Talk:Change of basis

This article leaves much to be desired

It is too obstruse and esoteric to be useful. This is a simple topic, can be presented with very simple examples in low-dimensional cases with basic linear algebra, and then more abstract and less relevant (to the majority of users) bits can then be given to expand on the basic simple points. Really what a tragedy when a beautiful topic is been presented in such a way as to elicit a gag reflex in 99/100 readers of the site. Truly frustrating. — Preceding unsigned comment added by 152.16.191.74 (talk) 07:59, 3 August 2015 (UTC)

Something missing?

Is there a phrase missing from the first sentence of https://en.wikipedia.org/wiki/Change_of_basis#Three_dimensions? --Ott0 (talk) 06:52, 2 November 2015 (UTC)

Yes, I think. I added the missing words "let R". Dexo568 (talk) 22:37, 4 November 2015 (UTC)

Preliminary notions statement seems cyclical

In the second line of this section, namely: If T: Rⁿ → R^m is a linear transformation...

What then is the matrix of T(e_j)? — Preceding unsigned comment added by 128.84.126.62 (talk) 09:01, 23 December 2016 (UTC)

General case, derived using Clifford algebra

A new section, "General case, derived using Clifford algebra," was recently inserted by Twy2008 (talk). I recommend removing it in its entirety, as it is far too complicated for this article and hence, not helpful.—Anita5192 (talk) 03:34, 27 January 2020 (UTC)

Okay, I know how these kinds of talks go and they lead to removal. So, I removed it from the page. I have put a copy of it below for reference or in case someone would like to improve it and maybe re-add it in some revised form.Twy2008 (talk) 05:37, 27 January 2020 (UTC)

General case, derived using Clifford algebra

In Clifford algebra (or Geometric Algebra), we can write any vector $\mathbf {a}$ in an $n$-dimensional real Euclidean space $\mathbb {R} ^{n}$ as $\mathbf {a} =a^{i}\mathbf {e} _{i}=a_{i}\mathbf {e} ^{i}$, $1\leq i\leq n$ (Einstein summation assumed), with scalar coefficients (coordinates) $a^{i}=a_{i}$ on a standard orthonormal basis $E=\{\mathbf {e} _{i}\}$ with reciprocal basis $\{\mathbf {e} ^{i}=\mathbf {e} _{i}\}$. On the standard basis $E$, the inner (dot) products are $[\mathbf {e} _{i}\cdot \mathbf {e} ^{j}]=[\delta _{i}^{j}]=\mathbf {I} _{n}$, where $\delta _{i}^{j}$ is the Kronecker delta and $\mathbf {I} _{n}$ is the $n\times n$ identity matrix. Generally, on any basis $W=\{\mathbf {w} _{i}\}$ with reciprocal basis $\mathbf {w} ^{i}$, we have $[\mathbf {w} _{i}\cdot \mathbf {w} ^{j}]=[\delta _{i}^{j}]=\mathbf {I} _{n}$. Although it is possible to generalize $\mathbb {R} ^{n}$ to $\mathbb {F} ^{n}$, the $n$-dimensional vector space over any field $\mathbb {F}$ with inner product, only $\mathbb {R} ^{n}$ is considered in the following. A standard orthonormal basis $E=\{\mathbf {e} _{i}\}$ is always assumed to exist since one can always be found by using the Gram-Schmidt process. In fact, the linear (matrix) algebra representation of column vectors $[a^{i}]$ and their dual row vectors $[a_{i}]$ often assumes an orthonormal basis where $\mathbf {e} _{i}=\mathbf {e} ^{i}$ and $a^{i}=a_{i}$ and $\mathbf {a} \cdot \mathbf {a} =a^{i}a_{i}$. However, the general case cannot assume an orthonormal basis and the inner product $\mathbf {a} \cdot \mathbf {b}$ on a general basis $U=\{\mathbf {u} _{i}\}$ is $\mathbf {a} \cdot \mathbf {b} =(\mathbf {a} \cdot \mathbf {u} ^{i})(\mathbf {b} \cdot \mathbf {u} _{i})=(\mathbf {a} \cdot \mathbf {u} _{i})(\mathbf {b} \cdot \mathbf {u} ^{i})=a^{i}b_{i}=a_{i}b^{i}$. We can define other bases $U,V,\ldots$ for $\mathbb {R} ^{n}$ with any sets of $n$ linearly independent vectors as $U=\{\mathbf {u} _{i}=u^{ik}\mathbf {e} _{k}\}$, $V=\{\mathbf {v} _{i}=v^{ik}\mathbf {e} _{k}\}$, etc., where a set of vectors $\{\mathbf {u} _{i}\}$ is linearly independent iff $(\mathbf {u} _{1}\wedge \mathbf {u} _{2}\wedge \cdots \wedge \mathbf {u} _{n})\neq 0$, or iff the determinant of the matrix of basis vectors as columns is non-zero $|\mathbf {u} _{1}\mathbf {u} _{2}\cdots \mathbf {u} _{n}|\neq 0$. For any basis vector $\mathbf {u} _{i}$, we can compute its reciprocal basis vector as $\mathbf {u} ^{i}=(-1)^{i-1}(\mathbf {u} _{1}\wedge \cdots \wedge ()_{i}\wedge \cdots \wedge \mathbf {u} _{n})\cdot (\mathbf {u} _{1}\wedge \mathbf {u} _{2}\wedge \cdots \wedge \mathbf {u} _{n})^{-1}$ such that $\mathbf {u} _{i}\cdot \mathbf {u} ^{i}=(\mathbf {u} _{1}\wedge \mathbf {u} _{2}\wedge \cdots \wedge \mathbf {u} _{n})(\mathbf {u} _{1}\wedge \mathbf {u} _{2}\wedge \cdots \wedge \mathbf {u} _{n})^{-1}=1$ or $\mathbf {u} ^{i}\cdot \mathbf {u} _{i}=(\mathbf {u} ^{1}\wedge \mathbf {u} ^{2}\wedge \cdots \wedge \mathbf {u} ^{n})(\mathbf {u} ^{1}\wedge \mathbf {u} ^{2}\wedge \cdots \wedge \mathbf {u} ^{n})^{-1}=1$ (no summation) and $[\mathbf {u} _{i}\cdot \mathbf {u} ^{j}]=[\delta _{i}^{j}]=\mathbf {I} _{n}$. Using basis vectors $\mathbf {u} _{i}$ and their reciprocal basis vectors $\mathbf {u} ^{i}$, the general case of change of basis, from any basis to the basis $U$, is expressed as $\mathbf {a} =a^{i}\mathbf {e} _{i}=a_{i}\mathbf {e} ^{i}=(\mathbf {a} \cdot \mathbf {u} ^{i})\mathbf {u} _{i}=(\mathbf {a} \cdot \mathbf {u} _{i})\mathbf {u} ^{i}$. Change of basis is a passive transformation that does not change the vector $\mathbf {a}$, but only rewrites $\mathbf {a}$ in terms of another basis $\mathbf {u} _{i}$ or its reciprocal basis $\mathbf {u} ^{i}$. Change of basis to a basis $V$, etc., is again $\mathbf {a} =(\mathbf {a} \cdot \mathbf {v} ^{i})\mathbf {v} _{i}=(\mathbf {a} \cdot \mathbf {v} _{i})\mathbf {v} ^{i}$, etc. Now, we must convert this Clifford algebra expression, for general change of basis of $\mathbf {a}$ from a basis $U=\{\mathbf {u} _{i}\}$ to a basis $V=\{\mathbf {v} _{i}\}$, into matrix algebra forms. To do this, we must, in effect, convert from basis $U$ back to the standard basis $E$, and then convert from standard basis $E$ to basis $V$.

