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Archive 1 |
Endomorphism f from V into V, where V is a vector space. is it nessary f to be one-one mapping? — Preceding unsigned comment added by 202.141.141.40 ( talk) 07:35, 20 September 2003 (UTC)
I moved a pagraph to the bottom of the article, with the edit summary
Let me clarify myself. First of all, I definitely agree with what the paragraph says, that linear transformations are not necessarily continuous. But the problem is the following. Linear trasformations are about vector spaces. That vector space can be the reals, the complex numbers, a vector space of over a field finite characteristic, over a field which is a Galois extension, etc. All that matters in a vector space is addition and multiplication by scalar.
As such, inserting in the middle of that article a paragraph operators on a Banach space (or if you wish, a linear topological space) was wrong. It distracts the reader from the main point, which is the linearity, addition and multiplication by scalars. To talk about continuity you need topology, you need a norm. It is a totally different realm than the one of a vector space. That's why inserting that continuity paragraph was out of place. It has to of course be mentioned somewhere, but since all the other topics in this article are closely bound together, I put this periferial one at the bottom. Oleg Alexandrov 18:34, 24 September 2005 (UTC)
Why vector spaces? I wandered over to this page from preadditive category which tells me that "composition of morphisms is bilinear over the integers"; and that page says it means " linear in both of its arguments." Of course, the integers are not a field, and the abelian group of homomorphisms do not form a vector space. On the other hand, any abelian group is a Z-module. -- 192.75.48.150 20:47, 3 August 2007 (UTC)
I'm taking multivariate calc at the moment, and our book uses the term 'Linear Transformation' as synonymous with 'Linear Function'. Unfortunately, it gives no explanation as to WHY linear functions are also called transformations. Could someone in the know please make an addition to address this? Celemourn 14:37, 4 October 2007 (UTC)
I've discussed this with some of the professors at my university if they could come up with a linear mapping satisfying that:
f(x+y)=f(x)+f(y)
But that f(a*x)!=af(x)
They couldn't think of one, but I assume that there is one? Snailwalker | talk 16:31, 7 March 2008 (UTC)
The complex numbers is a vector space over itself, so take f:C->C to be complex conjugation. Then f(a+b)=f(a)+f(b), but -1 = f(i*i) != i*f(i) = 1. JackSchmidt ( talk) 16:47, 7 March 2008 (UTC)
The logic of the second part of this sentence is not clear. Why should they call homogeneous something which is by definition both additive and homogeneous? It seems (incompletely) redundant. Either we call it "additive homogeneus map" or simply "linear map". Are you sure that the expression "homogeneous linear map" is used in the literature? Paolo.dL ( talk) 18:38, 28 June 2008 (UTC)
Why are the examples talking about eigenvectors and eigenvalues? Is this not a distraction? --anon
Currently, the following is in the examples section:
The map x to x^2 is nonlinear because it does not satisfy homogeneity or additivity. Would it not be more "enlightening" to add the example:
I suggest this because x^2 is automatically parsed as non-linear by anyone who has plotted a function in 2D space, however x+1 is a nice line in 2D space, yet is a non-linear transformation. Raazer 19:48, 12 October 2007 (UTC)
From my browsing I can't find any page which talks about the area of closed shapes under a linear transformation. Unless anyone could direct me to such a page I believe that this article would be the place to put such a comment. I do know that the area of the image of a closed shape under a linear transformation T:(x,y)→(ax+by,cx+dy) is:
The result gives coordinates in the destination vector space. It is important to stress this as the original linear map f(v) can be a m-by-n matrix applied to a vector space with n-by-1 matrices as elements which results in a m-by-1 matrix. The application of the matrix from a matrix representation of this linear map gives a m-by-1 matrix too but this one is not f(v). 193.174.53.122 ( talk) 10:31, 25 February 2009 (UTC)
the article states that the formula dim ker f + dim im f = dim V is only valid if V is finite dimensional. Is that really true? Can someone give me a counterexample? -July 2, 2005 01:19 (UTC)
I removed the following text from the eigenvalue article as extraneous. But just in case it is not all repeated here (I have not yet checked) I wanted to preserve it:
where x and y are any two vectors of the vector space L and α is any scalar.
Such a function is variously called a linear transformation, linear operator, or linear endomorphism on the space L. — Dhollm ( talk) 21:01, 15 July 2010 (UTC)
I wonder if this article should be entitled Linear map rather than Linear transformation.
