How to determine if a transformation is linear

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Homework Statement:
Check if the transformation from R^3 to R^3 is linear
Relevant Equations:
Homogen,additiv properties
Hello!

I need to check if this transformation (not sure if it is the right word in English) from ## R^3 to R^3 ## is linear

f(x1,x2,x3) = f(sin(x1),x2+x3,0). Now we are given that the transformation is linear if this you can prove this statement.

$$f(\lambda * u + \mu * v) = \lambda * f(u) + \mu * f(v)$$

Now what confuses me here is what is my u and what is my v? I'd though my u was (x1,x2,x3) and v sas the left brackes so (sin(x1),x2+x3,0) but that doenst seem to make much sence.I've searhed online a bit and I've found no examples of this property that was given to us in class but rathere these propperties.
Homogenous:
$$f(\alpha x) = \alpha f(x)$$
Additive:
$$f(\alpha x + y ) = \alpha f(x) + f(y)$$

Now I am not sure how to approach this,proofs are definetly not my thing and I am stuck on what is my u,v or in the cases i found online my,alpha y?
 

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  • #2
Gaussian97
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You should show some effort before we can help you, what have you done so far?
 
  • #3
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You should show some effort before we can help you, what have you done so far?
Okay I am not sure but latex simply wont cooperate with me, I've thought about assuming what my v and u are I though my u is (x1,x2,x3) and v (x1,x2,x3),represent them as a vector and plug in the formula and I get this (x1+sin(x1) = (x1+sin(x1)
(x2+x2+x3) = (x2+x2+x3)
(x3 ) ( x3 )

Where these brackes are supposed to represent a matrix, so I get that these two equals and so are linear,but I am somehow not sure that is 100% right.
 
  • #4
PeroK
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Okay I am not sure but latex simply wont cooperate with me, I've thought about assuming what my v and u are I though my u is (x1,x2,x3) and v (x1,x2,x3),represent them as a vector and plug in the formula and I get this (x1+sin(x1) = (x1+sin(x1)
(x2+x2+x3) = (x2+x2+x3)
(x3 ) ( x3 )

Where these brackes are supposed to represent a matrix, so I get that these two equals and so are linear,but I am somehow not sure that is 100% right.
##u## and ##v## are any two vectors: in this case, any two vectors in ##\mathbb R^3##. You can't assume that ##u = v##, so you need ##u = (x_1, x_2, x_3)## and ##v = (y_1, y_2, y_3)##, for example.
 
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  • #5
Gaussian97
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Well, probably
Now we are given that the transformation is linear if this you can prove this statement.

$$f(\lambda * u + \mu * v) = \lambda * f(u) + \mu * f(v)$$
Are you sure that this is what has been given to you? There is not any other piece of information?
As you say the previous statement is meaningless unless you define what ##\lambda, \mu, u## and ##v## are.
The correct statement that probably you have is
$$f(\lambda u + \mu v) = \lambda f(u) + \mu f(v), \quad \forall \lambda, \mu \in \mathbb{R},\quad \forall u, v \in \mathbb{R}^3$$

(x1+sin(x1) = (x1+sin(x1)
(x2+x2+x3) = (x2+x2+x3)
(x3 ) ( x3 )
I don't really understand what you did here.
 
  • #6
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Well, probably

Are you sure that this is what has been given to you? There is not any other piece of information?
As you say the previous statement is meaningless unless you define what ##\lambda, \mu, u## and ##v## are.
The correct statement that probably you have is
$$f(\lambda u + \mu v) = \lambda f(u) + \mu f(v), \quad \forall \lambda, \mu \in \mathbb{R},\quad \forall u, v \in \mathbb{R}^3$$


I don't really understand what you did here.
Oh yes I forgot to add, u,v € V and ## \lamba, \mu € R

So I am allowed to input any Real number for lambda and mu and for u and v any two vectors (like PeroK) said and than I should check according to the formula?
 
