Linear Transformations : Kernel(T) and being One-To-One using polynomial space
$begingroup$
I have this linear Transformation -
$ T : mathbb{P}^2 → mathbb{R}^2 $
in which T is defined by : $T(ax^2+bx+c)$ = $(a+c,2b)$
If we try to find the kernel of this Ker(T) = $0$,
so we set $(a+c,2b)$ = $0$ to get the values of b = $0$, and c = -a.
If we put this back into the quadratic polynomial we have:
$ax^2+bx+c$ = $ax^2+ 0x+(-a)$ = $ax^2-a$
So T($ax^2-a$) = $0$, and we can choose infinitely many values of 'a', so that means we get many of these polynomial vector types being sent to the $0$ vector in the output space.
So i can choose 2 values of 'a' to show that we have more than one input that maps to output.
This shows that this is not One-To-One.
BUT i am having difficulty with the basic definition of when it is one-to-one.
I read that only ONE vector can go to the zero vector in the output space.
But does that mean it can be any one vector?
OR
Since this is a linear transformation, is the only vector allowed be the Zero vector in the Input space and it goes to the Zero vector in the output space.
So my understanding of One-to-One is not that clear when we look at it from the perspective of the Kernel.
Input on this is greatly appreciated!
linear-algebra linear-transformations
asked Dec 22 '18 at 5:38
PaluPalu
3562822
$endgroup$
add a comment |
$begingroup$
I have this linear Transformation -
$ T : mathbb{P}^2 → mathbb{R}^2 $
in which T is defined by : $T(ax^2+bx+c)$ = $(a+c,2b)$
If we try to find the kernel of this Ker(T) = $0$,
so we set $(a+c,2b)$ = $0$ to get the values of b = $0$, and c = -a.
If we put this back into the quadratic polynomial we have:
$ax^2+bx+c$ = $ax^2+ 0x+(-a)$ = $ax^2-a$
So T($ax^2-a$) = $0$, and we can choose infinitely many values of 'a', so that means we get many of these polynomial vector types being sent to the $0$ vector in the output space.
So i can choose 2 values of 'a' to show that we have more than one input that maps to output.
This shows that this is not One-To-One.
BUT i am having difficulty with the basic definition of when it is one-to-one.
I read that only ONE vector can go to the zero vector in the output space.
But does that mean it can be any one vector?
OR
Since this is a linear transformation, is the only vector allowed be the Zero vector in the Input space and it goes to the Zero vector in the output space.
So my understanding of One-to-One is not that clear when we look at it from the perspective of the Kernel.
Input on this is greatly appreciated!
linear-algebra linear-transformations
asked Dec 22 '18 at 5:38
PaluPalu
3562822
$endgroup$
add a comment |
$begingroup$
I have this linear Transformation -
$ T : mathbb{P}^2 → mathbb{R}^2 $
in which T is defined by : $T(ax^2+bx+c)$ = $(a+c,2b)$
If we try to find the kernel of this Ker(T) = $0$,
so we set $(a+c,2b)$ = $0$ to get the values of b = $0$, and c = -a.
If we put this back into the quadratic polynomial we have:
$ax^2+bx+c$ = $ax^2+ 0x+(-a)$ = $ax^2-a$
So T($ax^2-a$) = $0$, and we can choose infinitely many values of 'a', so that means we get many of these polynomial vector types being sent to the $0$ vector in the output space.
So i can choose 2 values of 'a' to show that we have more than one input that maps to output.
This shows that this is not One-To-One.
BUT i am having difficulty with the basic definition of when it is one-to-one.
I read that only ONE vector can go to the zero vector in the output space.
But does that mean it can be any one vector?
OR
Since this is a linear transformation, is the only vector allowed be the Zero vector in the Input space and it goes to the Zero vector in the output space.
So my understanding of One-to-One is not that clear when we look at it from the perspective of the Kernel.
Input on this is greatly appreciated!
linear-algebra linear-transformations
asked Dec 22 '18 at 5:38
PaluPalu
3562822
$endgroup$
I have this linear Transformation -
$ T : mathbb{P}^2 → mathbb{R}^2 $
in which T is defined by : $T(ax^2+bx+c)$ = $(a+c,2b)$
If we try to find the kernel of this Ker(T) = $0$,
so we set $(a+c,2b)$ = $0$ to get the values of b = $0$, and c = -a.
