[Proof]. A linear operator T is injective iff $N(T)= {0}$
up vote
1
down vote
favorite
Part 1
Let $T:X rightarrow Y$ be a linear operator which is injective,
Since $T0=0$ for $0 in X$. Pick $x in X$ s.t $x neq 0$ this would imply that $Tx neq 0$ (as $T$ is injective) then the set of all vector in $X$ which makes $Tx=0$ is the singelton set ${0}$ $implies$ $N(T)={0}$.
Conversely,
Let $N(T)={0}$ $implies$ [ $Tx=0 implies x=0$ for $x in X$ ] now if we pick $x_o neq 0 implies x_o notin N(T) implies Tx_o neq T0 implies$ $T$ is injective.
functional-analysis proof-verification proof-writing linear-transformations
asked Nov 17 at 23:32
HybridAlien
2008
add a comment |
up vote
1
down vote
favorite
Part 1
Let $T:X rightarrow Y$ be a linear operator which is injective,
Since $T0=0$ for $0 in X$. Pick $x in X$ s.t $x neq 0$ this would imply that $Tx neq 0$ (as $T$ is injective) then the set of all vector in $X$ which makes $Tx=0$ is the singelton set ${0}$ $implies$ $N(T)={0}$.
Conversely,
Let $N(T)={0}$ $implies$ [ $Tx=0 implies x=0$ for $x in X$ ] now if we pick $x_o neq 0 implies x_o notin N(T) implies Tx_o neq T0 implies$ $T$ is injective.
functional-analysis proof-verification proof-writing linear-transformations
asked Nov 17 at 23:32
HybridAlien
2008
1
$T$ injective means that $Tx = Ty$ implies $x=y$ for all $x,y$.
– Henno Brandsma
Nov 17 at 23:40
True, also T is injective means that $x neq y$ implies $Tx neq Ty$ for all x,y (the contapositive statement to yours)
– HybridAlien
Nov 18 at 0:06
2
But you only consider the case $x=0$ and $y neq 0$, not arbitrary $x,y$. So your proof does not suffice.
– Henno Brandsma
Nov 18 at 5:53
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Part 1
Let $T:X rightarrow Y$ be a linear operator which is injective,
Since $T0=0$ for $0 in X$. Pick $x in X$ s.t $x neq 0$ this would imply that $Tx neq 0$ (as $T$ is injective) then the set of all vector in $X$ which makes $Tx=0$ is the singelton set ${0}$ $implies$ $N(T)={0}$.
Conversely,
Let $N(T)={0}$ $implies$ [ $Tx=0 implies x=0$ for $x in X$ ] now if we pick $x_o neq 0 implies x_o notin N(T) implies Tx_o neq T0 implies$ $T$ is injective.
functional-analysis proof-verification proof-writing linear-transformations
asked Nov 17 at 23:32
HybridAlien
2008
Part 1
Let $T:X rightarrow Y$ be a linear operator which is injective,
Since $T0=0$ for $0 in X$. Pick $x in X$ s.t $x neq 0$ this would imply that $Tx neq 0$ (as $T$ is injective) then the set of all vector in $X$ which makes $Tx=0$ is the singelton set ${0}$ $implies$ $N(T)={0}$.
Conversely,
Let $N(T)={0}$ $implies$ [ $Tx=0 implies x=0$ for $x in X$ ] now if we pick $x_o neq 0 implies x_o notin N(T) implies Tx_o neq T0 implies$ $T$ is injective.
functional-analysis proof-verification proof-writing linear-transformations
functional-analysis proof-verification proof-writing linear-transformations
asked Nov 17 at 23:32
HybridAlien
2008
asked Nov 17 at 23:32
HybridAlien
2008
asked Nov 17 at 23:32
HybridAlien
2008
asked Nov 17 at 23:32
HybridAlien
2008
asked Nov 17 at 23:32
HybridAlien
2008
2008
1
$T$ injective means that $Tx = Ty$ implies $x=y$ for all $x,y$.