  • #7
PeroK
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Oh yes I forgot to add, u,v € V and ## \lamba, \mu € R

So I am allowed to input any Real number for lambda and mu and for u and v any two vectors (like PeroK) said and than I should check according to the formula?
It looks like you are struggling with the basic concepts of these definitions. It's not what you are allowed to do, it's what you must check. To show the that a transformation is linear, you have to demonstrate that the linear condition is satisfied for any choice of vectors and any choice of scalars.

Note that to show that a transformation is not linear, you only have to find a single choice of vectors and scalars for which the linear condition fails.
 
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  • #8
Gaussian97
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Yes, but it's not just that you are allowed to put any real value for ##\lambda## and ##\mu## and any vector ##u## and ##v##.

You must prove the equation for all those possible values.
 
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  • #9
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Mhmmm okay, I'll give it a shot
 
  • #10
PeroK
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Mhmmm okay, I'll give it a shot
What do you think? Is ##f## linear or not? Are you going to try to prove that it's linear or show that it's not?

Any ideas before you dive in?
 
  • #11
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Well I've looked at this a bit more and my plan is to prove that it is not linear.Since we have sin in there and that is not linear in itself I'd assume the entire thing is not linear.Since showing that only x1 is not linear the rest (x2,x3) should also not be linear.So I can focus on proving that sin(x1),and find a value where the linear statement will not be valid.
 
  • #12
PeroK
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I agree: ##\sin x## is not a linear function.
 
  • #13
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Okay so I've done it but I am not sure if it is 100% right.Since I think sin(x) is the problem I will try to focus on it to prove that is not linear.

$$ f(\lambda (x1) = f(\lambda x1) = (sin(\lambda x1) $$ Now this has to be qual to this

$$\lambda f(sin(x1)) = \lambda (sin(x1) = (\lambda(sin(x1) $$ Now this is not linear,but what I am not so sure is if I did the first part correctly.Im not sure if i pull in the lambda will it change the argument of sin or not.Since my assumption was it was not linear I'd expect that the argument is the one that makes the diffrence.
 
  • #14
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Okay so I've done it but I am not sure if it is 100% right.Since I think sin(x) is the problem I will try to focus on it to prove that is not linear.

$$ f(\lambda (x1) = f(\lambda x1) = (sin(\lambda x1) $$ Now this has to be qual to this

$$\lambda f(sin(x1)) = \lambda (sin(x1) = (\lambda(sin(x1) $$ Now this is not linear,but what I am not so sure is if I did the first part correctly.Im not sure if i pull in the lambda will it change the argument of sin or not.Since my assumption was it was not linear I'd expect that the argument is the one that makes the diffrence.
Yes, you have done that part correctly and it will affect the argument. That looks fine to me - it can be seen that the equality does not hold true for all ## x_1 ##
 
  • #15
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Yes, you have done that part correctly and it will affect the argument. That looks fine to me - it can be seen that the equality does not hold true for all ## x_1 ##
Perfect,thank you for your help! (all of you) :)
 
  • #16
Gaussian97
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Okay so I've done it but I am not sure if it is 100% right.Since I think sin(x) is the problem I will try to focus on it to prove that is not linear.

$$ f(\lambda (x1) = f(\lambda x1) = (sin(\lambda x1) $$ Now this has to be qual to this

$$\lambda f(sin(x1)) = \lambda (sin(x1) = (\lambda(sin(x1) $$ Now this is not linear,but what I am not so sure is if I did the first part correctly.Im not sure if i pull in the lambda will it change the argument of sin or not.Since my assumption was it was not linear I'd expect that the argument is the one that makes the diffrence.
Well, first of all, why is there a ##f(\sin{x_1})##? I assume it is an error?
Also how do you know that ##\sin{(\lambda x_1)} \neq \lambda \sin{(x_1)}##?
 
