Diagonalization of a projection
$begingroup$
If I have a projection $T$ on a finite dimensional vector space $V$, how do I show that $T$ is diagonalizable?
linear-algebra diagonalization projection-matrices
edited Aug 17 '16 at 17:58
Rodrigo de Azevedo
13.1k41960
asked Oct 19 '11 at 2:29
smanoossmanoos
2,2481228
$endgroup$
add a comment |
$begingroup$
If I have a projection $T$ on a finite dimensional vector space $V$, how do I show that $T$ is diagonalizable?
linear-algebra diagonalization projection-matrices
edited Aug 17 '16 at 17:58
Rodrigo de Azevedo
13.1k41960
asked Oct 19 '11 at 2:29
smanoossmanoos
2,2481228
$endgroup$
$begingroup$
Have you determined the eigenvectors and eigenvalues of T?
$endgroup$
– Adam Saltz
Oct 19 '11 at 2:33
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No. As a matter of fact I do not know the matrix for T.
$endgroup$
– smanoos
Oct 19 '11 at 2:35
$begingroup$
Ok. What is your definition of projection?
$endgroup$
– Adam Saltz
Oct 19 '11 at 2:48
$begingroup$
If $V$ is the direct sum of $M$ and $N$, so that every $z$ in $V$ may be written uniquely in the form $z=x+y$ with $x$ in $M$ and $y$ in $N$, the projection on $M$ along $N$ is the transformation $T$ defined by $Tz=x$. That is the definition of projection.
$endgroup$
– smanoos
Oct 19 '11 at 2:54
$begingroup$
Sometimes the definition is taken to be that $T$ is a projection precisely if $T^2=T$, i.e. $T$ is idempotent. That's actually exactly equivalent.
$endgroup$
– Michael Hardy
Oct 19 '11 at 21:26
add a comment |
$begingroup$
If I have a projection $T$ on a finite dimensional vector space $V$, how do I show that $T$ is diagonalizable?
linear-algebra diagonalization projection-matrices
edited Aug 17 '16 at 17:58
Rodrigo de Azevedo
13.1k41960
asked Oct 19 '11 at 2:29
smanoossmanoos
2,2481228
$endgroup$
If I have a projection $T$ on a finite dimensional vector space $V$, how do I show that $T$ is diagonalizable?
linear-algebra diagonalization projection-matrices
linear-algebra diagonalization projection-matrices
edited Aug 17 '16 at 17:58
Rodrigo de Azevedo
13.1k41960
asked Oct 19 '11 at 2:29
smanoossmanoos
2,2481228
edited Aug 17 '16 at 17:58
Rodrigo de Azevedo
13.1k41960
asked Oct 19 '11 at 2:29
smanoossmanoos
2,2481228
edited Aug 17 '16 at 17:58
Rodrigo de Azevedo
13.1k41960
edited Aug 17 '16 at 17:58
Rodrigo de Azevedo
13.1k41960
edited Aug 17 '16 at 17:58
Rodrigo de Azevedo
13.1k41960
13.1k41960
asked Oct 19 '11 at 2:29
smanoossmanoos
2,2481228
asked Oct 19 '11 at 2:29
smanoossmanoos
2,2481228
asked Oct 19 '11 at 2:29
smanoossmanoos
2,2481228
2,2481228
$begingroup$
Have you determined the eigenvectors and eigenvalues of T?
$endgroup$
– Adam Saltz
Oct 19 '11 at 2:33
$begingroup$
No. As a matter of fact I do not know the matrix for T.
$endgroup$
– smanoos
Oct 19 '11 at 2:35
$begingroup$
Ok. What is your definition of projection?
$endgroup$
– Adam Saltz
Oct 19 '11 at 2:48
$begingroup$
If $V$ is the direct sum of $M$ and $N$, so that every $z$ in $V$ may be written uniquely in the form $z=x+y$ with $x$ in $M$ and $y$ in $N$, the projection on $M$ along $N$ is the transformation $T$ defined by $Tz=x$. That is the definition of projection.
$endgroup$
– smanoos
Oct 19 '11 at 2:54
$begingroup$
Sometimes the definition is taken to be that $T$ is a projection precisely if $T^2=T$, i.e. $T$ is idempotent. That's actually exactly equivalent.
