basis as $e_i'=\dfrac{e_i}{\|e_i\|}$.

Answer 2

Even in an inner product space you cannot assume that a given basis is orthonormal. In $\Bbb R^2$ the vectors $(2,3)$ and $(1,4)$ make a fine basis. From them you can create an orthonormal basis if you want. If you want an orthornormal basis you need to ask that $e_i\cdot e_j=\delta_{ij}$. You only asked that all the basis vectors be unit vectors, but not that they be pairwise normal. In a normed space you do not have the concept of vectors being orthogonal. You can still ask that a basis have all the vectors of unit length. If it does not, you can divide each by its length to get a basis that does.

Answer 3

If $\mathcal{B}$ is a base, you cannot assume that the vectors in $\mathcal{B}$ are orthonormal. By going through Gram-Schmidt Process, you are working with another set of vectors. This new set of vectors are no longer elements in $\mathcal{B}$.