– Henno Brandsma
Nov 17 at 23:40
True, also T is injective means that $x neq y$ implies $Tx neq Ty$ for all x,y (the contapositive statement to yours)
– HybridAlien
Nov 18 at 0:06
2
But you only consider the case $x=0$ and $y neq 0$, not arbitrary $x,y$. So your proof does not suffice.
– Henno Brandsma
Nov 18 at 5:53
add a comment |
1
$T$ injective means that $Tx = Ty$ implies $x=y$ for all $x,y$.
– Henno Brandsma
Nov 17 at 23:40
True, also T is injective means that $x neq y$ implies $Tx neq Ty$ for all x,y (the contapositive statement to yours)
– HybridAlien
Nov 18 at 0:06
2
But you only consider the case $x=0$ and $y neq 0$, not arbitrary $x,y$. So your proof does not suffice.
– Henno Brandsma
Nov 18 at 5:53
1
1
$T$ injective means that $Tx = Ty$ implies $x=y$ for all $x,y$.
– Henno Brandsma
Nov 17 at 23:40
$T$ injective means that $Tx = Ty$ implies $x=y$ for all $x,y$.
– Henno Brandsma
Nov 17 at 23:40
True, also T is injective means that $x neq y$ implies $Tx neq Ty$ for all x,y (the contapositive statement to yours)
– HybridAlien
Nov 18 at 0:06
True, also T is injective means that $x neq y$ implies $Tx neq Ty$ for all x,y (the contapositive statement to yours)
– HybridAlien
Nov 18 at 0:06
2
2
But you only consider the case $x=0$ and $y neq 0$, not arbitrary $x,y$. So your proof does not suffice.
– Henno Brandsma
Nov 18 at 5:53
But you only consider the case $x=0$ and $y neq 0$, not arbitrary $x,y$. So your proof does not suffice.
– Henno Brandsma
Nov 18 at 5:53
add a comment |
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
You do need linearity somewhere.
If $T$ is injective, then $T(0)=0$ and so $x neq 0$ implies $T(x) neq T(0)=0$.
This means that $N(T)={0}$.
Suppose that $N(T)={0}$, then suppose $T(x) = T(y)$. Then $0=T(x)-T(y) = T(x-y)$ by linearity and so $x-y = 0$, as $x-y in N(T)={0}$. Hence $x=y$ and $T$ is indeed injective.
answered Nov 17 at 23:40
Henno Brandsma
102k345111
If $T$ is linear then $T0=0$. So I used the linearity
– HybridAlien
Nov 18 at 0:13
1
@HybridAlien Well, $T(x)=x^2$ (between the reals) also has $N(T)={0}$ but is not injective. It has $T(0)=0$, so you do need to use more of $T$.
– Henno Brandsma
Nov 18 at 5:57
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
You do need linearity somewhere.
If $T$ is injective, then $T(0)=0$ and so $x neq 0$ implies $T(x) neq T(0)=0$.
This means that $N(T)={0}$.
Suppose that $N(T)={0}$, then suppose $T(x) = T(y)$. Then $0=T(x)-T(y) = T(x-y)$ by linearity and so $x-y = 0$, as $x-y in N(T)={0}$. Hence $x=y$ and $T$ is indeed injective.
answered Nov 17 at 23:40
Henno Brandsma
102k345111
If $T$ is linear then $T0=0$. So I used the linearity
– HybridAlien
Nov 18 at 0:13
1
@HybridAlien Well, $T(x)=x^2$ (between the reals) also has $N(T)={0}$ but is not injective. It has $T(0)=0$, so you do need to use more of $T$.
– Henno Brandsma
Nov 18 at 5:57
add a comment |
up vote
5
down vote
accepted
You do need linearity somewhere.
If $T$ is injective, then $T(0)=0$ and so $x neq 0$ implies $T(x) neq T(0)=0$.
This means that $N(T)={0}$.
