Linear transformation help?

In summary: L(alpha A + beta B) = +(alpha A + I + beta B, I) and +(alpha A + I + beta B, I) = alpha A + beta B + 2*Ialpha L(A) + beta L(B) = alpha (A + I) + beta (B + I) and alpha (A + I) + beta (B + I) = alpha A + beta B + I (alpha + beta)
  • #1
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c . L(A) = A + I

L(alpha A + beta B) = +(alpha A + I + beta B, I) and +(alpha A + I + beta B, I) = alpha A + beta B + 2*I

alpha L(A) + beta L(B) = alpha (A + I) + beta (B + I) and alpha (A + I) + beta (B + I) = alpha A + beta B + I (alpha + beta)

Are these steps correct for the linear transformation?

Since they aren't equal, it isn't a transformation.
 
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  • #2


They are equal only if alpha+beta=2. But yes, in general they aren't equal. L(A)=A+I is not linear. It is a transformation, it's just not linear. You could also check this by seeing if L(0)=0. It's not. BTW what does the "c ." part mean?
 
  • #3


I type up my homework in maple and that was part c of #7.
 
  • #4


And when I copy and paste from maple into the forum weird things happen like the common where there should be a plus and a plus outside () and alpha symbols become words.
 
  • #5


Dustinsfl said:
c . L(A) = A + I
Is there any significance to the c here?
Dustinsfl said:
L(alpha A + beta B) = +(alpha A + I + beta B, I) and +(alpha A + I + beta B, I) = alpha A + beta B + 2*I

alpha L(A) + beta L(B) = alpha (A + I) + beta (B + I) and alpha (A + I) + beta (B + I) = alpha A + beta B + I (alpha + beta)

Are these steps correct for the linear transformation?
It's hard for me to tell what you are doing. Are you trying to show that the transformation L is a linear transformation?
Dustinsfl said:
Since they aren't equal, it isn't a transformation.
It is a transformation, but it may or may not be a linear transformation, which is what I think you're trying to establish.

Presumably A is a square matrix of some size. It would be good for you to include information such as this so we don't have to guess at what you're trying to do and the problem you're working on.

Given that L(A) = A + I, then what you need to do is establish the usual properties for linear transformations: L(A + B) = L(A) + L(B) and L(cA) = cL(A).

I completely don't understand what you're trying to do here:
Dustinsfl said:
L(alpha A + beta B) = +(alpha A + I + beta B, I) and +(alpha A + I + beta B, I) = alpha A + beta B + 2*I

What is the + right after the =?
What does (alpha A + I + beta B, I) mean? You seem to be thinking in terms of vector coordinates, which are not at all appropriate in this problem (if I understand what the problem is).

This is pretty straightforward substitution. If L(A) = A + I, what is L(A + B)? Is L(A + B) = L(A) + L(B)?
 
  • #6


Ok, but it doesn't look linear to me. Do you agree?
 
  • #7


I know it isn't but when I do the transformation steps with matrices, I never get the algebra right for some reason so that is what I was concerned with. Vectors and polynomials are no problem but for some reason the matrices don't work the same for me.
 
  • #8


Dustinsfl said:
I know it isn't but when I do the transformation steps with matrices, I never get the algebra right for some reason so that is what I was concerned with. Vectors and polynomials are no problem but for some reason the matrices don't work the same for me.

Ok, but there's no explicit matrix here. I'm glad you know it's not linear. Just trust your feelings. L(x)=x+1 where x is real isn't linear either, for exactly the same reason. Because if it were, then L(0)=L(0+0)=L(0)+L(0) which means L(0) must be 0. And L(0)=1.
 
  • #9


Dustinsfl said:
I know it isn't but when I do the transformation steps with matrices, I never get the algebra right for some reason so that is what I was concerned with. Vectors and polynomials are no problem but for some reason the matrices don't work the same for me.

The algebra is very simple for this problem, so you would have to really work at it to get it wrong. Take a stab at it - show that L(A + B) [itex]\neq[/itex] L(A) + L(B).
 

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