Show that the direct sum of a kernel of a projection and its image create the originating vector space. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Let $T: V rightarrow W $ be a linear mapping such that $T^2=T$, show that $V = M oplus N$.Vector space as direct sum of kernel and imageKernel and Image of linear transformation are supplementaryDirect product of the kernel and image of a linear transformationlinear map between vector spaces/ Direct sum/ Kernel and imageFinding the kernel and image of a linear transformation over the field $Bbb Z_2$Central idempotent vector space endomorphismsProblem related to linear transformation and its image and kernelKernel and Image of an orthogonal projectionHow do I show that is is orthogonal to every vector in the plane?Diagonalisable iff kernel and image forms direct sumWhat can be said about linear maps where range and kernel are in direct sum?Calculate the dimensions and basis of the kernel and image (derivated function with matrix)Proving equalities on the kernel and image of a linear transformationFinding bases for Ker(T) and Rng(T)
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Show that the direct sum of a kernel of a projection and its image create the originating vector space.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Let $T: V rightarrow W $ be a linear mapping such that $T^2=T$, show that $V = M oplus N$.Vector space as direct sum of kernel and imageKernel and Image of linear transformation are supplementaryDirect product of the kernel and image of a linear transformationlinear map between vector spaces/ Direct sum/ Kernel and imageFinding the kernel and image of a linear transformation over the field $Bbb Z_2$Central idempotent vector space endomorphismsProblem related to linear transformation and its image and kernelKernel and Image of an orthogonal projectionHow do I show that is is orthogonal to every vector in the plane?Diagonalisable iff kernel and image forms direct sumWhat can be said about linear maps where range and kernel are in direct sum?Calculate the dimensions and basis of the kernel and image (derivated function with matrix)Proving equalities on the kernel and image of a linear transformationFinding bases for Ker(T) and Rng(T)
$begingroup$
I got the following question as my homework.
Given $V$ is a vector space with $P in operatornameEnd V$. $P circ P = P$ ("P is idempotent"). Show that $V = operatornameKer P oplus operatornameIm P$.
One $P$ I can imagine is a projection from 3d-space to plane, that just sets some coordinates to zero. For example $beginpmatrix x \ y \ zendpmatrix mapsto beginpmatrixx \ y \ 0endpmatrix$. Then $operatornameKer P$ would give the line $beginpmatrix 0 \ 0 \ zendpmatrix$ and $operatornameIm P$ would contain all $beginpmatrixx \ y \ 0endpmatrix$. So the result of $operatornameKer P oplus operatornameIm P$ is of course $V$.
But how do I prove that in a mathematical way?
linear-algebra
edited Dec 18 '12 at 22:15
Tom Oldfield
9,56812058
asked Dec 18 '12 at 21:12
iblueiblue
6271721
$endgroup$
add a comment |
$begingroup$
I got the following question as my homework.
Given $V$ is a vector space with $P in operatornameEnd V$. $P circ P = P$ ("P is idempotent"). Show that $V = operatornameKer P oplus operatornameIm P$.
One $P$ I can imagine is a projection from 3d-space to plane, that just sets some coordinates to zero. For example $beginpmatrix x \ y \ zendpmatrix mapsto beginpmatrixx \ y \ 0endpmatrix$. Then $operatornameKer P$ would give the line $beginpmatrix 0 \ 0 \ zendpmatrix$ and $operatornameIm P$ would contain all $beginpmatrixx \ y \ 0endpmatrix$. So the result of $operatornameKer P oplus operatornameIm P$ is of course $V$.
But how do I prove that in a mathematical way?
linear-algebra
edited Dec 18 '12 at 22:15
Tom Oldfield
9,56812058
asked Dec 18 '12 at 21:12
iblueiblue
6271721
$endgroup$
$begingroup$
very light hint: think about the transformation $1 - P$. What happens when you square it? What happens when you multiply it by $P$? What happens when you add $P$?
$endgroup$
– user29743
Dec 18 '12 at 21:16
$begingroup$
(by 1 i mean the identity)
$endgroup$
– user29743
Dec 18 '12 at 21:17
$begingroup$
$1 - P$ seems to give $operatornameker P$, $1 - P circ 1 - P = 1 - P$. And when I add $P$ to $1 - P$, I get $1$, but how does that help me?
