In mathematics, particularly linear algebra, an orthonormal basis for an inner product space V with finite dimension is a basis for V whose vectors are orthonormal, that is, they are all unit vectors and orthogonal to each other.^{[1]}^{[2]}^{[3]} For example, the standard basis for a Euclidean space R^{n} is an orthonormal basis, where the relevant inner product is the dot product of vectors. The image of the standard basis under a rotation or reflection (or any orthogonal transformation) is also orthonormal, and every orthonormal basis for R^{n} arises in this fashion.
For a general inner product space V, an orthonormal basis can be used to define normalized orthogonal coordinates on V. Under these coordinates, the inner product becomes a dot product of vectors. Thus the presence of an orthonormal basis reduces the study of a finitedimensional inner product space to the study of R^{n} under dot product. Every finitedimensional inner product space has an orthonormal basis, which may be obtained from an arbitrary basis using the Gram–Schmidt process.
In functional analysis, the concept of an orthonormal basis can be generalized to arbitrary (infinitedimensional) inner product spaces.^{[4]} Given a preHilbert space H, an orthonormal basis for H is an orthonormal set of vectors with the property that every vector in H can be written as an infinite linear combination of the vectors in the basis. In this case, the orthonormal basis is sometimes called a Hilbert basis for H. Note that an orthonormal basis in this sense is not generally a Hamel basis, since infinite linear combinations are required. Specifically, the linear span of the basis must be dense in H, but it may not be the entire space.
If we go on to Hilbert spaces, a nonorthonormal set of vectors having the same linear span as an orthonormal basis may not be a basis at all. For instance, any squareintegrable function on the interval [−1, 1] can be expressed (almost everywhere) as an infinite sum of Legendre polynomials (an orthonomal basis), but not necessarily as an infinite sum of the monomials x^{n}.
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Transcription
Let's say I've got me a set of vectors. So let me call my set B. And let's say I have the vectors v1, v2, all the way through vk. Now let's say this isn't just any set of vectors. There's some interesting things about these vectors. The first thing is that all of these guys have length of 1. So we could say the length of vector vi is equal to 1 for i is equal to well we could say between 1 and k or i is equal to 1, 2, all the way to k. All of these guys have length equal 1. Or another way to say it is that the square of their lengths are 1. The square of a vi whose length is equal to 1. Or vi dot vi is equal to 1 for i is any of these guys. Any i can be 1, 2, 3, all the way to k. So that's the first interesting thing about it. Let me write it in regular words. All the vectors in B have length 1. Or another way to say is that they've all been normalized. That's another way to say that is that they have all been normalized. Or they're all unit vectors. Normalized vectors are vectors that you've made their lengths 1. You're turned them into unit vectors. They have all been normalized. So that's the first interesting thing about my set, B. And then the next interesting thing about my set B is that all of the vectors are orthogonal to each other. So if you dot it with itself, if you dot a vector with itself, you get length 1. But if you take a vector and dot it with any other vector if you take vi and you were to dot it with vj. So if you took v2 and dotted it with v1, it's going to be equal to 0 for i does not equal j. All of these guys are orthogonal. Let me write that down. All of the vectors are orthogonal to each other. And of course they're not orthogonal to themselves because they all have length 1. So if you take the dot product with itself, you get 1. If you take a dot product with some other guy in your set you're going to get 0. Maybe I can write it this way. vi dot vj for all the members of the set is going to be equal to 0 for i does not equal j. And then if these guys are the same vector I'm dotting with myself I'm going to have length 1. So it would equal length 1 for i is equal to j. So I've got a special set. All of these guys have length 1 and they're all orthogonal with each other. They're normalized and they're all orthogonal. And we have a special word for this. This is called an orthonormal set. So B is an orthonormal set. Normal for normalized. Everything is orthogonal. They're all orthogonal relative to each other. And everything has been normalized. Everything has length 1. Now, the first interesting thing about an orthonormal set is that it's also going to be a linearly independent set. So if B is orthonormal, B is also going to be linearly independent. And how can I show that to you? Well let's assume that it isn't linearly independent. Let me take vi, let me take vj that are members of my set. And let's assume that i does not equal j. Now, we already know that it's an orthonormal set. So vi dot vj is going to be