KerF and ImF of linear copying

  • Thread starter Physicsissuef
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That's not what we got before! Last time we found a basis for the image with dimension 2! What went wrong? Nothing. Last time we were looking at an entirely different linear transformation. This time we have a different linear transformation. When you have different linear transformations, you will, usually, have different dimensions for the image and kernel. This time, we have exactly the same image and kernel as before: kernel dimension 2, image dimension 1. That's because we have exactly the same linear transformation. T<x, y, z>= <x- y+ z, 2x- 2y+ 2z, -x+ y-
  • #1

Homework Statement



Find just one base of the set of pictures and one base of the kernel.

[tex]f:\mathbb{R}^3\rightarrow\mathbb{R}^3\ ,f(x,y,z)=(x+2y-z,y+z,x+y-2z)[/tex]

Homework Equations



[tex]Im F=\{f(v)| v\in V\}[/tex]

[tex]ker F=\{v\in V | F(v)=0\}[/tex]

The Attempt at a Solution



Actually, I don't know the principle of solving this tasks. My textbook is rubbish. It doesn't say anything about this problem, or how to solve it. Thanks for your time, and help.
 
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  • #2
I find it very difficult to believe your book does not have anything about the definitions of "image" (what you are calling a "picture"- translation problem obviously. "Picture" is a very specific word, "image" has a much more general meaning that can include "picture".) or "kernel" (exactly the right word)- and that's all you should need. You should not expect your textbook to tell you how to solve every problem it has- that defeats the purpose of "education"- which is not just memorizing formulas or how to solve specific problems.

f(x,y,z)= (x+ 2y- z, y+ z, x+ y- 2z) has kernel given by x+ 2y- z= 0, y+ z= 0, x+y- 2z= 0 (that's directly from the definition of "kernel"). Solve those equations for x, y, and z to see what value give <0, 0, 0>. Adding the first two equations, x+ 3y= 0. Adding twice the second equation to the third, x+ 3y= 0. Those two equations are the same and are satisfied by any (x, y, z) such that x= -3y and, from the second equation, z= -y. In particular if we take y= 1, then x= -3, z= 1 so <-3, 1, -1> is a vector in this one dimensional subspace. Since that has dimension 1, you should know from the "dimension theorem" (you may know it by another name), that the image must have dimension 3- = 2.

To find the image, look at <x+ 2y- z, y+ z, x+y- 2z> and think about the values of those components for different values of the three variables, x, y, and z. As I have suggested before, let x= 1, y= z= 0, and you get <1, 0, 1>. Let x= z= 0, y= 1 and you get <2, 1, 1>. Let x= y= 0, z= 1 and you get <-1, 1, -2>. But notice that <-1, 1, -2>= <1, 0, 1>- 2<2, 1, 1>! (Which makes nonsense of what I said about this choice of "1" and "0" for each variable guarantee independent vectors!). Since the third vector is a linear combination of the first two, {<1, 0, 1>, <2, 1, 1>} is a basis for the image of the linear transformation- whidh is, as we checked before, dimension 2.
 
  • #3
I'll show you sometimes my book, you'll see. There are just equations, and not any examples. Ok, I understand all this. But I have few questions. For example,
How you know that the vector is one dimensional subspace, so you don't get 2-nd series of values for x,y or y,z or x,z?
 
  • #4
For the kernel, solving the equations defining the kernel, I arrived at the two independent equations x+ 3y= 0 and y+ z= 0. That told me that I could choose any one of x, y, or z however I wished and solve for the other 2. I have "one degree of freedom" and so the subspace is one dimensional. Of course, if all three equations had been independent, I could have solved for the unique solution: x= y= z= 0: in that case the linear transformation would be "one-to-one".

It is possible, say the linear transformation were T<x,y,z>= <x- y+ z, 2x- 2y+ 2z, -x+ y- z>, that I would find that all 3 equations, x- y+ z= 0, 2x- 2y+ 2z= 0, -x+ y- z= 0 were just multiples of one another: they all reduce to x- y+ z= 0. In general, one equation allows us to solve for one of the variables and reduce the dimension by one. Since there is only one independent equation here, the dimension of the kernel is 3- 1= 2.

More precisely, I can solve for y= x+ z so I am free to choose any values I want for x or z and then solve for y: If x= 1 and z= 0 (I'm a simple soul- I like simple numbers) y= 1 so we have <1, 1, 0>. If x= 0 and z= 1, y= 1 so <0, 1, 1>. A basis for the kernel is {<1, 1, 0>, <0, 1, 1>}. Any vector in that space can be written as a<1, 1, 0>+ b<0, 1, 1>= <a, a+b, b>. I'll leave it to you to see that the linear transformation applied to <a, a+ b, b> gives <0, 0, 0>.