If we put this back into the quadratic polynomial we have:
$ax^2+bx+c$ = $ax^2+ 0x+(-a)$ = $ax^2-a$
So T($ax^2-a$) = $0$, and we can choose infinitely many values of 'a', so that means we get many of these polynomial vector types being sent to the $0$ vector in the output space.
So i can choose 2 values of 'a' to show that we have more than one input that maps to output.
This shows that this is not One-To-One.
BUT i am having difficulty with the basic definition of when it is one-to-one.
I read that only ONE vector can go to the zero vector in the output space.
But does that mean it can be any one vector?
OR
Since this is a linear transformation, is the only vector allowed be the Zero vector in the Input space and it goes to the Zero vector in the output space.
So my understanding of One-to-One is not that clear when we look at it from the perspective of the Kernel.
Input on this is greatly appreciated!
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Dec 22 '18 at 5:38
PaluPalu
3562822
asked Dec 22 '18 at 5:38
PaluPalu
3562822
asked Dec 22 '18 at 5:38
PaluPalu
3562822
asked Dec 22 '18 at 5:38
PaluPalu
3562822
asked Dec 22 '18 at 5:38
PaluPalu
3562822
3562822
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Try proving:
A linear function is one to one if and only if it's kernel is a trivial vector space(contains only zero).
Hints:
$impliedby$ take two distinct vectors and try finding the difference in their image.
answered Dec 22 '18 at 5:55
mm-crjmm-crj
423213
$endgroup$
1
$begingroup$
So hence, based on your statement, that would mean that if i found a non-trivial vector that maps to the Zero vector in the output space, then that is enough to show that the linear transformation is not one-to-one. Let me know if i got this.
$endgroup$
– Palu
Dec 22 '18 at 6:04
1
$begingroup$
Yes, because the zero of the domain maps to the zero of the image anyway due to linearity.
$endgroup$
– mm-crj
Dec 22 '18 at 6:09
$begingroup$
OK, so you clarified my issue now. Thanks for all your help 'mm-crj', i will give your solution as THE solution for this problem.
$endgroup$
– Palu
Dec 22 '18 at 6:11
$begingroup$
Glad that I could help.
$endgroup$
– mm-crj
Dec 22 '18 at 6:13
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Try proving:
A linear function is one to one if and only if it's kernel is a trivial vector space(contains only zero).
Hints:
$impliedby$ take two distinct vectors and try finding the difference in their image.
answered Dec 22 '18 at 5:55
mm-crjmm-crj
423213
$endgroup$
1
$begingroup$
So hence, based on your statement, that would mean that if i found a non-trivial vector that maps to the Zero vector in the output space, then that is enough to show that the linear transformation is not one-to-one. Let me know if i got this.
$endgroup$
– Palu
Dec 22 '18 at 6:04
1
$begingroup$
Yes, because the zero of the domain maps to the zero of the image anyway due to linearity.
$endgroup$
– mm-crj
Dec 22 '18 at 6:09
$begingroup$
OK, so you clarified my issue now. Thanks for all your help 'mm-crj', i will give your solution as THE solution for this problem.
$endgroup$
– Palu
Dec 22 '18 at 6:11
$begingroup$
Glad that I could help.
$endgroup$
– mm-crj
Dec 22 '18 at 6:13
add a comment |
$begingroup$
Try proving:
A linear function is one to one if and only if it's kernel is a trivial vector space(contains only zero).
Hints:
$impliedby$ take two distinct vectors and try finding the difference in their image.
answered Dec 22 '18 at 5:55
mm-crjmm-crj
423213
$endgroup$
1
$begingroup$
So hence, based on your statement, that would mean that if i found a non-trivial vector that maps to the Zero vector in the output space, then that is enough to show that the linear transformation is not one-to-one. Let me know if i got this.
$endgroup$
– Palu
Dec 22 '18 at 6:04
1
$begingroup$
Yes, because the zero of the domain maps to the zero of the image anyway due to linearity.
