Mixing
Prove that linear operator T is the projection Operator
$begingroup$
I got question prove or disprove
Let be $T:mathbb R^n to mathbb R^n$ linear oprator and let be $Usubsetmathbb R^n$ hyperplane.
If known that $Image(T)=U$ and also that for every $uin U$, $T(u)=u$.
So $T$ is projection operator on $U$.
Thank you for the help :)
linear-transformations projection
edited Jan 26 at 14:11
Mundron Schmidt
7,5042729
asked Jan 26 at 14:06
Or Yehuda Ben ShimolOr Yehuda Ben Shimol
132
$endgroup$
add a comment |
$begingroup$
I got question prove or disprove
Let be $T:mathbb R^n to mathbb R^n$ linear oprator and let be $Usubsetmathbb R^n$ hyperplane.
If known that $Image(T)=U$ and also that for every $uin U$, $T(u)=u$.
So $T$ is projection operator on $U$.
Thank you for the help :)
linear-transformations projection
edited Jan 26 at 14:11
Mundron Schmidt
7,5042729
asked Jan 26 at 14:06
Or Yehuda Ben ShimolOr Yehuda Ben Shimol
132
$endgroup$
add a comment |
$begingroup$
I got question prove or disprove
Let be $T:mathbb R^n to mathbb R^n$ linear oprator and let be $Usubsetmathbb R^n$ hyperplane.
If known that $Image(T)=U$ and also that for every $uin U$, $T(u)=u$.
So $T$ is projection operator on $U$.
Thank you for the help :)
linear-transformations projection
edited Jan 26 at 14:11
Mundron Schmidt
7,5042729
asked Jan 26 at 14:06
Or Yehuda Ben ShimolOr Yehuda Ben Shimol
132
$endgroup$
I got question prove or disprove
Let be $T:mathbb R^n to mathbb R^n$ linear oprator and let be $Usubsetmathbb R^n$ hyperplane.
If known that $Image(T)=U$ and also that for every $uin U$, $T(u)=u$.
So $T$ is projection operator on $U$.
Thank you for the help :)
linear-transformations projection
linear-transformations projection
edited Jan 26 at 14:11
Mundron Schmidt
7,5042729
asked Jan 26 at 14:06
Or Yehuda Ben ShimolOr Yehuda Ben Shimol
132
edited Jan 26 at 14:11
Mundron Schmidt
7,5042729
asked Jan 26 at 14:06
Or Yehuda Ben ShimolOr Yehuda Ben Shimol
132
edited Jan 26 at 14:11
Mundron Schmidt
7,5042729
edited Jan 26 at 14:11
Mundron Schmidt
7,5042729
edited Jan 26 at 14:11
Mundron Schmidt
7,5042729
7,5042729
asked Jan 26 at 14:06
Or Yehuda Ben ShimolOr Yehuda Ben Shimol
132
asked Jan 26 at 14:06
Or Yehuda Ben ShimolOr Yehuda Ben Shimol
132
asked Jan 26 at 14:06
Or Yehuda Ben ShimolOr Yehuda Ben Shimol
132
132
add a comment |
add a comment |
1 Answer
1
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$begingroup$
I assume, the projection operator is defined as $T^2=T$.
Consider $vin mathbb R^n$. Can you think of a way, why $T^2v=Tv$ holds? It is maybe easier to see if you write
$$
T^2v=Tv Leftrightarrow T(T(v)) = T(v).
$$
You don't to it normally this way, but let us make it one step more obvious:
We define $w=Tv$. Now we get
$$
T(T(v))=T(v) Leftrightarrow T(w)=w.
$$
Consider, that $w$ is an element of the image of $T$.
And that this holds, is writen in your condition.
answered Jan 26 at 14:15
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
$begingroup$
I didn't understand, can you give me an example maybe ?
$endgroup$
– Or Yehuda Ben Shimol
Jan 26 at 15:43
$begingroup$
I rewrote it a bit. But the step to see, why $T^2=T$ holds if $T$ is the identity on its image, which is a hyperplane, is a very small one... It isn't even worth to be called a proof.
$endgroup$
– Mundron Schmidt
Jan 26 at 16:10
$begingroup$
Thank you very much !!!
$endgroup$
– Or Yehuda Ben Shimol
Jan 26 at 19:13
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
I assume, the projection operator is defined as $T^2=T$.
Consider $vin mathbb R^n$. Can you think of a way, why $T^2v=Tv$ holds? It is maybe easier to see if you write
$$
T^2v=Tv Leftrightarrow T(T(v)) = T(v).
$$
You don't to it normally this way, but let us make it one step more obvious:
We define $w=Tv$. Now we get
$$
T(T(v))=T(v) Leftrightarrow T(w)=w.
$$
Consider, that $w$ is an element of the image of $T$.
And that this holds, is writen in your condition.
answered Jan 26 at 14:15
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
$begingroup$
I didn't understand, can you give me an example maybe ?
$endgroup$
– Or Yehuda Ben Shimol
Jan 26 at 15:43
$begingroup$
I rewrote it a bit. But the step to see, why $T^2=T$ holds if $T$ is the identity on its image, which is a hyperplane, is a very small one... It isn't even worth to be called a proof.
