Linear Transformations Confusion

  • #1
Screwdriver
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Homework Statement



Not really a problem per se; more of an issue with some aspects of linear transformations. We've learned that a linear combination of linear transformations is defined as follows:

[tex](c_1T_1+c_2T_2)(\vec{x})=c_1T_1(\vec{x})+c_2T_2(\vec{x})\,\,\,\,\,\, \vec{x}\varepsilon \mathbb{R}^n\,\,\,c\varepsilon \mathbb{R}[/tex]

And if T1, T2 are linear, then so is c1T1 + c2T2 and

[tex][c_1T_1+c_2T_2]=c_1[T_1]+c_2[T_2][/tex]

Where [T] is the standard matrix of T and c is some constant.

The problem is, is that I don't know why T1 and T2 being linear implies that c1T1 + c2T2 is linear other than by noting that multiplying the transformations by c and adding them together is a linear combination, but that doesn't seem to be a very good proof. The second part there would seem to follow from this though just based on the fact that transforming some vector x is the same as multiplying it by the standard matrix of T.

Also, linear transformations don't make much sense in general. I was told that they're basically the same things as functions, but then all of a sudden we're adding them together and multiplying them left and right which never happened with f(x). Does that matter? Can I just think of those Ts as fs? If so, I think that will help a lot because, as it is, the T seems very strange and foreign to me.

I would appreciate any tips on such matters :smile:
 

Answers and Replies

  • #2
Dick
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What a transformation T being linear actually means is that T(a*x+b*y)=a*T(x)+b*T(y) where a and b are scalars and x and y are vectors. From that you shouldn't have any problems showing c1*T1+c2*T2 is linear if T1 and T2 are. I think you are confusing the consequences of a theorem with the definition of linear.
 
  • #3
Screwdriver
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Thank you for your help.

So if T1 and T2 are linear then:

[tex]T_1(p\vec{x}+q\vec{x})= pT_1(\vec{x})+qT_1(\vec{x})[/tex]

and

[tex]T_2(r\vec{x}+s\vec{x})= rT_2(\vec{x})+sT_2(\vec{x})[/tex]

Where c, p, q, r and s are all constants.

I want to show that:

[tex]c_1T_1(p\vec{x}+q\vec{x}) + c_2T_2(r\vec{x}+s\vec{x}) = c_1pT_1(\vec{x})+c_1qT_1(\vec{x})+c_1rT_2(\vec{x})+c_1sT_2(\vec{x})[/tex]

If that's the case, then I'm pretty much done, since that result comes just by subbing in the definition of a linear transformation for T1 and T2 and distributing the c.

I really want to know about the function thing though; can I interpret transformations as functions, ie. think of T as f?
 
Last edited:
  • #4
Dick
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Sure you can think of them as functions. It's sort of implicit in the notation T(x), yes? What might be confusing you is that the product of two linear transformations T1 and T2 acting on a vector x is T1(T2(x)). It's really a composition of functions. That's not the same as the product of two real functions which would be f1(x)*f2(x). They are different things.
 
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  • #5
Screwdriver
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That's a huge relief :tongue:.

Yes, the T(x) notation makes sense. The problem is that we use square brackets a lot in the notion and I'm never quite sure whether or not it means the standard matrix or just square brackets for the sake of square brackets. I'm also not sure whether or not it matters since

[tex](T)(S)=T \circ S[/tex]

but also

[tex][T]=[T \circ S][/tex]

That might just be a coincidence (or completely wrong)?
 
  • #6
Dick
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I think it's probably a formal distinction between the linear transformation and its matrix. But in the end it doesn't make much difference, does it? The composition of two matrices is just the matrix product.
 
  • #7
Screwdriver
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I see now. Thank you so much :smile:
 

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