Talk:Linear map/Archive 1

Preliminary remarks
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)


 * Nope. — Preceding unsigned comment added by 128.111.88.128 (talk) 00:12, 25 May 2004 (UTC)


 * Part about solving a system of equations ("f(x)=0 is called...") should probably be moved somewhere else. What about other articles about that topic? — Preceding unsigned comment added by 212.193.10.2 (talk) 09:52, 10 May 2005 (UTC)

Clarification of my move
I moved a pagraph to the bottom of the article, with the edit summary


 * moved the continuity section to the very bottom. Linear transformations are about vector spaces. For continuity, you need a Banach space. So, continuity is not the primary concern of this article.

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)


 * Ok, that is fine.--Patrick 00:07, 25 September 2005 (UTC)

Modules
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)

Clarification Request
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)

An example of a mapping where the additive property is satisfied but not the homogene
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)
 * Thanks for the help. Snailwalker | talk 17:12, 7 March 2008 (UTC)

Removed unclear sentence about alternative name

 * "It immediately follows from the definition that f(0) = 0. Hence linear maps are sometimes called homogeneous linear maps (see linear function)."

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)

Examples
Why are the examples talking about eigenvectors and eigenvalues? Is this not a distraction? --anon
 * I agree. I cut that out. Oleg Alexandrov (talk) 15:32, 20 July 2006 (UTC)

Example Suggestion x->x+1 is a non-linear map
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)
 * For real numbers, the map $$x\mapsto x^2$$ is not linear.
 * For real numbers, the map $$x\mapsto x+1$$ is not linear.


 * Yeah its got mi fooled. Why is it non linear?--ProperFraction (talk) 02:08, 8 August 2008 (UTC)


 * If f(x) = x+1, then f(0) + f(0) is not equal to f(0 + 0). JackSchmidt (talk) 03:08, 8 August 2008 (UTC)

Mapping of closed shapes
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:
 * Area of the original shape × $$\begin{vmatrix}

a & b \\ c & d \end{vmatrix}$$ —Preceding unsigned comment added by 124.168.226.120 (talk) 09:54, 14 November 2008 (UTC)

Result of using matrix representation
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)

dimension formula only in finite dimensions?
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)


 * The formula either doesn't make sense (initially) or isn't very useful (if you define addition on infinite cardinals somehow) in the infinite dimensional case. By and large, you're likely to get formulae such as $$ \infty + 4 = \infty $$, which are not very useful. If you're really prepared for some heavy mathematics, however, you might want to check out Fredholm operators which have an associated index, a finite number that in the case where V, W are finite dimensional is just dimV - dimU (by the formula in the article). The index has many mystical properties I wot not of. Ben 13:02, 10 August 2006 (UTC)


 * The original unsigned remark was valid; the sum formula is correct for infinite cardinals. Someone has already removed the restriction to finite dimensions in the meantime. Joriki (talk) 05:31, 31 July 2009 (UTC)

Material from eigenvalue article
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:


 * Linear transformations of a vector space, such as rotation, reflection, stretching, compression, shear or any combination of these, may be visualized by the effect they produce on vectors. In other words, they are vector functions. More formally, in a vector space L, a vector function A is defined if for each vector x of L there corresponds a unique vector y = A(x) of L. For the sake of brevity, the parentheses around the vector on which the transformation is acting are often omitted. A vector function A is linear if it has the following two properties:

where x and y are any two vectors of the vector space L and α is any scalar.
 * Additivity: A(x + y) = Ax + Ay
 * Homogeneity: A(αx) = αAx
 * See ; ; ; ; Rowland, Todd and Weisstein, Eric W. Linear transformation From MathWorld − A Wolfram Web Resource

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)

Title
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)


 * I object! The traditional terminology is "linear transformation" and this is still very widely used.  Also, the term "map" could just as well be "mapping".  How to choose one?  The term "operator" usually means it is an endomorphism.  The term "transformation" does not.  The article should be moved back to the old title.  I say this based on teaching linear algebra.
 * But are there other opinions, pro or con? Zaslav (talk) 07:28, 26 February 2011 (UTC)