$endgroup$
– iblue
Dec 18 '12 at 21:22
$begingroup$
sorry I know it is off-topic, but am i the only person for which mathjax does not render latex on stackexchange ?
$endgroup$
– nicolas
Mar 1 '14 at 13:48
$begingroup$
@nicolas: you should post on meta describing your problem. This is not the place for this kind of question.
$endgroup$
– robjohn♦
Mar 2 '14 at 4:44
add a comment |
$begingroup$
I got the following question as my homework.
Given $V$ is a vector space with $P in operatornameEnd V$. $P circ P = P$ ("P is idempotent"). Show that $V = operatornameKer P oplus operatornameIm P$.
One $P$ I can imagine is a projection from 3d-space to plane, that just sets some coordinates to zero. For example $beginpmatrix x \ y \ zendpmatrix mapsto beginpmatrixx \ y \ 0endpmatrix$. Then $operatornameKer P$ would give the line $beginpmatrix 0 \ 0 \ zendpmatrix$ and $operatornameIm P$ would contain all $beginpmatrixx \ y \ 0endpmatrix$. So the result of $operatornameKer P oplus operatornameIm P$ is of course $V$.
But how do I prove that in a mathematical way?
linear-algebra
edited Dec 18 '12 at 22:15
Tom Oldfield
9,56812058
asked Dec 18 '12 at 21:12
iblueiblue
6271721
$endgroup$
I got the following question as my homework.
Given $V$ is a vector space with $P in operatornameEnd V$. $P circ P = P$ ("P is idempotent"). Show that $V = operatornameKer P oplus operatornameIm P$.
One $P$ I can imagine is a projection from 3d-space to plane, that just sets some coordinates to zero. For example $beginpmatrix x \ y \ zendpmatrix mapsto beginpmatrixx \ y \ 0endpmatrix$. Then $operatornameKer P$ would give the line $beginpmatrix 0 \ 0 \ zendpmatrix$ and $operatornameIm P$ would contain all $beginpmatrixx \ y \ 0endpmatrix$. So the result of $operatornameKer P oplus operatornameIm P$ is of course $V$.
But how do I prove that in a mathematical way?
linear-algebra
linear-algebra
edited Dec 18 '12 at 22:15
Tom Oldfield
9,56812058
asked Dec 18 '12 at 21:12
iblueiblue
6271721
edited Dec 18 '12 at 22:15
Tom Oldfield
9,56812058
asked Dec 18 '12 at 21:12
iblueiblue
6271721
edited Dec 18 '12 at 22:15
Tom Oldfield
9,56812058
edited Dec 18 '12 at 22:15
Tom Oldfield
9,56812058
edited Dec 18 '12 at 22:15
Tom Oldfield
9,56812058
9,56812058
asked Dec 18 '12 at 21:12
iblueiblue
6271721
asked Dec 18 '12 at 21:12
iblueiblue
6271721
asked Dec 18 '12 at 21:12
iblueiblue
6271721
6271721
$begingroup$
very light hint: think about the transformation $1 - P$. What happens when you square it? What happens when you multiply it by $P$? What happens when you add $P$?
$endgroup$
– user29743
Dec 18 '12 at 21:16
$begingroup$
(by 1 i mean the identity)
$endgroup$
– user29743
Dec 18 '12 at 21:17
$begingroup$
$1 - P$ seems to give $operatornameker P$, $1 - P circ 1 - P = 1 - P$. And when I add $P$ to $1 - P$, I get $1$, but how does that help me?
$endgroup$
– iblue
Dec 18 '12 at 21:22
$begingroup$
sorry I know it is off-topic, but am i the only person for which mathjax does not render latex on stackexchange ?
$endgroup$
– nicolas
Mar 1 '14 at 13:48
$begingroup$
@nicolas: you should post on meta describing your problem. This is not the place for this kind of question.
$endgroup$
– robjohn♦
Mar 2 '14 at 4:44
add a comment |
$begingroup$
very light hint: think about the transformation $1 - P$. What happens when you square it? What happens when you multiply it by $P$? What happens when you add $P$?
