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Preamble

I know from linear algebra that any vector in a vector space can be written as a linear combination of basis vectors. However, in physics, unit vectors are used as basis vectors, which brings me to my question. Note: I do not have the math tools to formally prove any of this... I just need an explanation

The question

Which is correct (and why):

  • Are all unit vectors basis vectors?
  • Are all basis vector unit vectors?
  • Are unit vectors a subset of basis vectors?
  • Are basis vectors a subset of unit vectors?

In a nutshell if the set of all basis vectors was b and the set of all unit vectors was u, what is the relation between the two sets? (i.e. b ___ u)

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    $\begingroup$ Do you know what a basis of a vector space is? Read it up and everything should get clear. $\endgroup$ – dtell Jun 29 '17 at 23:57
  • $\begingroup$ Unit vector = vector of length 1. Basis of a vector space = set of linearly independent vectors spanning the vector space. $\endgroup$ – dromastyx Jun 30 '17 at 0:02
  • $\begingroup$ how does that relate to basis $\endgroup$ – Edward Jun 30 '17 at 0:02
  • $\begingroup$ For every given basis one can construct a basis of unit (and orthogonal as well) vectors using Gram-Schmidt process. Therefore, physicists say, without loss of generality, let assume that the basis is $\textbf{orthonormal}$. $\endgroup$ – user 1987 Jun 30 '17 at 0:24
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The term unit vector is well-defined: a vector of length one. The term basis vector is taken out of context, and doesn't make sense on its own.

A basis of a vector space $V$ is a set of vectors $\left\{v_1,\dots, v_n\right\}$ such each vector $v\in V$ can be written uniquely as a linear combination of $v_1,\dots, v_n$. So if you read “basis vector”, it means “member of the aforementioned basis,” and which basis is being referenced comes from the context.

  • Are all unit vectors basis vectors? No. The vector $e_1 = (1,0,0)$ in $\mathbb{R}^3$ is a unit vector. But on its own it's not a “basis vector.” There is a basis of $\mathbb{R}^3$ which contains $e_1$—for instance $\left\{e_1, e_2, e_3\right\}$. In general, for any unit vector $v$ we can find a basis of $V$ that has $v$ as a member.

  • Are all basis vectors unit vectors? The best way to interpret this question is: Does every basis consist of unit vectors? The answer is no. For instance $\left\{e_1, e_2, e_1+e_3\right\}$ is a basis of $\mathbb{R}^3$ too, and not all of its elements are unit vectors.

  • Are unit vectors a subset of basis vectors? No. However, there is an additional property of vectors that's related. Vectors $v$ and $w$ are orthogonal if their inner (dot) product $v\cdot w= 0$. A set of vectors is orthonormal if every single vector in the set is a unit vector, and any pair of vectors in the set are orthogonal. An orthonormal set of $n$ vectors in an $n$-dimensional vector space is automatically a basis.

  • Are basis vectors a subset of unit vectors? No. However, if you start with a basis $v_1, \dots, v_n$, there exists an orthonormal basis $u_1, \dots, u_n$ such that for each $k$ between $1$ and $n$, the span of $u_1, \dots, u_k$ is the same as the span of $v_1, \dots, v_k$. The algorithm which constructs $u_1, \dots, u_n$ is called the Gram-Schmidt process. Because of this property, the physicists might assume that any given basis is orthonormal.

I hope that clarifies some of these terms.

Edit You asked for some more explanation on the third point. The standard basis $\hat{\mathbf x}, \hat{\mathbf y}, \hat{\mathbf z}$ (using your notation) of $\mathbb{R}^n$ is orthonormal. Put your hand in the right-hand-rule position with these three vectors, and rotate your hand. The vectors change, but not the property that each of them is unit length, and that each pair are perpendicular/orthogonal.

Now let $\mathbf{v}$ be any vector. If $\mathbf{v}$ can be written as a linear combination $a \hat{\mathbf x} + b \hat{\mathbf y} + c \hat{\mathbf z}$, then what are $a$, $b$, and $c$? Use the orthonormality property: \begin{align*} \mathbf{v}\cdot\hat{\mathbf x} &= (a \hat{\mathbf x} + b \hat{\mathbf y} + c \hat{\mathbf z})\hat{\mathbf x}\\ &= a \hat{\mathbf x}\cdot \hat{\mathbf x} + b \hat{\mathbf y}\cdot \hat{\mathbf x} + c \hat{\mathbf z}\cdot \hat{\mathbf x} \\ &= a (1) + b(0) + c(0) = a \end{align*} [Since $\hat{\mathbf x}$ is a unit vector, $\hat{\mathbf x}\cdot\hat{\mathbf x} = \left\Vert\hat{\mathbf x}\right\Vert^2 = 1$.] You can do the same for $\hat{\mathbf y}$ and $\hat{\mathbf z$}$. Therefore the coefficients of $\mathbf{v}$ can be recovered by the dot products: $$ \mathbf{v} = (\mathbf{v}\cdot\hat{\mathbf x})\hat{\mathbf x} + (\mathbf{v}\cdot\hat{\mathbf y})\hat{\mathbf x} + (\mathbf{v}\cdot\hat{\mathbf z})\hat{\mathbf x} $$

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    $\begingroup$ Thanks you! Can you please explain "A set of vectors is orthonormal if every single vector in the set is a unit vector, and any pair of vectors in the set are orthogonal. An orthonormal set of nn vectors in an nn-dimensional vector space is automatically a basis" more because I think this is why physics does this: a = x'' x_hat + y'' y_hat + z'' z_hat (a linear combination of acceleration for example) $\endgroup$ – Edward Jun 30 '17 at 15:47
  • $\begingroup$ @EdwardSmith see edits. $\endgroup$ – Matthew Leingang Jul 1 '17 at 10:18
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Being a "basis vector" is not a property of vectors on their own, so your questions have no clear answer.

Take, for instance, the real plane. One basis here would be $\{(1,0),(1,1)\}$. So in this case, those two would be basis vectors.

On the other hand, if you take $\{(2,2),(1,1)\}$, then this set of vectors forms no basis, and thus there's no reason to call either a "basis vector".

In general, a basis is something that you can chose for any given vector space - any set of vectors that is both linearly independant (no linear combination of them except with all zero coefficients adds to 0) and spans the vector space (any vector can be expressed as linear combination).

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