Of course, any subspace has an infinite number of possible bases. We could as easily have solved x-y+z= 0 for z= y- x. Now taking x= 1, y= 0, then z= -1 and if we take x= 0, y= 1, then z= 1. We get {<1, 0, -1> and <0, 1, 1> which is another basis for the same subspace. You can prove that by showing that <1, 0, -1> and <0, 1, 1> can be written as linear combinations of <1, 1, 0> and <0, 1, 1> and vice-versa.

How about the image of this linear transformation? T<x, y, z= <x- y+ z, 2x- 2y+ 2z, -x+ y- z> so, taking x= 1, y= z= 0, T<1, 0, 0>= <1, 2, ->, taking x= z= 0, y= 1, T<0, 1, 0>= <-1, -2, 1> and taking x= y= 0, z= 1, T<0, 0, 1>= <1, 2, -1>. It's easy to see that those are all multiples of one another so we really have just one independent vector. We could use anyone of those as the single vector in the basis. The image has dimension 1 and again we see that the dimension of the kernel plus the dimension of the image is the dimension of the whole space, 3.

By the way do you see how my recommendation of using "x= 1, y= z= 0, then x= z= 0, y= 1, etc." just gives the standard basis: <1, 0, 0>, <0, 1, 0>, <0, 0, 1>?
 
  • #5
Yes, nice recommendation. Again thank you very much for the help. You're one of the rarely good and patient persons. Thanks for your efforts to help me, even with my problem with English language.
 
  • #6
I have one task in my book:
[tex]f:\mathbb{R}^3\rightarrow\mathbb{R}^2\ ,f(x,y,z)=(x-2y,x-2y)[/tex]
Just tell me please, do I solved the problem correctly?
First for kerF.
x-2y=0
x-2y=0

x=2y
y=1
(2,1)
So the base is one dimensional.
3-1=2
The image will be 2 dimensional.
x=1
y=0
<1,1,0>
x=-1
y=1
<-1,-1,0>
x=3
y=0
<3,3,0>
But <3,3,0>=3<1,1,0>
<1,1,0>=-1<-1,-1,0>
The dimension will be 1 here?
What is the problem?
 
  • #7
Physicsissuef said:
I have one task in my book:
[tex]f:\mathbb{R}^3\rightarrow\mathbb{R}^2\ ,f(x,y,z)=(x-2y,x-2y)[/tex]
Just tell me please, do I solved the problem correctly?
First for kerF.
x-2y=0
x-2y=0

x=2y
y=1
(2,1)
So the base is one dimensional.
3-1=2
The image will be 2 dimensional.
x=1
y=0
<1,1,0>
x=-1
y=1
<-1,-1,0>
x=3
y=0
<3,3,0>
But <3,3,0>=3<1,1,0>
<1,1,0>=-1<-1,-1,0>
The dimension will be 1 here?
What is the problem?
When I did this about 2 hours ago, I did it in a rush and was completely wrong.

In looking at the kernel, you ignored z! Yes, to be in the kernel, x must equal 2y and so <2, 1, 0> is a vector in the kernel. But that is true for any z. In particular, < 0, 0, 1> is in the kernel also. A basis is {<2, 1, 0>, <0, 0, 1>} (or {<2, 1, 0>, <2, 1, 1> etc.) and the kernel has dimension 2.

Yes, it should be obvious that no matter what x, y, and z are the result of the linear transformation is <a, a> since the two components are the same. A basis for the image is {<1, 1>} and the image is 1 dimensional: 2+ 1= 3, the dimension of the domain space.
 
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  • #8
HallsofIvy said:
When I did this about 2 hours ago, I did it in a rush and was completely wrong.

In looking at the kernel, you ignored z! Yes, to be in the kernel, x must equal 2y and so <2, 1, 0> is a vector in the kernel. But that is true for any z. In particular, < 0, 0, 1> is in the kernel also. A basis is {<2, 1, 0>, <0, 0, 1>} (or {<2, 1, 0>, <2, 1, 1> etc.) and the kernel has dimension 2.

Yes, it should be obvious that no matter what x, y, and z are the result of the linear transformation is <a, a> since the two components are the same. A basis for the image is {<1, 1>} and the image is 1 dimensional: 2+ 1= 3, the dimension of the domain space.
Can I implement here the "dimension theorem" and how?? Is it correct, if I write for the image {1,1,0} instead of {1,1}? Can the image be {1,1,1}?
 