$endgroup$
– mm-crj
Dec 22 '18 at 6:09
$begingroup$
OK, so you clarified my issue now. Thanks for all your help 'mm-crj', i will give your solution as THE solution for this problem.
$endgroup$
– Palu
Dec 22 '18 at 6:11
$begingroup$
Glad that I could help.
$endgroup$
– mm-crj
Dec 22 '18 at 6:13
add a comment |
$begingroup$
Try proving:
A linear function is one to one if and only if it's kernel is a trivial vector space(contains only zero).
Hints:
$impliedby$ take two distinct vectors and try finding the difference in their image.
answered Dec 22 '18 at 5:55
mm-crjmm-crj
423213
$endgroup$
Try proving:
A linear function is one to one if and only if it's kernel is a trivial vector space(contains only zero).
Hints:
$impliedby$ take two distinct vectors and try finding the difference in their image.
answered Dec 22 '18 at 5:55
mm-crjmm-crj
423213
answered Dec 22 '18 at 5:55
mm-crjmm-crj
423213
answered Dec 22 '18 at 5:55
mm-crjmm-crj
423213
answered Dec 22 '18 at 5:55
mm-crjmm-crj
423213
423213
1
$begingroup$
So hence, based on your statement, that would mean that if i found a non-trivial vector that maps to the Zero vector in the output space, then that is enough to show that the linear transformation is not one-to-one. Let me know if i got this.
$endgroup$
– Palu
Dec 22 '18 at 6:04
1
$begingroup$
Yes, because the zero of the domain maps to the zero of the image anyway due to linearity.
$endgroup$
– mm-crj
Dec 22 '18 at 6:09
$begingroup$
OK, so you clarified my issue now. Thanks for all your help 'mm-crj', i will give your solution as THE solution for this problem.
$endgroup$
– Palu
Dec 22 '18 at 6:11
$begingroup$
Glad that I could help.
$endgroup$
– mm-crj
Dec 22 '18 at 6:13
add a comment |
1
$begingroup$
So hence, based on your statement, that would mean that if i found a non-trivial vector that maps to the Zero vector in the output space, then that is enough to show that the linear transformation is not one-to-one. Let me know if i got this.
$endgroup$
– Palu
Dec 22 '18 at 6:04
1
$begingroup$
Yes, because the zero of the domain maps to the zero of the image anyway due to linearity.
$endgroup$
– mm-crj
Dec 22 '18 at 6:09
$begingroup$
OK, so you clarified my issue now. Thanks for all your help 'mm-crj', i will give your solution as THE solution for this problem.
$endgroup$
– Palu
Dec 22 '18 at 6:11
$begingroup$
Glad that I could help.
$endgroup$
– mm-crj
Dec 22 '18 at 6:13
1
1
$begingroup$
So hence, based on your statement, that would mean that if i found a non-trivial vector that maps to the Zero vector in the output space, then that is enough to show that the linear transformation is not one-to-one. Let me know if i got this.
$endgroup$
– Palu
Dec 22 '18 at 6:04
$begingroup$
So hence, based on your statement, that would mean that if i found a non-trivial vector that maps to the Zero vector in the output space, then that is enough to show that the linear transformation is not one-to-one. Let me know if i got this.
$endgroup$
– Palu
Dec 22 '18 at 6:04
1
1
$begingroup$
Yes, because the zero of the domain maps to the zero of the image anyway due to linearity.
$endgroup$
– mm-crj
Dec 22 '18 at 6:09
$begingroup$
Yes, because the zero of the domain maps to the zero of the image anyway due to linearity.
$endgroup$
– mm-crj
Dec 22 '18 at 6:09
$begingroup$
OK, so you clarified my issue now. Thanks for all your help 'mm-crj', i will give your solution as THE solution for this problem.
$endgroup$
– Palu
Dec 22 '18 at 6:11
$begingroup$
OK, so you clarified my issue now. Thanks for all your help 'mm-crj', i will give your solution as THE solution for this problem.
$endgroup$
– Palu
Dec 22 '18 at 6:11
$begingroup$
Glad that I could help.
$endgroup$
– mm-crj
Dec 22 '18 at 6:13
$begingroup$
Glad that I could help.
$endgroup$
– mm-crj
Dec 22 '18 at 6:13
add a comment |
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