Suppose that $N(T)={0}$, then suppose $T(x) = T(y)$. Then $0=T(x)-T(y) = T(x-y)$ by linearity and so $x-y = 0$, as $x-y in N(T)={0}$. Hence $x=y$ and $T$ is indeed injective.
answered Nov 17 at 23:40
Henno Brandsma
102k345111
If $T$ is linear then $T0=0$. So I used the linearity
– HybridAlien
Nov 18 at 0:13
1
@HybridAlien Well, $T(x)=x^2$ (between the reals) also has $N(T)={0}$ but is not injective. It has $T(0)=0$, so you do need to use more of $T$.
– Henno Brandsma
Nov 18 at 5:57
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
You do need linearity somewhere.
If $T$ is injective, then $T(0)=0$ and so $x neq 0$ implies $T(x) neq T(0)=0$.
This means that $N(T)={0}$.
Suppose that $N(T)={0}$, then suppose $T(x) = T(y)$. Then $0=T(x)-T(y) = T(x-y)$ by linearity and so $x-y = 0$, as $x-y in N(T)={0}$. Hence $x=y$ and $T$ is indeed injective.
answered Nov 17 at 23:40
Henno Brandsma
102k345111
You do need linearity somewhere.
If $T$ is injective, then $T(0)=0$ and so $x neq 0$ implies $T(x) neq T(0)=0$.
This means that $N(T)={0}$.
Suppose that $N(T)={0}$, then suppose $T(x) = T(y)$. Then $0=T(x)-T(y) = T(x-y)$ by linearity and so $x-y = 0$, as $x-y in N(T)={0}$. Hence $x=y$ and $T$ is indeed injective.
answered Nov 17 at 23:40
Henno Brandsma
102k345111
answered Nov 17 at 23:40
Henno Brandsma
102k345111
answered Nov 17 at 23:40
Henno Brandsma
102k345111
answered Nov 17 at 23:40
Henno Brandsma
102k345111
102k345111
If $T$ is linear then $T0=0$. So I used the linearity
– HybridAlien
Nov 18 at 0:13
1
@HybridAlien Well, $T(x)=x^2$ (between the reals) also has $N(T)={0}$ but is not injective. It has $T(0)=0$, so you do need to use more of $T$.
– Henno Brandsma
Nov 18 at 5:57
add a comment |
If $T$ is linear then $T0=0$. So I used the linearity
– HybridAlien
Nov 18 at 0:13
1
@HybridAlien Well, $T(x)=x^2$ (between the reals) also has $N(T)={0}$ but is not injective. It has $T(0)=0$, so you do need to use more of $T$.
– Henno Brandsma
Nov 18 at 5:57
If $T$ is linear then $T0=0$. So I used the linearity
– HybridAlien
Nov 18 at 0:13
If $T$ is linear then $T0=0$. So I used the linearity
– HybridAlien
Nov 18 at 0:13
1
1
@HybridAlien Well, $T(x)=x^2$ (between the reals) also has $N(T)={0}$ but is not injective. It has $T(0)=0$, so you do need to use more of $T$.
– Henno Brandsma
Nov 18 at 5:57
@HybridAlien Well, $T(x)=x^2$ (between the reals) also has $N(T)={0}$ but is not injective. It has $T(0)=0$, so you do need to use more of $T$.
– Henno Brandsma
Nov 18 at 5:57
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002949%2fproof-a-linear-operator-t-is-injective-iff-nt-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$T$ injective means that $Tx = Ty$ implies $x=y$ for all $x,y$.
– Henno Brandsma
Nov 17 at 23:40
True, also T is injective means that $x neq y$ implies $Tx neq Ty$ for all x,y (the contapositive statement to yours)
– HybridAlien
Nov 18 at 0:06
2
But you only consider the case $x=0$ and $y neq 0$, not arbitrary $x,y$. So your proof does not suffice.
– Henno Brandsma
Nov 18 at 5:53