  • #9
It is possible, say the linear transformation were T<x,y,z>= <x- y+ z, 2x- 2y+ 2z, -x+ y- z>, that I would find that all 3 equations, x- y+ z= 0, 2x- 2y+ 2z= 0, -x+ y- z= 0 were just multiples of one another: they all reduce to x- y+ z= 0. In general, one equation allows us to solve for one of the variables and reduce the dimension by one. Since there is only one independent equation here, the dimension of the kernel is 3- 1= 2.
You have some mistake here. It should be 3-2=1. In the last example of mine, if I use the dimension theorem, should I get [tex]dimR^2=2 or R^3=3[/tex]?
 
  • #10
Physicsissuef said:
Can I implement here the "dimension theorem" and how?? Is it correct, if I write for the image {1,1,0} instead of {1,1}? Can the image be {1,1,1}?
You can't write the image as "the subspace spanned by <1, 1, 0> because the image is a subset of R2. And, by the way, the image is not "{1, 1, 1}" because that is a single vector. The image, a subspace of R2, is the space spanned by the vector <1, 1>: all multiples of that vector or all vectors of the form <a, a>

Physicsissuef said:
You have some mistake here. It should be 3-2=1. In the last example of mine, if I use the dimension theorem, should I get [tex]dimR^2=2 or R^3=3[/tex]?

No, there is not a mistake. The kernel is all <x, y, z> such that T<x, y, z>= <0, 0, 0>. As I showed, the all reduces to single equation x- y+ z= 0. Since R3 has dimension 3, that one equation reduce the dimension of the kernel to 3- 1= 2. We could also do this: if x= 1, y= 0, z satisfies 1- 0+ z= 0 so z= -1. <1, 0, -1> is a vector in the kernel. If x= 0, y= 1, z satisfies 0- 1+ z= 0 so z= 1. <0, 1, 1> is a vector in the kernel and {<1, 0, -1> , <0, 1, 1>} is a basis for the kernel. Again, it has dimension 2.

The dimension of the kernel is 2 so the dimension of the image is 3- 2= 1. In fact, it is easy to see (taking x= 1, y= 0, z= 0 for example) that a vector in the image is <1, 2, -1> and you should be able to see that if you use any other values for x, y, and z, you get multiples of that. The image is spanned by the single vector <1, 2, -1>. {<1, 2, -1>} is a basis for the image which has dimension 1.

The "dimension theorem" says that if T is a linear transformation from U to V, the dimension of the kernel of T plus the dimension of the image of T is equal to the dimension of U. In your example, that was 3.
 
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  • #11
And how we will use the dimension theorem on the last example?
[tex]f:\mathbb{R}^3\rightarrow\mathbb{R}^2\ ,f(x,y,z)=(x-2y,x-2y)[/tex]
 
  • #12
Physicsissuef said:
My textbook is rubbish.

I have to ask, what textbook are you using? That is, what is the title and author?

The reason I ask is that if I know the level of the book, I might have a link to a free online book that is better.
 
  • #13
Physicsissuef said:
And how we will use the dimension theorem on the last example?
[tex]f:\mathbb{R}^3\rightarrow\mathbb{R}^2\ ,f(x,y,z)=(x-2y,x-2y)[/tex]

You really don't need the "dimension theorem" except as a check.

I have already shown that the kernel has basis {<2, 1, 0>, <0, 0, 1>} and so has dimension 2, and that the image has basis {< 1, 1>} and so has dimension 1. We can use that fact that dim of ker (f)+ dim of img(f)= 2+ 1= 3 = dimension of R3 as a check.
 
  • #14
HallsofIvy said:
You really don't need the "dimension theorem" except as a check.

I have already shown that the kernel has basis {<2, 1, 0>, <0, 0, 1>} and so has dimension 2, and that the image has basis {< 1, 1>} and so has dimension 1. We can use that fact that dim of ker (f)+ dim of img(f)= 2+ 1= 3 = dimension of R3 as a check.

But in previous posts you said me that I should use R2 (dimension 2) instead of R3, because of the linear copying. Why now R3?
Tom, I use textbook in my native language, which is called,
Linear Algebra and Analytic Geometry for 3rd grade high school. In my country, there are (now), 1-9 grades primary school, and 1-4 grades high school. Here are some screen shots:
http://aycu40.webshots.com/image/46399/2000404273812954927_rs.jpg"
http://aycu03.webshots.com/image/46762/2000444365217069959_rs.jpg"
http://aycu27.webshots.com/image/43466/2000425691560932769_rs.jpg"
 
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  • #15
Where did I say that?

In post 10 of this thread I said:
"The "dimension theorem" says that if T is a linear transformation from U to V, the dimension of the kernel of T plus the dimension of the image of T is equal to the dimension of U. In your example, that was 3."
 