equal to 0. They are orthogonal. These are two vectors in my set. Now, let's assume that they are linearly dependent. I want to prove that they are linearly independent and the way I'm going to prove that is by assuming they are linearly dependent and then arriving at a contradiction. So let's assume that vi and vj are linearly dependent. Well then that means that I can represent one of these guys as a scalar multiple the other. And I can pick either way. So let's just say, for the sake of argument, that I can represent vi let's say that vi is equal to sum scalar c times vj. That's what linear dependency means. That one of them can be represented as a scalar multiple of the other. Well if this is true, then I can just substitute this back in for vi. And what do I get? I get c times vj which is just another way of writing vi because I assumed linear dependence. That dot vj has got to be equal to 0. This guy was vi. This is vj. They are orthogonal to each other. But this right here is just equal to c times vj dot vj which is just equal to c times the length of vj squared. And that has to equal 0. They are orthogonal so that has to equal 0. Which implies that the length of vj has to be equal to 0. If we assume that this is some nonzero multiple, and this has to be some nonzero multiple I should have written it there c does not equal 0. Why does this have to be a nonzero multiple? Because these were both nonzero vectors. This is a nonzero vector. So this guy can't be 0. This guy has length 1. So if this is a nonzero vector, there's no way that I can just put a 0 here. Because if I put a 0 then I would get a 0 vector. So c can't be 0. So if c isn't 0, then this guy right here has to be 0. And so we get that the length of vj is 0. Which we know is false. The length of vj is 1. This is an orthonormal set. The length of all of the members of B are 1. So we reach a contradiction. This is our contradiction. Vj is not the 0 vector. It has length 1. Contradiction. So if you have a bunch of vectors that are orthogonal and they're nonzero, they have to be linearly independent. Which is pretty interesting. So if I have this set, this orthonormal set right here, it's also a set of linearly independent vectors, so it can be a basis for a subspace. So let's say that B is the basis for some subspace, v. Or we could say that v is equal to the span of v1, v2, all the way to vk. Then we called B if it was just a set, we'd call it a orthonormal set, but it can be an orthonormal basis when it's spans some subspace. So we can write, we can say that B is an orthonormal basis for v. Now everything I've done is very abstract, but let me do some quick examples for you. Just so you understand what an orthonormal basis looks like with real numbers. So let's say I have two vectors. Let's say I have the vector, v1, that is say we're dealing in R3 so it's 1/3, 2/3, 2/3 and 2/3. And let's say I have another vector, v2, that is equal to 2/3, 1/3, and minus 2/3. And let's say that B is the set of v1 and v2. So the first question is, is what are the lengths of these guys? So let's take the length. The length of v1 squared is just v1 dot v1. Which is just 1/3 squared, which is just 1 over 0. Plus 2/3 squared, which is 4/9. Plus 2/3 squared, which is 4/9. Which is equal to 1. So if the length squared is 1, then that tells us that the length of our first vector is equal to 1. If the square of the length is 1, you take the square root, so the length is 1. What about vector 2? Well the length of vector 2 squared is equal to v2 dot v2. Which is equal to let's see, two 2/3 squared is 4/9 plus 1/3 squared is 1/9. Plus 2/3 squared is 4/9. So that is 9/9, which is equal to 1. Which tells us that the length of v2, the length of vector v2 is equal to 1. So we know that these guys are definitely normalized. We can call this a normalized set. But is it an orthonormal set? Are these guys orthogonal to each other? And to test that out we just take their dot product. So v1 dot v2 is equal to 1/3 times 2/3, which is 2/9. Plus 2/3 times 1/3, which is 2/9. Plus 2/3 times the minus 2/3. That's minus 4/9. 2 plus 2 minus 4 is 0. So it equals 0. So these guys are indeed orthogonal. So B is an orthonormal set. And if I have some subspace, let's say that B is equal to the span of v1 and v2, then we can say that the basis for v, or we could say that B is an orthonormal basis. for V.
Contents
Examples
 The set of vectors {e_{1} = (1, 0, 0), e_{2} = (0, 1, 0), e_{3} = (0, 0, 1)} (the standard basis) forms an orthonormal basis of R^{3}.
 Proof: A straightforward computation shows that the inner products of these vectors equals zero, ⟨e_{1}, e_{2}⟩ = ⟨e_{1}, e_{3}⟩ = ⟨e_{2}, e_{3}⟩ = 0 and that each of their magnitudes equals one, e_{1} = e_{2} = e_{3} = 1. This means that {e_{1}, e_{2}, e_{3}} is an orthonormal set. All vectors (x, y, z) in R^{3} can be expressed as a sum of the basis vectors scaled
 so {e_{1}, e_{2}, e_{3}} spans R^{3} and hence must be a basis. It may also be shown that the standard basis rotated about an axis through the origin or reflected in a plane through the origin forms an orthonormal basis of R^{3}.