In linear algebra, the standard orthonormal basis $E=\{\mathbf {e} _{i}=\mathbf {e} ^{i}\}$, $1\leq i\leq n$, of $\mathbb {R} ^{n}$ has each vector $\mathbf {e} _{i}$ represented as an $n$-element column vector $[\mathbf {e} _{i}]=[(0)_{1}(0)_{2}\cdots (1)_{i}\cdots (0)_{n}]^{\mathrm {T} }$ where the $i$th row is $1$ and elsewhere $0$, and $\mathbf {A} ^{\mathrm {T} }$ is the transpose of matrix $\mathbf {A}$. We can express vector $\mathbf {a}$ on the standard basis $E$ as the column vector $[\mathbf {a} ]=[\mathbf {a} \cdot \mathbf {e} ^{i}]=[a^{i}]$. Then, we can express $\mathbf {a} =a^{i}\mathbf {e} _{i}=(\mathbf {a} \cdot \mathbf {u} ^{i})\mathbf {u} _{i}=(\mathbf {a} \cdot \mathbf {u} _{i})\mathbf {u} ^{i}$ on the basis $U=\{\mathbf {u} _{i}\}$ as the column vector $[\mathbf {a} \cdot \mathbf {u} ^{i}]$ of coordinates on $\mathbf {u} _{i}$. Each basis vector $\mathbf {u} _{i}$ is assumed to be in terms of the standard basis as $\mathbf {u} _{i}=u^{ik}\mathbf {e} _{k}$. Each reciprocal basis vector $\mathbf {u} ^{i}$ can be expressed as $\mathbf {u} ^{j}=u_{jk}\mathbf {e} ^{k}$, where the coordinates $u_{jk}$ of $\mathbf {u} ^{j}$ are $u_{jk}=\mathbf {e} _{k}\cdot \mathbf {u} ^{j}=(\mathbf {u} _{1}\wedge \cdots \wedge (\mathbf {e} _{k})_{j}\wedge \cdots \wedge \mathbf {u} _{n})\cdot (\mathbf {u} _{1}\wedge \mathbf {u} _{2}\wedge \cdots \wedge \mathbf {u} _{n})^{-1}=|\mathbf {u} _{1}\cdots (\mathbf {e} _{k})_{j}\cdots \mathbf {u} _{n}|/|\mathbf {u} _{1}\mathbf {u} _{2}\cdots \mathbf {u} _{n}|$, which is an expression of Cramer's rule that constitutes a method of finding the inverse matrix $[u^{ik}]^{-1}=[u_{jk}]^{\mathrm {T} }=[u_{kj}]$ since $[u^{ik}][u_{kj}]=[\mathbf {u} _{i}\cdot \mathbf {u} ^{j}]=[\delta _{i}^{j}]=\mathbf {I} _{n}$. The column vector $[\mathbf {a} ]=[a^{k}]$, on the standard basis $E$, is transformed onto the basis $U$ as $[\mathbf {u} ^{j}\cdot \mathbf {a} ]=[u_{jk}][a^{k}]$ and transformed back onto the standard basis $E$ as $[u^{ki}]([u_{jk}][a^{k}])=[u_{jk}][u^{ki}][a^{k}]=\mathbf {I} [a^{k}]=[a^{k}]$. The general case of the transformation of vector $\mathbf {a}$ from basis $U$ to basis $V$ is $[v_{jk}][u^{ki}]([u_{jk}][a^{k}])=[v^{ik}]^{-\mathrm {T} }[u^{ki}]([u_{jk}][a^{k}])=[v^{ki}]^{-1}[u^{ki}]([u_{jk}][a^{k}])$, where $[u^{ki}]$ is the matrix of column vectors $[u^{ki}]=[\mathbf {u} _{1}\mathbf {u} _{2}\cdots \mathbf {u} _{n}]$, $1\leq i\leq n$, and $[v^{ki}]^{-1}$ is the inverse of the matrix of column vectors $[v^{ki}]^{-1}=[\mathbf {v} _{1}\mathbf {v} _{2}\cdots \mathbf {v} _{n}]^{-1}$. The transition matrix $T_{U\rightarrow V}$ for the general change of basis from the basis $U=\{\mathbf {u} _{i}=u^{ik}\mathbf {e} _{k}\}$ to the basis $V=\{\mathbf {v} _{i}=v^{ik}\mathbf {e} _{k}\}$ is therefore $T_{U\rightarrow V}=[\mathbf {v} _{1}\mathbf {v} _{2}\cdots \mathbf {v} _{n}]^{-1}[\mathbf {u} _{1}\mathbf {u} _{2}\cdots \mathbf {u} _{n}]$. The matrix $T_{U\rightarrow E}=[\mathbf {u} _{1}\mathbf {u} _{2}\cdots \mathbf {u} _{n}]$ transforms coordinates from the basis $U$ to the standard basis $E$. The matrix $T_{V\rightarrow E}^{-1}=T_{E\rightarrow V}=[\mathbf {v} _{1}\mathbf {v} _{2}\cdots \mathbf {v} _{n}]^{-1}$ transforms coordinates from the standard basis $E$ to the basis $V$. Finally, $T_{U\rightarrow V}=T_{V\rightarrow E}^{-1}T_{U\rightarrow E}=[\mathbf {v} _{1}\mathbf {v} _{2}\cdots \mathbf {v} _{n}]^{-1}[\mathbf {u} _{1}\mathbf {u} _{2}\cdots \mathbf {u} _{n}]=[\mathbf {v} ^{1}\mathbf {v} ^{2}\cdots \mathbf {v} ^{n}]^{\mathrm {T} }[\mathbf {u} _{1}\mathbf {u} _{2}\cdots \mathbf {u} _{n}]$, so that we can either compute an inverse matrix using linear algebra methods or, equivalently, compute a matrix of reciprocal row vectors using Clifford (geometric) algebra methods. Note that, $T_{U\rightarrow E}[\mathbf {e} _{i}]=[\mathbf {u} _{i}]=[\mathbf {u} _{i}\cdot \mathbf {e} ^{k}]=[u^{ki}]$ and $T_{U\rightarrow E}[\mathbf {a} \cdot \mathbf {u} ^{k}]=[(\mathbf {u} _{k}\cdot \mathbf {e} ^{i})(\mathbf {a} \cdot \mathbf {u} ^{k})]=[\mathbf {e} ^{i}\cdot \mathbf {a} ]=[a^{i}]$ with Einstein summation over $k$.