  • #17
PeroK
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Okay so I've done it but I am not sure if it is 100% right.Since I think sin(x) is the problem I will try to focus on it to prove that is not linear.

$$ f(\lambda (x1) = f(\lambda x1) = (sin(\lambda x1) $$ Now this has to be qual to this

$$\lambda f(sin(x1)) = \lambda (sin(x1) = (\lambda(sin(x1) $$ Now this is not linear,but what I am not so sure is if I did the first part correctly.Im not sure if i pull in the lambda will it change the argument of sin or not.Since my assumption was it was not linear I'd expect that the argument is the one that makes the diffrence.
You've not provided a proof or demonstration that ##f## is not linear other than to state (without proof) that ##sine## is not linear.

That may be sufficient or not, depending on the expectations of the question setter.

In particular, you haven't produced any vectors or scalars where the non-linearity fails.

Moreover, what you've done doesn't demonstrate that you understand the concept of a counterexample.
 
  • #18
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I will try to focus on it to prove that is not linear.
$$ f(\lambda (x1) = f(\lambda x1) = (sin(\lambda x1) $$ Now this has to be qual to this

$$\lambda f(sin(x1)) = \lambda (sin(x1) = (\lambda(sin(x1) $$
Minor nit: all three expressions above have unbalanced parentheses.
Well, first of all, why is there a ##f(sin(x_1))##
To elaborate on the comments above, ##f## is a function whose inputs are vectors in ##\mathbb R^3##. For example, ##<\sin(x_1), x_2, x_3>##
 
  • #19
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Well I've tested that ## \lambda sin(x1)## gives a diffrent result than ##sin(\lambda x1) ## Also I've gotten a review on this question and it was sufficient,they were pleased with it.For a fact simply recognizing that sin is not linear was actually enough.Although I am interested now to see how a full written proof would look like.
 
  • #20
PeroK
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Well I've tested that ## \lambda sin(x1)## gives a diffrent result than ##sin(\lambda x1) ## Also I've gotten a review on this question and it was sufficient,they were pleased with it.For a fact simply recognizing that sin is not linear was actually enough.Although I am interested now to see how a full written proof would look like.
What did you use for ##\lambda## and ##x_1##?
 
  • #21
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Well I've tested that ## \lambda sin(x1)## gives a diffrent result than ##sin(\lambda x1) ## Also I've gotten a review on this question and it was sufficient,they were pleased with it.For a fact simply recognizing that sin is not linear was actually enough.Although I am interested now to see how a full written proof would look like.
Are you referring to the comment by Master1024 in this thread? You should note that you received several replies that said your proof was not a good one.

A proper proof would show that ##f(\lambda \vec u + \mu \vec v) \ne \lambda f(\vec u) + \mu f(\vec v)##, where ##\lambda## and ##\mu## are scalars, and ##\vec u## and ##\vec v## are vectors in ##\mathbb R^3##. This should be pretty straightforward.
 
  • #22
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No no, this was a (not sure how to say it in english),but like an exercise sheet that was not mandatory,but rather we do and simply discuss in class (online class). And I've showed it to the tutor and he said it was fine,in the sence it was important to see that sin is not linear.We very rarely do proofs and they are almost never in our exams/tests,hence this entire exercise sheet was simply there for the people curious.And I am pretty bad at proofs so I though I'd be fun.
 
  • #23
Office_Shredder
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To prove that sin is not linear you can just give an explicit counterexample

$$\sin(2*\pi/2)=? 2*\sin(\pi/2)$$

This is obviously false, the left side is 0 and the right side is 2.
 
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  • #24
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Apologies @Mark44 @PeroK for causing issues... it wasn't my intention. I didn't think it was necessary to write out any counter-example in my post as I thought the OP would just pick one when they write out their official solution for their class. I thought they were simply asking about the position of the ## \lambda ##. I do agree that writing a (counter-)example down is needed to justify that statement.

Apologies for the inconvenience.
 
  • #25
PeroK
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I don't think you caused any problems. It's sometimes difficult for the helpers to know what's expected in terms of rigour.

The problem for the OP, as I see it, is that if you have an easy question and just wave your arms and say it's obvious, then you don't develop the techniques to deal with harder problems. And, in particular, many students seem to struggle with the concept of finding a single counterexample. This seems like a missed opportunity to me.
 

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