 Proof: A straightforward computation shows that the inner products of these vectors equals zero, ⟨e_{1}, e_{2}⟩ = ⟨e_{1}, e_{3}⟩ = ⟨e_{2}, e_{3}⟩ = 0 and that each of their magnitudes equals one, e_{1} = e_{2} = e_{3} = 1. This means that {e_{1}, e_{2}, e_{3}} is an orthonormal set. All vectors (x, y, z) in R^{3} can be expressed as a sum of the basis vectors scaled
 Notice that an orthogonal transformation of the standard innerproduct space can be used to construct other orthogonal bases of .
 The set {f_{n} : n ∈ Z} with f_{n}(x) = exp(2πinx) forms an orthonormal basis of the space of functions with finite Lebesgue integrals, L^{2}([0,1]), with respect to the 2norm. This is fundamental to the study of Fourier series.
 The set {e_{b} : b ∈ B} with e_{b}(c) = 1 if b = c and 0 otherwise forms an orthonormal basis of ℓ^{2}(B).
 Eigenfunctions of a Sturm–Liouville eigenproblem.
 An orthogonal matrix is a matrix whose column vectors form an orthonormal set.
Basic formula
If B is an orthogonal basis of H, then every element x of H may be written as
When B is orthonormal, this simplifies to
and the square of the norm of x can be given by
Even if B is uncountable, only countably many terms in this sum will be nonzero, and the expression is therefore welldefined. This sum is also called the Fourier expansion of x, and the formula is usually known as Parseval's identity.
If B is an orthonormal basis of H, then H is isomorphic to ℓ^{ 2}(B) in the following sense: there exists a bijective linear map Φ : H → ℓ^{ 2}(B) such that
for all x and y in H.
Incomplete orthogonal sets
Given a Hilbert space H and a set S of mutually orthogonal vectors in H, we can take the smallest closed linear subspace V of H containing S. Then S will be an orthogonal basis of V; which may of course be smaller than H itself, being an incomplete orthogonal set, or be H, when it is a complete orthogonal set.
Existence
Using Zorn's lemma and the Gram–Schmidt process (or more simply wellordering and transfinite recursion), one can show that every Hilbert space admits a basis, but not orthonormal base^{[5]}; furthermore, any two orthonormal bases of the same space have the same cardinality (this can be proven in a manner akin to that of the proof of the usual dimension theorem for vector spaces, with separate cases depending on whether the larger basis candidate is countable or not). A Hilbert space is separable if and only if it admits a countable orthonormal basis. (One can prove this last statement without using the axiom of choice).
As a homogeneous space
The set of orthonormal bases for a space is a principal homogeneous space for the orthogonal group O(n), and is called the Stiefel manifold of orthonormal nframes.
In other words, the space of orthonormal bases is like the orthogonal group, but without a choice of base point: given an orthogonal space, there is no natural choice of orthonormal basis, but once one is given one, there is a onetoone correspondence between bases and the orthogonal group. Concretely, a linear map is determined by where it sends a given basis: just as an invertible map can take any basis to any other basis, an orthogonal map can take any orthogonal basis to any other orthogonal basis.
The other Stiefel manifolds for of incomplete orthonormal bases (orthonormal kframes) are still homogeneous spaces for the orthogonal group, but not principal homogeneous spaces: any kframe can be taken to any other kframe by an orthogonal map, but this map is not uniquely determined.
See also
References
 ^ Lay, David C. (2006). Linear Algebra and Its Applications (3rd ed.). Addison–Wesley. ISBN 0321287134.
 ^ Strang, Gilbert (2006). Linear Algebra and Its Applications (4th ed.). Brooks Cole. ISBN 0030105676.
 ^ Axler, Sheldon (2002). Linear Algebra Done Right (2nd ed.). Springer. ISBN 0387982582.
 ^ Rudin, Walter (1987). Real & Complex Analysis. McGrawHill. ISBN 0070542341.
 ^ Linear Functional Analysis Authors: Rynne, Bryan, Youngson, M.A. page 79