In linear algebra, the dot product $\mathbf {a} \cdot \mathbf {b} =a^{i}b_{i}=[a^{i}]^{\mathrm {T} }[b_{i}]$ of vector $[\mathbf {a} \cdot \mathbf {e} ^{i}]=[a^{i}]=[a_{i}]$ expressed as the column vector $[\mathbf {a} \cdot \mathbf {u} ^{i}]$ with coordinates on basis $U=\{\mathbf {u} _{i}\}$ and vector $[\mathbf {b} \cdot \mathbf {e} _{i}]=[b_{i}]=[b^{i}]$ expressed as the column vector $[\mathbf {b} \cdot \mathbf {v} ^{i}]$ with coordinates on basis $V=\{\mathbf {v} _{i}\}$, requires transforming the coordinates of $[\mathbf {a} \cdot \mathbf {u} ^{i}]$ onto an arbitrary basis $W=\{\mathbf {w} _{i}\}$ as $T_{U\rightarrow W}[\mathbf {a} \cdot \mathbf {u} ^{i}]=T_{W\rightarrow E}^{-1}T_{U\rightarrow E}[\mathbf {a} \cdot \mathbf {u} ^{i}]$ and transforming the coordinates of $[\mathbf {b} \cdot \mathbf {v} ^{i}]$ onto the reciprocal basis $\{\mathbf {w} ^{i}\}$ as $T_{W\rightarrow E}^{\mathrm {T} }T_{V\rightarrow E}[\mathbf {b} \cdot \mathbf {v} ^{i}]$. We can then form the dual of $[\mathbf {a} \cdot \mathbf {w} ^{i}]$ on $W$ as the row vector $(T_{W\rightarrow E}^{-1}T_{U\rightarrow E}[\mathbf {a} \cdot \mathbf {u} ^{i}])^{\mathrm {T} }$, and finally the dot product is $\mathbf {a} \cdot \mathbf {b} =(T_{W\rightarrow E}^{-1}T_{U\rightarrow E}[\mathbf {a} \cdot \mathbf {u} ^{i}])^{\mathrm {T} }T_{W\rightarrow E}^{\mathrm {T} }T_{V\rightarrow E}[\mathbf {b} \cdot \mathbf {v} ^{i}]=[\mathbf {a} \cdot \mathbf {u} ^{i}]^{\mathrm {T} }T_{U\rightarrow E}^{\mathrm {T} }T_{V\rightarrow E}[\mathbf {b} \cdot \mathbf {v} ^{i}]$. Therefore, the dot product is independent of the choice of basis $W$, or we may assume $W$ is the standard basis $W=E=\{\mathbf {e} _{i}\}$, where $\mathbf {e} _{i}=\mathbf {e} ^{i}$. We may now write the dot product as $\mathbf {a} \cdot \mathbf {b} =[\mathbf {a} \cdot \mathbf {u} ^{i}]^{\mathrm {T} }[\mathbf {u} _{1}\mathbf {u} _{2}\cdots \mathbf {u} _{n}]^{\mathrm {T} }[\mathbf {v} _{1}\mathbf {v} _{2}\cdots \mathbf {v} _{n}][\mathbf {b} \cdot \mathbf {v} ^{i}]=[\mathbf {a} \cdot \mathbf {u} ^{i}]^{\mathrm {T} }[\mathbf {u} _{i}\cdot \mathbf {v} _{j}][\mathbf {b} \cdot \mathbf {v} ^{i}]$. If $U=V$, then $G_{ij}=[\mathbf {v} _{i}\cdot \mathbf {v} _{j}]$ is called the Gram matrix. Furthermore, if $U=V=E$, then $G_{ij}=\mathbf {I} _{n}$.