$endgroup$
– user29743
Dec 18 '12 at 21:16
$begingroup$
(by 1 i mean the identity)
$endgroup$
– user29743
Dec 18 '12 at 21:17
$begingroup$
$1 - P$ seems to give $operatornameker P$, $1 - P circ 1 - P = 1 - P$. And when I add $P$ to $1 - P$, I get $1$, but how does that help me?
$endgroup$
– iblue
Dec 18 '12 at 21:22
$begingroup$
sorry I know it is off-topic, but am i the only person for which mathjax does not render latex on stackexchange ?
$endgroup$
– nicolas
Mar 1 '14 at 13:48
$begingroup$
@nicolas: you should post on meta describing your problem. This is not the place for this kind of question.
$endgroup$
– robjohn♦
Mar 2 '14 at 4:44
$begingroup$
very light hint: think about the transformation $1 - P$. What happens when you square it? What happens when you multiply it by $P$? What happens when you add $P$?
$endgroup$
– user29743
Dec 18 '12 at 21:16
$begingroup$
very light hint: think about the transformation $1 - P$. What happens when you square it? What happens when you multiply it by $P$? What happens when you add $P$?
$endgroup$
– user29743
Dec 18 '12 at 21:16
$begingroup$
(by 1 i mean the identity)
$endgroup$
– user29743
Dec 18 '12 at 21:17
$begingroup$
(by 1 i mean the identity)
$endgroup$
– user29743
Dec 18 '12 at 21:17
$begingroup$
$1 - P$ seems to give $operatornameker P$, $1 - P circ 1 - P = 1 - P$. And when I add $P$ to $1 - P$, I get $1$, but how does that help me?
$endgroup$
– iblue
Dec 18 '12 at 21:22
$begingroup$
$1 - P$ seems to give $operatornameker P$, $1 - P circ 1 - P = 1 - P$. And when I add $P$ to $1 - P$, I get $1$, but how does that help me?
$endgroup$
– iblue
Dec 18 '12 at 21:22
$begingroup$
sorry I know it is off-topic, but am i the only person for which mathjax does not render latex on stackexchange ?
$endgroup$
– nicolas
Mar 1 '14 at 13:48
$begingroup$
sorry I know it is off-topic, but am i the only person for which mathjax does not render latex on stackexchange ?
$endgroup$
– nicolas
Mar 1 '14 at 13:48
$begingroup$
@nicolas: you should post on meta describing your problem. This is not the place for this kind of question.
$endgroup$
– robjohn♦
Mar 2 '14 at 4:44
$begingroup$
@nicolas: you should post on meta describing your problem. This is not the place for this kind of question.
$endgroup$
– robjohn♦
Mar 2 '14 at 4:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Take $x in V$. Since $P=P^2$ we must have $Px=P^2x$ and so $P(x-Px)=0$. Hence $x-Px=xi$ for some $xi in operatornameKerP$. Thus $x = Px + xi$. This shows that $V=operatornameImP + operatornameKerP$. Now take $y in operatornameImP cap operatornameKerP$. Since $y in operatornameImP$ we have $y=Pz$ for some $z in V$. Applying $P$ to both sides we get $Py=P^2z$. But $y in operatornameKerP$, hence $0=Py=P^2z=Pz=y$. This shows that $operatornameImP cap operatornameKerP=0$ and so we have $V=operatornameImP oplus operatornameKerP$.
edited Mar 27 at 6:26
Carlos Jiménez
2,2841621
answered Dec 18 '12 at 22:03
ManosManos
14.1k33288
$endgroup$
$begingroup$
Thank you! I don't understand the last step. Why $operatornameIm P cap operatornameKer P = 0 Rightarrow V=operatornameImP oplus operatornameKerP$?
$endgroup$
– iblue
Dec 18 '12 at 22:27
$begingroup$
Because this is the definition of a vector space being the direct sum of two subspaces. The subspaces must span the whole space and have zero intersection.
$endgroup$
– Manos
Dec 18 '12 at 22:28
$begingroup$
I need to clarify that strictly speaking what you have written is not true. What is true is $Im(P) cap Ker(P)=0$ and $V=Im(P)+Ker(P)$ imply $V=Im(P)oplus Ker(P)$.
$endgroup$
– Manos
Dec 18 '12 at 23:29
$begingroup$
We should be writing $operatornameim P cap operatornameker P = 0$ rather than just $0$.
