dimension (linear application)

[bluemath]

Hello,

If we have a linear application from Rⁿ to R , what is the dimension of this application ? One or n ?

Suppose we have : f R³ ----> R : (x,y,z) ---> 2x + 3x - z

The dimension of this application is for me one because we have only one vector in the image. And this is the same for all application from Rⁿ to R
Is that correct ?

Thanks

 

[daon2]

Hello,

If we have a linear application from Rⁿ to R , what is the dimension of this application ? One or n ?

Suppose we have : f R³ ----> R : (x,y,z) ---> 2x + 3x - z

The dimension of this application is for me one because we have only one vector in the image. And this is the same for all application from Rⁿ to R
Is that correct ?

Thanks

What is an application?

edit: after some visits to google, it appears application just a synonym for function/transformation, and best as I can come up with is that the dimension of a transformation is an abuse of terminology and actually means rank of a transformation, i.e. the dimension of the image.

You might find the following self-evident fact useful:

If \(\displaystyle T:X\to Y\) is a linear transformation of vector spaces then

\(\displaystyle \text{dim} \left(T(X)\right) \le \min\{\text{dim}X,\text{dim}Y\}\)

In fact, it is easy to see in your case

\(\displaystyle f(x,y,z) = \left[\begin{matrix} 2 & 3 & -1\end{matrix}\right]\left[\begin{matrix} x \\ y \\ z\end{matrix}\right]\)

So the matrix representation of \(\displaystyle f\) has size \(\displaystyle 1\times 3\) and is not the zero matrix. Therefore its rank is 1.

 

[bluemath]

Yes, thanks for your answer.

So we can generalize for Rⁿ to R I presum. The rank of the image will be always one then ?

 

[daon2]

Yes, thanks for your answer.

So we can generalize for Rⁿ to R I presum. The rank of the image will be always one then ?

Yes, unless it is the zero-transformation

 

[Ishuda]

Hello,

If we have a linear application from Rⁿ to R , what is the dimension of this application ? One or n ?

Suppose we have : f R³ ----> R : (x,y,z) ---> 2x + 3x - z

The dimension of this application is for me one because we have only one vector in the image. And this is the same for all application from Rⁿ to R
Is that correct ?

Thanks

My formal math is, for the most part, from some 50 or so years ago so please forgive me if 'm wrong when I say that a function (application) itself does not have a dimension. The domain, R³ in your example, of the function has a dimension, 3 in your example, and the range (co-domain, image), R in your example, of the function has a dimension, 1 in your example, but not the function itself.

 

[Ishuda]

Yes, thanks for your answer.

So we can generalize for Rⁿ to R I presum. The rank of the image will be always one then ?

Yes, unless it is the zero-transformation

Possibly. It will depend on the function. As daon2 implied, if it the zero transformation it will have rank zero but some function might be unable to be put into a simple matrix form so what would you call the rank then? For example:
f: R³ ----> R : (x,y,z) ---> 2x² + 3\(\displaystyle \Gamma(y/z)\)
where \(\displaystyle \Gamma\) is the Gamma function. Note that this is from (part of) R³ into R and not (part of) R³ onto R.

However, if you do restrict the (non zero) transformation to a linear one as you do in your original post, then the rank of the corresponding matrix may be 1 but the dimension of the image may be zero [think of mapping to any constant].

 

[daon2]

Possibly. It will depend on the function. As daon2 implied, if it the zero transformation it will have rank zero but some function might be unable to be put into a simple matrix form so what would you call the rank then? For example:
f: R³ ----> R : (x,y,z) ---> 2x² + 3\(\displaystyle \Gamma(y/z)\)
where \(\displaystyle \Gamma\) is the Gamma function. Note that this is from (part of) R³ into R and not (part of) R³ onto R.

However, if you do restrict the (non zero) transformation to a linear one as you do in your original post, then the rank of the corresponding matrix may be 1 but the dimension of the image may be zero [think of mapping to any constant].

Linear transformation (almost) by definition requires that the image of zero is zero. So the only constant linear transformation is the zero transformation. The type of "linear" transformation that allows non-zero constant images is called an Affine transformation

 

[bluemath]

Possibly. It will depend on the function. As daon2 implied, if it the zero transformation it will have rank zero but some function might be unable to be put into a simple matrix form so what would you call the rank then? For example:
f: R³ ----> R : (x,y,z) ---> 2x² + 3\(\displaystyle \Gamma(y/z)\)
.

We are in the case of a linear application then your example is not possible actually.

PS You don't use this name in english " linear application " ? We say in french " application linéaire " and this is not exactly the def for a linear function.(an application is an injective function actually)

 

[daon2]

We are in the case of a linear application then your example is not possible actually.

PS You don't use this name in english " linear application " ? We say in french " application linéaire " and this is not exactly the def for a linear function.(an application is an injective function actually)

This is the first I have heard of application being used in place of function/mapping/transformation. The term application usually is used, in my experience, in using a mathematical model for a purpose. If you search the (US) google for linear application it will give mostly results on applications of linear transformations.

 

[Ishuda]

We are in the case of a linear application then your example is not possible actually.

PS You don't use this name in english " linear application " ? We say in french " application linéaire " and this is not exactly the def for a linear function.(an application is an injective function actually)

There seems to be some confusion, at least to me, in what the original question was really asking. As I understand it, an injective function, is one-to-one. See
http://www.mathsisfun.com/sets/injective-surjective-bijective.html
for example which says "A function is a way of matching the members of a set "A" to a set "B": ... Injective means that every member of "A" has its own unique matching member in "B"." Thus, it will not map two different elements in the domain to the same element in the range. Your example function in the original post
f R³ ----> R : (x,y,z) ---> 2x + 3y - z
(assuming you meant the second x to be a y) then is not a linear applicaion, as you have now defined it, since
f(1, 1, 1) = f(0,0,-3) = ...


By your example, you implied that a linear application might not be one-to-one (injective). That is, that it was more a linear function which was not necessarily injective, nor surjective, nor bijective. In fact, IMO, it also left open the possibility the linear application may just be a linear function which can map a multiple dimensional space to a point. Thus my comments.

In what I have learned, in a strict sense, a linear map [linear application] must satisfy two conditions:
f(x+y) = f(x) + f(y)
f(ax) = a f(x)
where x and y may be thought of as vectors and a as a scalar. If the spaces involved are finite dimensional, the linear map can be represented as a matrix. So, as you mentioned, a constant mapping is not allowed unless the constant is zero nor is the example I gave involving the Gamma function. But then, that is the reason I said "if you do restrict the (non zero) transformation to a linear one ..."

 

[bluemath]

Ok then I will use " linear map " in the future. Sorry for this translation's misstake.