General case, using tensor algebra

Here is another proposed section. Like the other proposed section of mine, I guess that it also will not be wanted, but I offer it here for discussion. I think that this is better than the other proposal (which was about the general case using Clifford algebra and linear algebra). Maybe this general case using tensor algebra can be edited and included on the wiki someday. Change of basis is very important in tensor algebra and it seems like it should have a good discussion on this page. Twy2008 (talk) 12:18, 20 February 2020 (UTC)

General case, using tensor algebra

Introduction: Change of basis is fundamental to tensor algebra. The following presents a notation for change of basis using tensor algebra. It is primarily a notational problem to express tensor algebra clearly without seeming too complicated, but this is hard to avoid entirely and there is no exact standard for the notation. Anyone seeking to learn anything about advanced mathematics must accept and learn "complicated" notations sometimes. So please, try to bare with it, and try not to reject it immediately just because it seems "too complicated" for an encyclopedic article! The Einstein summation convention is used throughout.

Define the standard basis $E=\{\mathbf {e} _{i^{E}}\}$ for real ($n=p+q$)-dimensional pseudo-Euclidean space $\mathbb {R} ^{p,q}$ with reciprocal basis $E^{\ast }=\{\mathbf {e} ^{i^{E}}\}$ using the basis index set $\{i^{E},i_{1}^{E},i_{2}^{E},\ldots :1_{x}^{E}\leq i_{x}^{E}\leq n_{x}^{E}\}$ [see Notation (indices), below]. Let $\mathbf {e} ^{i^{E}}=\mathbf {e} _{i^{E}}$ for $1\leq i\leq p$, and $\mathbf {e} ^{i^{E}}=-\mathbf {e} _{i^{E}}$ for $p+1\leq i\leq p+q$. The covariant metric tensor for $E$ is

$g_{i_{1}^{E}i_{2}^{E}}=\mathbf {e} _{i_{1}^{E}}\cdot \mathbf {e} _{i_{2}^{E}}=\left\{{\begin{array}{lll}0&:&i_{1}^{E}\neq i_{2}^{E}\\1&:&1\leq (i_{1}^{E}=i_{2}^{E})\leq p\\-1&:&p+1\leq (i_{1}^{E}=i_{2}^{E})\leq p+q\end{array}}\right.$

and

$\mathbf {e} _{i_{1}^{E}}\cdot \mathbf {e} ^{i_{2}^{E}}=\delta _{i_{1}^{E}}^{i_{2}^{E}}=\left\{{\begin{array}{lll}0&:&i_{1}^{E}\neq i_{2}^{E}\\1&:&i_{1}^{E}=i_{2}^{E}\end{array}}\right.,$

where $\delta _{j}^{i}$ is the Knocker delta. The contravariant metric tensor is $g^{i_{1}^{E}i_{2}^{E}}=\mathbf {e} ^{i_{1}^{E}}\cdot \mathbf {e} ^{i_{2}^{E}}$.

The vector $\mathbf {a}$ on $E$ is $\mathbf {a} =a^{i^{E}}\mathbf {e} _{i^{E}}=(\mathbf {a} \cdot \mathbf {e} ^{i^{E}})\mathbf {e} _{i^{E}}$, and $\mathbf {a}$ on $E^{\ast }$ is $\mathbf {a} =a_{i^{E}}\mathbf {e} ^{i^{E}}=(\mathbf {a} \cdot \mathbf {e} _{i^{E}})\mathbf {e} ^{i^{E}}$. An index is lowered, and its vector is brought onto the reciprocal basis, as $g_{i_{1}^{E}i_{2}^{E}}a^{i_{1}^{E}}=(\mathbf {e} _{i_{1}^{E}}\cdot \mathbf {e} _{i_{2}^{E}})(\mathbf {a} \cdot \mathbf {e} ^{i_{1}^{E}})=\mathbf {e} _{i_{2}^{E}}\cdot \mathbf {a} =a_{i_{2}^{E}}$. An index is raised onto the basis as $g^{i_{1}^{E}i_{2}^{E}}a_{i_{1}^{E}}=(\mathbf {e} ^{i_{1}^{E}}\cdot \mathbf {e} ^{i_{2}^{E}})(\mathbf {a} \cdot \mathbf {e} _{i_{1}^{E}})=\mathbf {e} ^{i_{2}^{E}}\cdot \mathbf {a} =a^{i_{2}^{E}}$.

Now, define another general basis $U=\{\mathbf {u} _{i^{U}}\}$ for $\mathbb {R} ^{p,q}$ with reciprocal basis $\{\mathbf {u} ^{i^{U}}\}$ using the basis index set $\{i^{U},i_{1}^{U},i_{2}^{U},\ldots \}$.

Notation (indices): The superscript $U$ on $i_{x}^{U}$ is the part of the index name that indicates the basis of the index set. Index name $i_{x}^{E}$ and index name $i_{x}^{U}$ are completely different index names that run through $1_{x}^{E}\ldots n_{x}^{E}$ on basis $E$ and $1_{x}^{U}\ldots n_{x}^{U}$ on basis $U$, independently (they would not indicate an index pair for contraction). The index names $i_{x}$, where $x$ is any convenient integer (or omitted), are used to avoid the use of too many letters such as $i$, $j\sim i_{1}$, $k\sim i_{2}$ etc. Therefore, index names with different values of the subscript $x$ are also independent indices. The symbolic index name $i_{x}^{U}$ has the numerical value $i$ and can be abridged to $i^{U}$ or $i_{x}$ or $i$ when the particular basis and/or vector do not need to be indicated, depending on the usage. Basically, the whole index name $i_{x}^{U}$ is a value that runs from $i_{x}^{U}=1$ to $i_{x}^{U}=n$ which can also be expressed as $1_{x}^{U}\ldots n_{x}^{U}$ to mean the same thing. The contraction (or inner product) of two tensor indices, representing two vectors, cannot be performed until both indices are changed onto the same basis and also raised or lowered into a contra/co-variant pair. The contraction of any two basis vectors forms a metric tensor entry, possibly of mixed-bases. The contractions of a set of basis vectors with the same (or different) set of basis vectors forms the metric tensor (or the mixed-basis metric tensor between two different bases).

