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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

3 Answers

3 Answers

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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