Both terms are in common use, but the term transformation suggests a specialization to the case of endomorphisms (see function), just as the term Linear operator suggests an infinite dimensional context. Geometry guy 20:10, 9 February 2007 (UTC)
No objections received, so I've moved it, fixed the double redirects, and edited the article. Geometry guy 19:25, 13 February 2007 (UTC)
![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 |
Endomorphism f from V into V, where V is a vector space. is it nessary f to be one-one mapping? — Preceding unsigned comment added by 202.141.141.40 ( talk) 07:35, 20 September 2003 (UTC)
I moved a pagraph to the bottom of the article, with the edit summary
Let me clarify myself. First of all, I definitely agree with what the paragraph says, that linear transformations are not necessarily continuous. But the problem is the following. Linear trasformations are about vector spaces. That vector space can be the reals, the complex numbers, a vector space of over a field finite characteristic, over a field which is a Galois extension, etc. All that matters in a vector space is addition and multiplication by scalar.
As such, inserting in the middle of that article a paragraph operators on a Banach space (or if you wish, a linear topological space) was wrong. It distracts the reader from the main point, which is the linearity, addition and multiplication by scalars. To talk about continuity you need topology, you need a norm. It is a totally different realm than the one of a vector space. That's why inserting that continuity paragraph was out of place. It has to of course be mentioned somewhere, but since all the other topics in this article are closely bound together, I put this periferial one at the bottom. Oleg Alexandrov 18:34, 24 September 2005 (UTC)
Why vector spaces? I wandered over to this page from preadditive category which tells me that "composition of morphisms is bilinear over the integers"; and that page says it means " linear in both of its arguments." Of course, the integers are not a field, and the abelian group of homomorphisms do not form a vector space. On the other hand, any abelian group is a Z-module. -- 192.75.48.150 20:47, 3 August 2007 (UTC)
I'm taking multivariate calc at the moment, and our book uses the term 'Linear Transformation' as synonymous with 'Linear Function'. Unfortunately, it gives no explanation as to WHY linear functions are also called transformations. Could someone in the know please make an addition to address this? Celemourn 14:37, 4 October 2007 (UTC)
I've discussed this with some of the professors at my university if they could come up with a linear mapping satisfying that:
f(x+y)=f(x)+f(y)
But that f(a*x)!=af(x)
They couldn't think of one, but I assume that there is one? Snailwalker | talk 16:31, 7 March 2008 (UTC)
The complex numbers is a vector space over itself, so take f:C->C to be complex conjugation. Then f(a+b)=f(a)+f(b), but -1 = f(i*i) != i*f(i) = 1. JackSchmidt ( talk) 16:47, 7 March 2008 (UTC)
The logic of the second part of this sentence is not clear. Why should they call homogeneous something which is by definition both additive and homogeneous? It seems (incompletely) redundant. Either we call it "additive homogeneus map" or simply "linear map". Are you sure that the expression "homogeneous linear map" is used in the literature? Paolo.dL ( talk) 18:38, 28 June 2008 (UTC)
Why are the examples talking about eigenvectors and eigenvalues? Is this not a distraction? --anon
Currently, the following is in the examples section:
The map x to x^2 is nonlinear because it does not satisfy homogeneity or additivity. Would it not be more "enlightening" to add the example:
I suggest this because x^2 is automatically parsed as non-linear by anyone who has plotted a function in 2D space, however x+1 is a nice line in 2D space, yet is a non-linear transformation. Raazer 19:48, 12 October 2007 (UTC)
From my browsing I can't find any page which talks about the area of closed shapes under a linear transformation. Unless anyone could direct me to such a page I believe that this article would be the place to put such a comment. I do know that the area of the image of a closed shape under a linear transformation T:(x,y)→(ax+by,cx+dy) is:
The result gives coordinates in the destination vector space. It is important to stress this as the original linear map f(v) can be a m-by-n matrix applied to a vector space with n-by-1 matrices as elements which results in a m-by-1 matrix. The application of the matrix from a matrix representation of this linear map gives a m-by-1 matrix too but this one is not f(v). 193.174.53.122 ( talk) 10:31, 25 February 2009 (UTC)
the article states that the formula dim ker f + dim im f = dim V is only valid if V is finite dimensional. Is that really true? Can someone give me a counterexample? -July 2, 2005 01:19 (UTC)
I removed the following text from the eigenvalue article as extraneous. But just in case it is not all repeated here (I have not yet checked) I wanted to preserve it:
where x and y are any two vectors of the vector space L and α is any scalar.
Such a function is variously called a linear transformation, linear operator, or linear endomorphism on the space L. — Dhollm ( talk) 21:01, 15 July 2010 (UTC)
I wonder if this article should be entitled Linear map rather than Linear transformation.
Both terms are in common use, but the term transformation suggests a specialization to the case of endomorphisms (see function), just as the term Linear operator suggests an infinite dimensional context. Geometry guy 20:10, 9 February 2007 (UTC)
No objections received, so I've moved it, fixed the double redirects, and edited the article. Geometry guy 19:25, 13 February 2007 (UTC)