Let $\mathbf {u} _{i^{U}}=(\mathbf {u} _{i^{U}}\cdot \mathbf {e} ^{i^{E}})\mathbf {e} _{i^{E}}=u_{i^{U}}^{i^{E}}\mathbf {e} _{i^{E}}$ and $\mathbf {u} ^{i^{U}}=(\mathbf {u} ^{i^{U}}\cdot \mathbf {e} _{i^{E}})\mathbf {e} ^{i^{E}}=u_{i^{E}}^{i^{U}}\mathbf {e} ^{i^{E}}$, where

${\begin{array}{lll}u_{i^{E}}^{i^{U}}&=&\mathbf {e} _{i^{E}}\cdot \mathbf {u} ^{i^{U}}\\&=&\mathbf {e} _{i^{E}}\cdot ((-1)^{i^{U}-1}(\mathbf {u} _{1^{U}}\wedge \cdots \wedge ()_{i^{U}}\wedge \cdots \wedge \mathbf {u} _{n^{U}})\cdot (\mathbf {u} _{1^{U}}\wedge \mathbf {u} _{2^{U}}\wedge \cdots \wedge \mathbf {u} _{n^{U}})^{-1})\\&=&(-1)^{i^{U}-1}(\mathbf {e} _{i^{E}}\wedge \mathbf {u} _{1^{U}}\wedge \cdots \wedge ()_{i^{U}}\wedge \cdots \wedge \mathbf {u} _{n^{U}})\cdot (\mathbf {u} _{1^{U}}\wedge \mathbf {u} _{2^{U}}\wedge \cdots \wedge \mathbf {u} _{n^{U}})^{-1}\\&=&(\mathbf {u} _{1^{U}}\wedge \cdots \wedge (\mathbf {e} _{i^{E}})_{i^{U}}\wedge \cdots \wedge \mathbf {u} _{n^{U}})\cdot (\mathbf {u} _{1^{U}}\wedge \mathbf {u} _{2^{U}}\wedge \cdots \wedge \mathbf {u} _{n^{U}})^{-1}\\&=&C_{i^{E}i^{U}}/\Delta .\end{array}}$

The last expression for $u_{i^{E}}^{i^{U}}$ is a form of Cramer's rule, which is the $i^{E}i^{U}$th cofactor $C_{i^{E}i^{U}}$ of $[u_{i^{U}}^{i^{E}}]$ divided by the determinant $\Delta =|u_{i^{U}}^{i^{E}}|$.

The covariant metric tensor on $U$ is

$g_{i_{1}^{U}i_{2}^{U}}=\mathbf {u} _{i_{1}^{U}}\cdot \mathbf {u} _{i_{2}^{U}}=u_{i_{1}^{U}}^{i_{1}^{E}}\mathbf {e} _{i_{1}^{E}}\cdot u_{i_{2}^{U}}^{i_{2}^{E}}\mathbf {e} _{i_{2}^{E}}=u_{i_{1}^{U}}^{i_{1}^{E}}u_{i_{2}^{U}}^{i_{2}^{E}}g_{i_{1}^{E}i_{2}^{E}}.$

The contravariant metric tensor on $U$ is

$g^{i_{1}^{U}i_{2}^{U}}=\mathbf {u} ^{i_{1}^{U}}\cdot \mathbf {u} ^{i_{2}^{U}}=u_{i_{1}^{E}}^{i_{1}^{U}}\mathbf {e} ^{i_{1}^{E}}\cdot u_{i_{2}^{E}}^{i_{2}^{U}}\mathbf {e} ^{i_{2}^{E}}=u_{i_{1}^{E}}^{i_{1}^{U}}u_{i_{2}^{E}}^{i_{2}^{U}}g^{i_{1}^{E}i_{2}^{E}}.$

The tensor $u_{i^{E}}^{i^{U}}$ is the mixed-basis metric tensor for change of basis of contravariant coordinates $a^{i^{E}}$:

$u_{i^{E}}^{i^{U}}a^{i^{E}}=(\mathbf {u} ^{i^{U}}\cdot \mathbf {e} _{i^{E}})(\mathbf {a} \cdot \mathbf {e} ^{i^{E}})=\mathbf {u} ^{i^{U}}\cdot \mathbf {a} =a^{i^{U}}.$

The tensor $u_{i^{U}}^{i^{E}}$ is the mixed-basis metric tensor for change of basis of covariant coordinates $a_{i^{E}}$:

$u_{i^{U}}^{i^{E}}a_{i^{E}}=(\mathbf {u} _{i^{U}}\cdot \mathbf {e} ^{i^{E}})(\mathbf {a} \cdot \mathbf {e} _{i^{E}})=\mathbf {u} _{i^{U}}\cdot \mathbf {a} =a_{i^{U}}.$

The inverse transformations are also possible since $u_{i^{E}}^{i^{U}}u_{i^{U}}^{i^{E}}=(\mathbf {u} ^{i^{U}}\cdot \mathbf {e} _{i^{E}})(\mathbf {u} _{i^{U}}\cdot \mathbf {e} ^{i^{E}})=1$ when contracting one pair of matching indices while holding the other pair of matching indices to a fixed value, or $u_{i^{E}}^{i_{1}^{U}}u_{i_{2}^{U}}^{i^{E}}=\delta _{i_{2}^{U}}^{i_{1}^{U}}$ and $u_{i_{1}^{E}}^{i^{U}}u_{i^{U}}^{i_{2}^{E}}=\delta _{i_{1}^{E}}^{i_{2}^{E}}$.