$endgroup$
– Jon Warneke
Oct 5 '16 at 19:18
add a comment |
$begingroup$
Hint: $V = operatornameKerP oplus operatornameImP$ iff every $vin V$ has a unique representation as $v = u+w$ for some $u in operatornameKerP, w in operatornameImP$ (If you haven't seen that already, it's not too hard to prove.)
How can you find such an expression for general $v$?
answered Dec 18 '12 at 22:13
Tom OldfieldTom Oldfield
9,56812058
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take $x in V$. Since $P=P^2$ we must have $Px=P^2x$ and so $P(x-Px)=0$. Hence $x-Px=xi$ for some $xi in operatornameKerP$. Thus $x = Px + xi$. This shows that $V=operatornameImP + operatornameKerP$. Now take $y in operatornameImP cap operatornameKerP$. Since $y in operatornameImP$ we have $y=Pz$ for some $z in V$. Applying $P$ to both sides we get $Py=P^2z$. But $y in operatornameKerP$, hence $0=Py=P^2z=Pz=y$. This shows that $operatornameImP cap operatornameKerP=0$ and so we have $V=operatornameImP oplus operatornameKerP$.
edited Mar 27 at 6:26
Carlos Jiménez
2,2841621
answered Dec 18 '12 at 22:03
ManosManos
14.1k33288
$endgroup$
$begingroup$
Thank you! I don't understand the last step. Why $operatornameIm P cap operatornameKer P = 0 Rightarrow V=operatornameImP oplus operatornameKerP$?
$endgroup$
– iblue
Dec 18 '12 at 22:27
$begingroup$
Because this is the definition of a vector space being the direct sum of two subspaces. The subspaces must span the whole space and have zero intersection.
$endgroup$
– Manos
Dec 18 '12 at 22:28
$begingroup$
I need to clarify that strictly speaking what you have written is not true. What is true is $Im(P) cap Ker(P)=0$ and $V=Im(P)+Ker(P)$ imply $V=Im(P)oplus Ker(P)$.
$endgroup$
– Manos
Dec 18 '12 at 23:29
$begingroup$
We should be writing $operatornameim P cap operatornameker P = 0$ rather than just $0$.
$endgroup$
– Jon Warneke
Oct 5 '16 at 19:18
add a comment |
$begingroup$
Take $x in V$. Since $P=P^2$ we must have $Px=P^2x$ and so $P(x-Px)=0$. Hence $x-Px=xi$ for some $xi in operatornameKerP$. Thus $x = Px + xi$. This shows that $V=operatornameImP + operatornameKerP$. Now take $y in operatornameImP cap operatornameKerP$. Since $y in operatornameImP$ we have $y=Pz$ for some $z in V$. Applying $P$ to both sides we get $Py=P^2z$. But $y in operatornameKerP$, hence $0=Py=P^2z=Pz=y$. This shows that $operatornameImP cap operatornameKerP=0$ and so we have $V=operatornameImP oplus operatornameKerP$.
edited Mar 27 at 6:26
Carlos Jiménez
2,2841621
answered Dec 18 '12 at 22:03
ManosManos
14.1k33288
$endgroup$
$begingroup$
Thank you! I don't understand the last step. Why $operatornameIm P cap operatornameKer P = 0 Rightarrow V=operatornameImP oplus operatornameKerP$?
$endgroup$
– iblue
Dec 18 '12 at 22:27
$begingroup$
Because this is the definition of a vector space being the direct sum of two subspaces. The subspaces must span the whole space and have zero intersection.
$endgroup$
– Manos
Dec 18 '12 at 22:28
$begingroup$
I need to clarify that strictly speaking what you have written is not true. What is true is $Im(P) cap Ker(P)=0$ and $V=Im(P)+Ker(P)$ imply $V=Im(P)oplus Ker(P)$.
$endgroup$
– Manos
Dec 18 '12 at 23:29
$begingroup$
We should be writing $operatornameim P cap operatornameker P = 0$ rather than just $0$.