$endgroup$
– Mundron Schmidt
Jan 26 at 16:10
$begingroup$
Thank you very much !!!
$endgroup$
– Or Yehuda Ben Shimol
Jan 26 at 19:13
add a comment |
$begingroup$
I assume, the projection operator is defined as $T^2=T$.
Consider $vin mathbb R^n$. Can you think of a way, why $T^2v=Tv$ holds? It is maybe easier to see if you write
$$
T^2v=Tv Leftrightarrow T(T(v)) = T(v).
$$
You don't to it normally this way, but let us make it one step more obvious:
We define $w=Tv$. Now we get
$$
T(T(v))=T(v) Leftrightarrow T(w)=w.
$$
Consider, that $w$ is an element of the image of $T$.
And that this holds, is writen in your condition.
answered Jan 26 at 14:15
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
$begingroup$
I didn't understand, can you give me an example maybe ?
$endgroup$
– Or Yehuda Ben Shimol
Jan 26 at 15:43
$begingroup$
I rewrote it a bit. But the step to see, why $T^2=T$ holds if $T$ is the identity on its image, which is a hyperplane, is a very small one... It isn't even worth to be called a proof.
$endgroup$
– Mundron Schmidt
Jan 26 at 16:10
$begingroup$
Thank you very much !!!
$endgroup$
– Or Yehuda Ben Shimol
Jan 26 at 19:13
add a comment |
$begingroup$
I assume, the projection operator is defined as $T^2=T$.
Consider $vin mathbb R^n$. Can you think of a way, why $T^2v=Tv$ holds? It is maybe easier to see if you write
$$
T^2v=Tv Leftrightarrow T(T(v)) = T(v).
$$
You don't to it normally this way, but let us make it one step more obvious:
We define $w=Tv$. Now we get
$$
T(T(v))=T(v) Leftrightarrow T(w)=w.
$$
Consider, that $w$ is an element of the image of $T$.
And that this holds, is writen in your condition.
answered Jan 26 at 14:15
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
I assume, the projection operator is defined as $T^2=T$.
Consider $vin mathbb R^n$. Can you think of a way, why $T^2v=Tv$ holds? It is maybe easier to see if you write
$$
T^2v=Tv Leftrightarrow T(T(v)) = T(v).
$$
You don't to it normally this way, but let us make it one step more obvious:
We define $w=Tv$. Now we get
$$
T(T(v))=T(v) Leftrightarrow T(w)=w.
$$
Consider, that $w$ is an element of the image of $T$.
And that this holds, is writen in your condition.
answered Jan 26 at 14:15
Mundron SchmidtMundron Schmidt
7,5042729
edited Jan 26 at 16:08
answered Jan 26 at 14:15
Mundron SchmidtMundron Schmidt
7,5042729
answered Jan 26 at 14:15
Mundron SchmidtMundron Schmidt
7,5042729
answered Jan 26 at 14:15
Mundron SchmidtMundron Schmidt
7,5042729
7,5042729
$begingroup$
I didn't understand, can you give me an example maybe ?
$endgroup$
– Or Yehuda Ben Shimol
Jan 26 at 15:43
$begingroup$
I rewrote it a bit. But the step to see, why $T^2=T$ holds if $T$ is the identity on its image, which is a hyperplane, is a very small one... It isn't even worth to be called a proof.
$endgroup$
– Mundron Schmidt
Jan 26 at 16:10
$begingroup$
Thank you very much !!!
$endgroup$
– Or Yehuda Ben Shimol
Jan 26 at 19:13
add a comment |
$begingroup$
I didn't understand, can you give me an example maybe ?
$endgroup$
– Or Yehuda Ben Shimol
Jan 26 at 15:43
$begingroup$
I rewrote it a bit. But the step to see, why $T^2=T$ holds if $T$ is the identity on its image, which is a hyperplane, is a very small one... It isn't even worth to be called a proof.
$endgroup$
– Mundron Schmidt
Jan 26 at 16:10
$begingroup$
Thank you very much !!!
$endgroup$
– Or Yehuda Ben Shimol
Jan 26 at 19:13
$begingroup$
I didn't understand, can you give me an example maybe ?
$endgroup$
– Or Yehuda Ben Shimol
Jan 26 at 15:43
$begingroup$
I didn't understand, can you give me an example maybe ?
$endgroup$
– Or Yehuda Ben Shimol
Jan 26 at 15:43
$begingroup$
I rewrote it a bit. But the step to see, why $T^2=T$ holds if $T$ is the identity on its image, which is a hyperplane, is a very small one... It isn't even worth to be called a proof.
$endgroup$
– Mundron Schmidt
Jan 26 at 16:10
$begingroup$
I rewrote it a bit. But the step to see, why $T^2=T$ holds if $T$ is the identity on its image, which is a hyperplane, is a very small one... It isn't even worth to be called a proof.
$endgroup$
– Mundron Schmidt
Jan 26 at 16:10
$begingroup$
Thank you very much !!!
$endgroup$
– Or Yehuda Ben Shimol
Jan 26 at 19:13
$begingroup$
Thank you very much !!!
$endgroup$
– Or Yehuda Ben Shimol
Jan 26 at 19:13
add a comment |
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