For references section:

Schouten, J. A. (1954), Ricci-Calculus (2nd ed.), Springer, doi:10.1007/978-3-662-12927-2

extra details

For the expressions of $u_{i^{E}}^{i^{U}}$, more details are given below to show that they are equivalent. It is just a transposed matrix inverse, since the matrix of cofactors is not transposed as it usually is in linear algebra for the matrix inverse formula. This is all just an exercise and too much detail for the article, unless it were placed into a hiddle element that can pop open when clicked. Now, I'll stop cluttering up the Talk page any further unless there is some discussion!

${\begin{array}{lll}u_{i^{E}}^{i^{U}}&=&\mathbf {e} _{i^{E}}\cdot \mathbf {u} ^{i^{U}}\\&=&(\mathbf {u} _{1^{U}}\wedge \cdots \wedge (\mathbf {e} _{i^{E}})_{i^{U}}\wedge \cdots \wedge \mathbf {u} _{n^{U}})\cdot (\mathbf {u} _{1^{U}}\wedge \mathbf {u} _{2^{U}}\wedge \cdots \wedge \mathbf {u} _{n^{U}})^{-1}\\&=&{\frac {(\mathbf {u} _{1^{U}}\wedge \cdots \wedge (\mathbf {e} _{i^{E}})_{i^{U}}\wedge \cdots \wedge \mathbf {u} _{n^{U}})(\mathbf {u} _{1^{U}}\wedge \mathbf {u} _{2^{U}}\wedge \cdots \wedge \mathbf {u} _{n^{U}})}{(\mathbf {u} _{1^{U}}\wedge \mathbf {u} _{2^{U}}\wedge \cdots \wedge \mathbf {u} _{n^{U}})(\mathbf {u} _{1^{U}}\wedge \mathbf {u} _{2^{U}}\wedge \cdots \wedge \mathbf {u} _{n^{U}})}}\\&=&{\frac {(\mathbf {u} _{n^{U}}\wedge \cdots \wedge (\mathbf {e} _{i^{E}})_{i^{U}}\wedge \cdots \wedge \mathbf {u} _{1^{U}})(\mathbf {u} _{1^{U}}\wedge \mathbf {u} _{2^{U}}\wedge \cdots \wedge \mathbf {u} _{n^{U}})}{(\mathbf {u} _{n^{U}}\wedge \mathbf {u} _{2^{U}}\wedge \cdots \wedge \mathbf {u} _{1^{U}})(\mathbf {u} _{1^{U}}\wedge \mathbf {u} _{2^{U}}\wedge \cdots \wedge \mathbf {u} _{n^{U}})}}\\&=&{\frac {(\mathbf {u} _{n^{U}}\wedge \cdots \wedge (\mathbf {e} _{i^{E}})_{i^{U}}\wedge \cdots \wedge \mathbf {u} _{1^{U}})(\mathbf {u} _{1^{U}}\wedge \mathbf {u} _{2^{U}}\wedge \cdots \wedge \mathbf {u} _{n^{U}})}{|g_{i_{1}^{U}i_{2}^{U}}|}}\\&=&{\frac {(\mathbf {u} _{n^{U}}\wedge \cdots \wedge (\mathbf {e} _{i^{E}})_{i^{U}}\wedge \cdots \wedge \mathbf {u} _{1^{U}})(\mathbf {u} _{1^{U}}\wedge \mathbf {u} _{2^{U}}\wedge \cdots \wedge \mathbf {u} _{n^{U}})}{|u_{i_{1}^{U}}^{i_{1}^{E}}u_{i_{2}^{U}}^{i_{2}^{E}}g_{i_{1}^{E}i_{2}^{E}}|}}\\&=&{\frac {\left|{\begin{array}{llll}u_{1^{U}}^{i_{1}^{E}}u_{1^{U}i_{1}^{E}}&u_{1^{U}}^{i_{1}^{E}}u_{2^{U}i_{1}^{E}}&\cdots &u_{1^{U}}^{i_{1}^{E}}u_{n^{U}i_{1}^{E}}\\u_{2^{U}}^{i_{1}^{E}}u_{1^{U}i_{1}^{E}}&\ddots &\ddots &\vdots \\\vdots &\ddots &\ddots &\vdots \\(e_{i^{E}}^{i_{1}^{E}})_{i^{U}}u_{1^{U}i_{1}^{E}}&(e_{i^{E}}^{i_{1}^{E}})_{i^{U}}u_{2^{U}i_{1}^{E}}&\cdots &(e_{i^{E}}^{i_{1}^{E}})_{i^{U}}u_{n^{U}i_{1}^{E}}\\\vdots &\vdots &\ddots &\vdots \\u_{n^{U}}^{i_{1}^{E}}u_{1^{U}i_{1}^{E}}&u_{n^{U}}^{i_{1}^{E}}u_{2^{U}i_{1}^{E}}&\cdots &u_{n^{U}}^{i_{1}^{E}}u_{n^{U}i_{1}^{E}}\end{array}}\right|}{|u_{i_{1}^{U}}^{i_{1}^{E}}u_{i_{2}^{U}i_{1}^{E}}|}}\\&=&{\frac {|C(u_{i_{1}^{U}}^{i_{1}^{E}})_{i^{E}i^{U}}u_{i_{2}^{U}i_{1}^{E}}|}{|u_{i_{1}^{U}}^{i_{1}^{E}}u_{i_{2}^{U}i_{1}^{E}}|}}\\&=&{\frac {|C(u_{i_{1}^{U}}^{i_{1}^{E}})_{i^{E}i^{U}}||u_{i_{2}^{U}i_{1}^{E}}|}{|u_{i_{1}^{U}}^{i_{1}^{E}}||u_{i_{2}^{U}i_{1}^{E}}|}}\\&=&{\frac {|C(u_{i_{1}^{U}}^{i_{1}^{E}})_{i^{E}i^{U}}|}{|u_{i_{1}^{U}}^{i_{1}^{E}}|}}=C_{i^{E}i^{U}}/\Delta \end{array}}$