$endgroup$
– Jon Warneke
Oct 5 '16 at 19:18
add a comment |
$begingroup$
Take $x in V$. Since $P=P^2$ we must have $Px=P^2x$ and so $P(x-Px)=0$. Hence $x-Px=xi$ for some $xi in operatornameKerP$. Thus $x = Px + xi$. This shows that $V=operatornameImP + operatornameKerP$. Now take $y in operatornameImP cap operatornameKerP$. Since $y in operatornameImP$ we have $y=Pz$ for some $z in V$. Applying $P$ to both sides we get $Py=P^2z$. But $y in operatornameKerP$, hence $0=Py=P^2z=Pz=y$. This shows that $operatornameImP cap operatornameKerP=0$ and so we have $V=operatornameImP oplus operatornameKerP$.
edited Mar 27 at 6:26
Carlos Jiménez
2,2841621
answered Dec 18 '12 at 22:03
ManosManos
14.1k33288
$endgroup$
Take $x in V$. Since $P=P^2$ we must have $Px=P^2x$ and so $P(x-Px)=0$. Hence $x-Px=xi$ for some $xi in operatornameKerP$. Thus $x = Px + xi$. This shows that $V=operatornameImP + operatornameKerP$. Now take $y in operatornameImP cap operatornameKerP$. Since $y in operatornameImP$ we have $y=Pz$ for some $z in V$. Applying $P$ to both sides we get $Py=P^2z$. But $y in operatornameKerP$, hence $0=Py=P^2z=Pz=y$. This shows that $operatornameImP cap operatornameKerP=0$ and so we have $V=operatornameImP oplus operatornameKerP$.
edited Mar 27 at 6:26
Carlos Jiménez
2,2841621
answered Dec 18 '12 at 22:03
ManosManos
14.1k33288
edited Mar 27 at 6:26
Carlos Jiménez
2,2841621
edited Mar 27 at 6:26
Carlos Jiménez
2,2841621
edited Mar 27 at 6:26
Carlos Jiménez
2,2841621
2,2841621
answered Dec 18 '12 at 22:03
ManosManos
14.1k33288
answered Dec 18 '12 at 22:03
ManosManos
14.1k33288
answered Dec 18 '12 at 22:03
ManosManos
14.1k33288
14.1k33288
$begingroup$
Thank you! I don't understand the last step. Why $operatornameIm P cap operatornameKer P = 0 Rightarrow V=operatornameImP oplus operatornameKerP$?
$endgroup$
– iblue
Dec 18 '12 at 22:27
$begingroup$
Because this is the definition of a vector space being the direct sum of two subspaces. The subspaces must span the whole space and have zero intersection.
$endgroup$
– Manos
Dec 18 '12 at 22:28
$begingroup$
I need to clarify that strictly speaking what you have written is not true. What is true is $Im(P) cap Ker(P)=0$ and $V=Im(P)+Ker(P)$ imply $V=Im(P)oplus Ker(P)$.
$endgroup$
– Manos
Dec 18 '12 at 23:29
$begingroup$
We should be writing $operatornameim P cap operatornameker P = 0$ rather than just $0$.
$endgroup$
– Jon Warneke
Oct 5 '16 at 19:18
add a comment |
$begingroup$
Thank you! I don't understand the last step. Why $operatornameIm P cap operatornameKer P = 0 Rightarrow V=operatornameImP oplus operatornameKerP$?
$endgroup$
– iblue
Dec 18 '12 at 22:27
$begingroup$
Because this is the definition of a vector space being the direct sum of two subspaces. The subspaces must span the whole space and have zero intersection.
$endgroup$
– Manos
Dec 18 '12 at 22:28
$begingroup$
I need to clarify that strictly speaking what you have written is not true. What is true is $Im(P) cap Ker(P)=0$ and $V=Im(P)+Ker(P)$ imply $V=Im(P)oplus Ker(P)$.
$endgroup$
– Manos
Dec 18 '12 at 23:29
$begingroup$
We should be writing $operatornameim P cap operatornameker P = 0$ rather than just $0$.
$endgroup$
– Jon Warneke
Oct 5 '16 at 19:18
$begingroup$
Thank you! I don't understand the last step. Why $operatornameIm P cap operatornameKer P = 0 Rightarrow V=operatornameImP oplus operatornameKerP$?
$endgroup$
– iblue
Dec 18 '12 at 22:27
$begingroup$
Thank you! I don't understand the last step. Why $operatornameIm P cap operatornameKer P = 0 Rightarrow V=operatornameImP oplus operatornameKerP$?
