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

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 $$i in {1,2, cdots , n}$$?

Mathematics Asked on November 18, 2021

A basis $${e_1,dots,e_n}$$ of an inner product space $$E$$ is orthonormal iff $$langle e_i,e_jrangle=delta_{ij}$$, 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 $$e_i'=dfrac{e_i}{|e_i|}$$.

Answered by user598858 on November 18, 2021

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 $$mathcal{B}$$.

Answered by Danny Pak-Keung Chan on November 18, 2021

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 $$e_icdot e_j=delta_{ij}$$. 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.

Answered by Ross Millikan on November 18, 2021

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