@Twy2008 I think in editing these sort of articles, we should keep in mind that a majority of people visiting this page are likely students who are not entirely familiar with the topic. The article itself is already notation-heavy and somewhat pedantic. It abuses set notation to denote sequences. It introduces a lot of symbols. I think it's well-organized in general, but the "preliminary notions" section seems unnecessary and delays the main point. Consider my addition to the intro: For example, if $A\in \mathbb {R} ^{n\times n}$ is a matrix whose columns comprise a basis of $\mathbb {R} ^{n}$, a vector $\mathbf {v}$ (in the standard basis) can also be expressed as a linear combination of $A$'s columns by the vector $A^{-1}\mathbf {v}$. By definition then, $A(A^{-1}\mathbf {v} )=(AA^{-1})\mathbf {v} =\mathbf {v}$. This summarizes the first two sections in two sentences. Can you summarize your addition without so much notation? AP295 (talk) 16:17, 13 January 2021 (UTC)

References

Article needs improvement.

The notation in this article needs some work. Sometimes it uses set notation for sequences, and sometimes it uses sequence-like notation for sets. I don't think there's any problem with defining "basis" as sequence of linearly independent spanning vectors rather than a set. This would also obviate the need for phrases like "ordered basis", which frequently appear in the article. Sequences are by definition ordered so editors should also avoid using redundant phrases like "ordered sequence".

Some of the examples also seem overly-complex. The example with Euler angles is unlikely to be helpful to a reader who's unfamiliar with the concept of a basis or change-of-basis. I like the organization though the content in the section "Preliminary notions" can be explained in-line with the rest of the content and does not need its own section in my opinion. Opinions? AP295 (talk) 14:25, 14 January 2021 (UTC)

I tried to clean up some of this. Some of it was clearly redundant. However, the order of the vectors in a basis is important, and my textbooks use the phrases "ordered set" and "ordered basis" instead of "sequence," so I removed the word "sequence."—Anita5192 (talk) 17:31, 14 January 2021 (UTC)

Thanks, though I think defining a "basis" as a sequence is more natural exactly because the order is important. In my book (that is, the book that I read), Linear Algebra Done Right, Axler uses "list", which is analogous to a sequence. A basis is linearly independent so by definition it contains no duplicate elements. Concatenating sequences is clearer on paper than taking the union of two bases, since the order of the union's elements would be ambiguous without any further explanation. Is there anything to be gained from calling them sets? AP295 (talk) 18:08, 14 January 2021 (UTC)

@Anita5192 The article also uses an unfortunate mix of latex and some other sort of markup (e.g. "&alpha") for math, which makes it a bit difficult to edit. Looking it over again, I really think it needs to be changed to "sequence". For example, the "standard basis" is not just a set of vectors, it's a sequence. Using sequence notation makes things much simpler notation-wise, and better lends itself to inductive arguments and things of that nature. For example, if you have two bases (a1,a2,a3) and (b1,b2,b3) of two orthogonal subspaces of R^6, then (a1,a2,a3,b1,b2,b3) is a basis for R^6, and the order is unambiguous despite not being implied by the subscripts alone. As another example, in concatenating (a1,a2,a3) and (a1), the obtained sequence is not a basis and we simply say "the sequence is not a basis". However, the union of the sets {a1,a2,a3}, {a1} would be a basis, and we'd have to say something like awkward like "the elements of {a1,a2,a3} and {a1} are not a linearly independent collection of vectors". I'll do it at some point if that's okay with you. AP295 (talk) 15:16, 15 January 2021 (UTC)

I am opposed to introducing the word "sequence," as that is not standard terminology in anything I have read. And I don't do much editing of mathematics typesetting, so I will leave that to someone else.—Anita5192 (talk) 16:49, 15 January 2021 (UTC)

The textbook I learned from uses sequences. From a notational and linguistic standpoint, does it not make more sense to call them sequences for the reasons that I've stated? The term "sequence" is very common in analysis and many other subjects, so I don't understand your objection. AP295 (talk) 23:45, 15 January 2021 (UTC)

This article did not state its main object, namely the basis-change formula. I have rewritten the article for fixing this, but most of the body still requires to be also rewritten. D.Lazard (talk) 19:59, 15 January 2021 (UTC)

It does, but it's hard to pick out. Your changes have a somewhat editorial tone. I think this is too much for the intro, and I'm not sure why you deleted my example. I thought it was a pretty clear way of introducing the idea. You use the word "basis-change matrix" as if the reader is supposed to know what that means already, and it makes your definition a bit circular. My example demonstrated that changing from the standard basis to basis B is accomplished by multiplying with the inverse of B, and to the standard basis from basis B by multiplication with B itself. From there the reader understands that they can change from one arbitrary basis A to another B by taking (B^-1 A). Is this not much clearer and more explicit? And yes, the matrix of a linear map is determined by the map and a choice of bases for the domain and codomain, and I considered adding this to the intro myself, but I'm not sure if the intro is the place for it. It's an important thing to know but it is covered in the article itself. It just needs to be re-worked for clarity. AP295 (talk) 23:48, 15 January 2021 (UTC)

@Anita5192, @D.Lazard Since the basis article also defines a basis as a set, I've opened up an RFC on that article's talk page. I believe this is an important distinction, and I invite you to comment if you have a legitimate objection to this and can explain it clearly. If you change your minds and agree, that would be just swell too. AP295 (talk) 01:43, 16 January 2021 (UTC)

@D.Lazard In your comment on the maintenance tag, you wrote "most of the article is devoted to an overdetailed description of the fundamental concepts of linear algebra which are much better described in the relevant articles. This must be restricted to the minimum that is required for this article, and regrouped in a specific section". I agree with some of this but actually I think this article does a better job of introducing the concept of a "basis" than the actual article. If the notation is improved I think the content should be moved rather than deleted. AP295 (talk) 20:42, 16 January 2021 (UTC)

I have fixed the indentation of the previous post.