  • #16
Oh, you're from Macedonia! Lovely country. I was there the summer of 2006.

Spent most of my time in Skopje, but managed a few days in Ohrid.
 
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  • #17
HallsofIvy said:
Oh, you're from Macedonia! Lovely country. I was there the summer of 2006.

Spent most of my time in Skopje, but managed a few days in Ohrid.

Thanks, I am glad, that you have visited Macedonia. What were you doing in Macedonia?
Btw- Isn't [tex]\mathbb{R}^3[/tex] linear copying into [tex]\mathbb{R}^2[/tex], so the dimension should be 2?
 
  • #18
Why we get <x,y> for image and <x,y,z> (three coordinates) for kernel?
 
  • #19
Look at the definitions! If T is a linear transformation from U to V, the kernel of T is the set of all vectors u in U such that Tu= 0. They must be in U so that T can be applied to them. The image of T is the set of all vectors v in V such that v= Tx for some x in U. They must be in V since they are the result of applying T to a vector.

If T is from R3 to R2, then the kernel is a subspace of R3 and the image is a subspace of R2.

I was visiting my finacee who worked in the U.S. Embassy there.
 
  • #20
Does [tex]0\in V[/tex], for the kernel is T(u)=0?
Ok, and for this example:
[tex]f:\mathbb{R}^3\rightarrow\mathbb{R}^3\ ,f(x,y,z)=(x-2y,x-3y,x+y-z)[/tex]
For kerF, (x-2y,x-3y,x+y-z)=(0,0,0)
x=0
y=0
z=0
Why in my textbook result, they wrote [itex]\varnothing[/itex] instead of (0,0,0)?
 
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  • #21
Please go back and review the basic definitions. There is nothing more important in learning mathematics than knowing the precise wording of the definitions!

Yes, of course 0 is in V. Having an additive identity, 0, is part of the definition of "vector space". The 0 vector is in every vector space.

The example you give here, f(x,y,z)= (x- 2y, x- 3y, x+ y- z) , has, as it's kernel, all (x,y,z) satisfying x- 2y= 0, x- 3y= 0 and x+ y- z= 0. Yes, by, perhaps, subtracting the second equation from the first to get y= 0, then putting that back into the first equation to get x= 0, finally putting x=y= 0 into the third equation to get z= 0, you find that the kernel consists of the single 0 vector {(0, 0, 0)}. [itex]\varnothing[/itex] is a standard notation for the "empty set". I don't know that it is commonly used to represent the 0 vector, but that isn't a large leap if they want to avoid confusing the 0 vector with the number 0. Look again. Are they using [itex]\varnothing[/itex] to represent the 0 itself or perhaps what they were saying is that a basis for the kernel is the empty set?

By the way, the fact that the kernel is "trivial", tells us that the linear transformation is "one-to-one": that the only way you can have f(x)= f(y) is if x= y. Also, it follows that the image of f is all of R3- that f is an "onto" function. The fact that it is both "one-to-one" and "onto" (for linear transformations, either one implies the other) means that f is "invertible". That is, if we have x-2y= a, x- 3y= b, x+ y- z= c, we could solve for x, y, and z so we can go from any point (a, b, c) in R3 back to its "preimage", the (x,y,z) such that f(x,y,z)= (a,b,c).
 
  • #22
HallsofIvy said:
Please go back and review the basic definitions. There is nothing more important in learning mathematics than knowing the precise wording of the definitions!

Yes, of course 0 is in V. Having an additive identity, 0, is part of the definition of "vector space". The 0 vector is in every vector space.

The example you give here, f(x,y,z)= (x- 2y, x- 3y, x+ y- z) , has, as it's kernel, all (x,y,z) satisfying x- 2y= 0, x- 3y= 0 and x+ y- z= 0. Yes, by, perhaps, subtracting the second equation from the first to get y= 0, then putting that back into the first equation to get x= 0, finally putting x=y= 0 into the third equation to get z= 0, you find that the kernel consists of the single 0 vector {(0, 0, 0)}. [itex]\varnothing[/itex] is a standard notation for the "empty set". I don't know that it is commonly used to represent the 0 vector, but that isn't a large leap if they want to avoid confusing the 0 vector with the number 0. Look again. Are they using [itex]\varnothing[/itex] to represent the 0 itself or perhaps what they were saying is that a basis for the kernel is the empty set?