$endgroup$
– iblue
Dec 18 '12 at 22:27
$begingroup$
Because this is the definition of a vector space being the direct sum of two subspaces. The subspaces must span the whole space and have zero intersection.
$endgroup$
– Manos
Dec 18 '12 at 22:28
$begingroup$
Because this is the definition of a vector space being the direct sum of two subspaces. The subspaces must span the whole space and have zero intersection.
$endgroup$
– Manos
Dec 18 '12 at 22:28
$begingroup$
I need to clarify that strictly speaking what you have written is not true. What is true is $Im(P) cap Ker(P)=0$ and $V=Im(P)+Ker(P)$ imply $V=Im(P)oplus Ker(P)$.
$endgroup$
– Manos
Dec 18 '12 at 23:29
$begingroup$
I need to clarify that strictly speaking what you have written is not true. What is true is $Im(P) cap Ker(P)=0$ and $V=Im(P)+Ker(P)$ imply $V=Im(P)oplus Ker(P)$.
$endgroup$
– Manos
Dec 18 '12 at 23:29
$begingroup$
We should be writing $operatornameim P cap operatornameker P = 0$ rather than just $0$.
$endgroup$
– Jon Warneke
Oct 5 '16 at 19:18
$begingroup$
We should be writing $operatornameim P cap operatornameker P = 0$ rather than just $0$.
$endgroup$
– Jon Warneke
Oct 5 '16 at 19:18
add a comment |
$begingroup$
Hint: $V = operatornameKerP oplus operatornameImP$ iff every $vin V$ has a unique representation as $v = u+w$ for some $u in operatornameKerP, w in operatornameImP$ (If you haven't seen that already, it's not too hard to prove.)
How can you find such an expression for general $v$?
answered Dec 18 '12 at 22:13
Tom OldfieldTom Oldfield
9,56812058
$endgroup$
add a comment |
$begingroup$
Hint: $V = operatornameKerP oplus operatornameImP$ iff every $vin V$ has a unique representation as $v = u+w$ for some $u in operatornameKerP, w in operatornameImP$ (If you haven't seen that already, it's not too hard to prove.)
How can you find such an expression for general $v$?
answered Dec 18 '12 at 22:13
Tom OldfieldTom Oldfield
9,56812058
$endgroup$
add a comment |
$begingroup$
Hint: $V = operatornameKerP oplus operatornameImP$ iff every $vin V$ has a unique representation as $v = u+w$ for some $u in operatornameKerP, w in operatornameImP$ (If you haven't seen that already, it's not too hard to prove.)
How can you find such an expression for general $v$?
answered Dec 18 '12 at 22:13
Tom OldfieldTom Oldfield
9,56812058
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Hint: $V = operatornameKerP oplus operatornameImP$ iff every $vin V$ has a unique representation as $v = u+w$ for some $u in operatornameKerP, w in operatornameImP$ (If you haven't seen that already, it's not too hard to prove.)
How can you find such an expression for general $v$?
answered Dec 18 '12 at 22:13
Tom OldfieldTom Oldfield
9,56812058
answered Dec 18 '12 at 22:13
Tom OldfieldTom Oldfield
9,56812058
answered Dec 18 '12 at 22:13
Tom OldfieldTom Oldfield
9,56812058
answered Dec 18 '12 at 22:13
Tom OldfieldTom Oldfield
9,56812058
9,56812058
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add a comment |
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$begingroup$
very light hint: think about the transformation $1 - P$. What happens when you square it? What happens when you multiply it by $P$? What happens when you add $P$?
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– user29743
Dec 18 '12 at 21:16
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(by 1 i mean the identity)
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– user29743
Dec 18 '12 at 21:17
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$1 - P$ seems to give $operatornameker P$, $1 - P circ 1 - P = 1 - P$. And when I add $P$ to $1 - P$, I get $1$, but how does that help me?
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– iblue
Dec 18 '12 at 21:22
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sorry I know it is off-topic, but am i the only person for which mathjax does not render latex on stackexchange ?
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– nicolas
Mar 1 '14 at 13:48
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@nicolas: you should post on meta describing your problem. This is not the place for this kind of question.
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– robjohn♦
Mar 2 '14 at 4:44