When an article does not describe well its subject, it is against Wikipedia policy to rewrite it elsewhere (see WP:Content fork). So, it is a very bad idea to introduce here the concept of basis. In any case, edits do not delete anything, as anything that is removed can be rtrieved from the history of the article. D.Lazard (talk) 10:43, 17 January 2021 (UTC)

"Moved" as in moved to Basis (linear algebra). AP295 (talk) 20:07, 20 January 2021 (UTC)

I have rewritten the lead to be conform to MOS:MATH#Article introduction. As the article topic cannot be understand without a minimal knowledge of the subject, the lead must recall only what is relly needed to understand the phrasing. This is the reason for the removal of most of the previous lead. The new lead is incomplete, as it does not summarize really the content of the sections. In fact this cannot be done before fixing the flaws mentioned in the maintenance tags. D.Lazard (talk) 10:43, 17 January 2021 (UTC)

Also, I have added the change-of-basis formula, because it is likely that many reader may come here to remember the exact form of the formula, in order to apply it correctly in their own application. So, it is important to give a direct access to this formula. D.Lazard (talk) 10:56, 17 January 2021 (UTC)

Suggestion

The ideal basis is the standard basis $\{e_{i}\}_{i=1}^{n}$ which has the property

$e_{i}\cdot e_{j}={\begin{cases}0\ \ i\neq j\\1\ \ i=j\end{cases}}.$

The cases where the inner product is zero mean that the vectors e_i and e_j are orthogonal to each other, while the cases $e_{i}\cdot e_{i}=1$ mean that the vectors are unit vectors. The standard basis is said to be orthonormal, meaning that the above property holds.

Now if $\{f_{i}\}_{i=1}^{n}$ is a set of linearly independent vectors, then the matrix M with these vectors as rows is an invertible matrix and has the property $e_{i}M=f_{i}$ for each i = 1,...,n.

To show that $\{f_{i}\}_{i=1}^{n}$ is a basis an arbitrary w in the vector space must be expressible in that basis. Let $v=wM^{-1}$ and express v in the standard basis: $v=\sum _{i=1}^{n}a_{i}e_{i}.$ Then

$w=vM=(\sum _{i=1}^{n}a_{i}e_{i})M=\sum _{i=1}^{n}a_{i}(e_{i}M)=\sum _{i=1}^{n}a_{i}f_{i}.$

One reason to make a change of basis may be to take a set of linearly independent vectors like {f_i} (above) and use it to construct an orthonormal basis. As indicated, this procedure is equivalent to finding the inverse of the matrix using the f’s as rows. The step-wise procedure uses elementary rotations and provides Gram-Schmidt orthogonalization.

Another example of a useful change of basis re-configures rectangular hyperbolas to establish the hyperbolic functions sinh and cosh. Change of basis is also used in abstract algebra to relate alternate representations of algebras.

Perhaps the article would improve with this approach. Rgdboer (talk) 03:38, 11 February 2021 (UTC)

How could this exist so long and nobody noticed that "basis" is used, but not defined, in the lede?

Well? Huh? Well?

Verdana♥Bold 12:20, 9 July 2021 (UTC)

A link is provided for a definition, and a characteristic property is given. This should suffice for people who have already heard of a basis. For others, this is unimportant, as it is irrealistic to try changing something that one does not know about. D.Lazard (talk) 14:32, 9 July 2021 (UTC)

Change of basis formulas are wrong

The change of basis formulas for linear maps, endomorphisms, and bilinear forms are wrong. For instance, for endomorphisms, it should be $P^{-1}MP$ and not $PMP^{-1}$. Right? Seub (talk) 00:50, 18 August 2021 (UTC)

 Fixed. Good point. A way to remember this formula is that the product of matrices must provide "new coordintes". So the matrix on the left must express "new" coordinates in terms of "old" ones. So, it must be the inverse of the change-of-basis matrix, which expresses the "old" coordintes in terms of the "new" ones. D.Lazard (talk) 07:56, 18 August 2021 (UTC)

Thanks for the correction. The new matrix of the bilinear form also needs to be changed: it is $P^{\mathsf {T}}\mathbf {B} P$ and not $P\mathbf {B} P^{\mathsf {T}}$. Seub (talk) 10:58, 18 August 2021 (UTC)

Seub,  Done. You could have done this by yourself. D.Lazard (talk) 14:27, 18 August 2021 (UTC)

Caption wrong for pic?

the second illustration picture, when you click it to view by itself, displays the caption "A vector (here in 3d, shown the purple arrow) can be represented in terms of two different bases (green and blue arrows), each basis vector is scalar-multiplied appropriately so they add to the vector."

But there are no green or blue items in that pic. 2601:80:4300:3AE0:74D3:9ED1:B92F:79C4 (talk) 03:07, 19 October 2022 (UTC)

I cannot display this caption. The only caption that I have found is "A vector represented by two different bases (purple and red arrows)". D.Lazard (talk) 08:33, 19 October 2022 (UTC)

I can see what the IP is referring to. When I click the image in the article, I am taken to the image. When I scroll down, I see the erroneous description.—Anita5192 (talk) 15:52, 19 October 2022 (UTC)