By the way, the fact that the kernel is "trivial", tells us that the linear transformation is "one-to-one": that the only way you can have f(x)= f(y) is if x= y. Also, it follows that the image of f is all of R3- that f is an "onto" function. The fact that it is both "one-to-one" and "onto" (for linear transformations, either one implies the other) means that f is "invertible". That is, if we have x-2y= a, x- 3y= b, x+ y- z= c, we could solve for x, y, and z so we can go from any point (a, b, c) in R3 back to its "preimage", the (x,y,z) such that f(x,y,z)= (a,b,c).

They say the basis is empty set. Also, I sow in "Vector subspace" thread, Tom is not getting <0,0,0> as dimension 1.
 
  • #23
Okay, so they are not saying that [itex]\varnothing[/itex] is the zero vector they are saying the basis is the empty set?
 
  • #24
HallsofIvy said:
Okay, so they are not saying that [itex]\varnothing[/itex] is the zero vector they are saying the basis is the empty set?

Yes.
 
  • #25
"<0,0,0> as dimension 1." Be careful about your wording. A single vector doesn't have a "dimension", only vector spaces have dimension (in Linear Algebra). As I have said before, the vector space containing only the single vector 0 has the empty set as its basis and so has dimension 0. I can't find anywhere in the thread about intersections where Tom says that space has dimension 1. He may have said (or implied) that you can determine the dimension of a space by counting the number of vectors in a basis, but, again, a basis for the space containing only the 0 vector is empty and so contains 0 vectors. The dimension of that space is 0. If any non-zero vector is in a vector space, then we can always have a basis containing at least that vector and so the dimension is at least 1.
 
  • #26
And is this possible?

[tex]\mathbb{R}^2: \mathbb{R}[/tex]

[tex]f:(x,y)=(x+y,z)[/tex]

the kernel will be:
y=-x
z=0
for x=1, will the base for the kernel be (1,-1)?
and the base for ImF will be (1)?
dim(kerF)+dim(ImF)=1+1=2

Right?
 
  • #27
Physicsissuef said:
And is this possible?

[tex]\mathbb{R}^2: \mathbb{R}[/tex]
What do you mean by this? f is a function from R2 to R?

[tex]f:(x,y)=(x+y,z)[/tex]
What is z? If f is from R2, then it must be f(x,y) and you cannot have another variable z in the definition of f! And, of course, if the value of the function is to be in R, it cannot have 2 components. You could define this as a function from R3 to R2.

the kernel will be:
y=-x
z=0
for x=1, will the base for the kernel be (1,-1)?
As I said before, you can't just ignore z. If this is from R3 to R2, then a basis for the kernel would be {(1, -1, 0)}.
and the base for ImF will be (1)?
The image is all things in R2 that can be written as (x+y, z) for some x and z- that can be anything. The image of f is all of R2 and its basis is {(1, 0), (0, 1)}.
dim(kerF)+dim(ImF)=1+1=2
The way I have changed your problem, dim(ker f)+ dim(im f)= 1+ 2= 3.

Right?
Perhaps you meant f:R2->R defined by f(x, y)= x+ y. Then the kernel is all (x,y) such that x+ y= 0 or y= -1. A basis for that subspace is {(1, -1)}. The image is all real numbers that can be written as x+y for some real numbers x and y. Obviously that is all real numbers (take y= 0, x any real number) so I am f= R and a basis is {(0)}. Now, dim(ker f)+ dim(im f)= 1+ 1= 2.
 
  • #28
So the correct one, should be
[tex]f: \mathbb {R}^3 \rightarrow \mathbb {R}^2[/tex]
So:
f(x,y,z)=(x+y,z)
right?
In the previous case I meant [tex]f: \mathbb {R}^2 \rightarrow \mathbb{R}[/tex] (I forgot to write \rightarrow in latex code).
 
  • #29
And what if [tex]f: \mathbb{R}^2 \rightarrow \mathbb{R}^2[/tex]
f(x.y)=(x+y,z)
is this possible now?
f(x,0)=(x+y,0)
amd
f(0,y)=(0,z)
so I can choose y and z any numbers right?
 
  • #30
Physicsissuef said:
And what if [tex]f: \mathbb{R}^2 \rightarrow \mathbb{R}^2[/tex]
f(x.y)=(x+y,z)
is this possible now?
f(x,0)=(x+y,0)
amd
f(0,y)=(0,z)
so I can choose y and z any numbers right?

"f(x,y)= (x+y,z)" makes no sense because "f(x,y)" does not tell us what z is! How would you find f(1, 1) or f(2, 3)? Perhaps, if you had already said that z= 5, say, then you could write f(x,y)= (x+y,z) which would simply be the same as f(x,y)= (x+y, 5). But if z is intended to be a variable, it's value MUST be given in the "variable list" f(x,y,z).
 

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