Talk:Change of basis

message: 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)
thread_title: This article leaves much to be desired

message: 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)
replies: text: Yes, I think. I added the missing words "let R". Dexo568 (talk) 22:37, 4 November 2015 (UTC)
thread_title:  Something missing? 

message: In the second line of this section, namely:
If T: Rn → Rm is a linear transformation... 
What then is the matrix of T(ej)? — Preceding unsigned comment added by 128.84.126.62 (talk) 09:01, 23 December 2016 (UTC)
thread_title:  Preliminary notions statement seems cyclical 

message: 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)
replies: text: 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)
thread_title:  General case, derived using Clifford algebra 

message: 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^U_1 i^U_2}
  |}\\
  & = & \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^E_1}_{i^U_1}
  u^{i^E_2}_{i^U_2} g_{i^E_1 i^E_2} |}\\
  & = & \frac{\left| \begin{array}{llll}
    u^{i^E_1}_{1^U} u_{1^U i^E_1} & u^{i^E_1}_{1^U} u_{2^U i^E_1} & \cdots &
    u^{i^E_1}_{1^U} u_{n^U i^E_1}\\
    u^{i^E_1}_{2^U} u_{1^U i^E_1} & \ddots & \ddots & \vdots\\
    \vdots & \ddots & \ddots & \vdots\\
    (e^{i^E_1}_{i^E})_{i^U} u_{1^U i^E_1} & (e^{i^E_1}_{i^E})_{i^U} u_{2^U
    i^E_1} & \cdots & (e^{i^E_1}_{i^E})_{i^U} u_{n^U i^E_1}\\
    \vdots & \vdots & \ddots & \vdots\\
    u^{i^E_1}_{n^U} u_{1^U i^E_1} & u^{i^E_1}_{n^U} u_{2^U i^E_1} & \cdots &
    u^{i^E_1}_{n^U} u_{n^U i^E_1}
  \end{array} \right|}{| u^{i^E_1}_{i^U_1} u_{i^U_2 i^E_1} |}\\
  & = & \frac{| C (u^{i_1^E}_{i_1^U})_{i^E i^U} u_{i^U_2 i^E_1} |}{|
  u^{i^E_1}_{i^U_1} u_{i^U_2 i^E_1} |}\\
  & = & \frac{| C (u^{i_1^E}_{i_1^U})_{i^E i^U} | | u_{i^U_2 i^E_1} |}{|
  u^{i^E_1}_{i^U_1} | | u_{i^U_2 i^E_1} |}\\
  & = & \frac{| C (u^{i_1^E}_{i_1^U})_{i^E i^U} |}{| u^{i^E_1}_{i^U_1} |} =
  C_{i^E i^U} / \Delta
\end{array}
: @ 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)
thread_title:  extra details 

message: 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 \ne j \\ 1 \ \ i = j \end{cases} .
The cases where the inner product is zero mean that the vectors ei and ej 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 = w M^{-1} and express v in the standard basis: v = \sum_{i=1}^n a_i e_i . Then
:w = v M = (\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 {fi} (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)
thread_title:  Suggestion 

message: Well? Huh? Well?
Verdana♥Bold 12:20, 9 July 2021 (UTC)
replies: text: 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)
thread_title:  How could this exist so long and nobody noticed that "basis" is used, but not defined, in the lede? 

message: The change of basis formulas for linear maps, endomorphisms, and bilinear forms are wrong. For instance, for endomorphisms, it should be P^{-1} M P and not P MP^{-1}. Right? Seub (talk) 00:50, 18 August 2021 (UTC)
replies: text: . 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)
replies: text: 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)
replies: text: , . You could have done this by yourself. D.Lazard (talk) 14:27, 18 August 2021 (UTC)
thread_title:  Change of basis formulas are wrong 

message: 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)
replies: text: 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)
replies: text: 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)
thread_title:  Caption wrong for pic?