$endgroup$
– Michael Hardy
Oct 19 '11 at 21:26
add a comment |
$begingroup$
Have you determined the eigenvectors and eigenvalues of T?
$endgroup$
– Adam Saltz
Oct 19 '11 at 2:33
$begingroup$
No. As a matter of fact I do not know the matrix for T.
$endgroup$
– smanoos
Oct 19 '11 at 2:35
$begingroup$
Ok. What is your definition of projection?
$endgroup$
– Adam Saltz
Oct 19 '11 at 2:48
$begingroup$
If $V$ is the direct sum of $M$ and $N$, so that every $z$ in $V$ may be written uniquely in the form $z=x+y$ with $x$ in $M$ and $y$ in $N$, the projection on $M$ along $N$ is the transformation $T$ defined by $Tz=x$. That is the definition of projection.
$endgroup$
– smanoos
Oct 19 '11 at 2:54
$begingroup$
Sometimes the definition is taken to be that $T$ is a projection precisely if $T^2=T$, i.e. $T$ is idempotent. That's actually exactly equivalent.
$endgroup$
– Michael Hardy
Oct 19 '11 at 21:26
$begingroup$
Have you determined the eigenvectors and eigenvalues of T?
$endgroup$
– Adam Saltz
Oct 19 '11 at 2:33
$begingroup$
Have you determined the eigenvectors and eigenvalues of T?
$endgroup$
– Adam Saltz
Oct 19 '11 at 2:33
$begingroup$
No. As a matter of fact I do not know the matrix for T.
$endgroup$
– smanoos
Oct 19 '11 at 2:35
$begingroup$
No. As a matter of fact I do not know the matrix for T.
$endgroup$
– smanoos
Oct 19 '11 at 2:35
$begingroup$
Ok. What is your definition of projection?
$endgroup$
– Adam Saltz
Oct 19 '11 at 2:48
$begingroup$
Ok. What is your definition of projection?
$endgroup$
– Adam Saltz
Oct 19 '11 at 2:48
$begingroup$
If $V$ is the direct sum of $M$ and $N$, so that every $z$ in $V$ may be written uniquely in the form $z=x+y$ with $x$ in $M$ and $y$ in $N$, the projection on $M$ along $N$ is the transformation $T$ defined by $Tz=x$. That is the definition of projection.
$endgroup$
– smanoos
Oct 19 '11 at 2:54
$begingroup$
If $V$ is the direct sum of $M$ and $N$, so that every $z$ in $V$ may be written uniquely in the form $z=x+y$ with $x$ in $M$ and $y$ in $N$, the projection on $M$ along $N$ is the transformation $T$ defined by $Tz=x$. That is the definition of projection.
$endgroup$
– smanoos
Oct 19 '11 at 2:54
$begingroup$
Sometimes the definition is taken to be that $T$ is a projection precisely if $T^2=T$, i.e. $T$ is idempotent. That's actually exactly equivalent.
$endgroup$
– Michael Hardy
Oct 19 '11 at 21:26
$begingroup$
Sometimes the definition is taken to be that $T$ is a projection precisely if $T^2=T$, i.e. $T$ is idempotent. That's actually exactly equivalent.
$endgroup$
– Michael Hardy
Oct 19 '11 at 21:26
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If $T$ is a projection, that means there's a subspace $W$ onto which it projects. It maps every vector in $W$ to itself. Therefore every vector in $W$ is an eigenvector with eigenvalue $1$. Every vector not in $W$ is mapped to a vector in $W$. Take any vector $v$ and write
$$
v = Tv + (v-Tv),
$$
so the first term $Tv$ is in $W$. It is easy to see that the second term, $v-Tv$, is in the kernel of $T$: the first term is mapped to $Tv$, and the second is mapped to $Tv-T^2v$. But since $Tv$ is in $W$, it must be fixed by $T$, so $T^2v=Tv$; thus $T(v-TV)=0$. In this way, every vector $v$ is written as the sum of a vector in $W$, which is an eigenvector with eigenvalue $1$, and a vector in the kernel of $T$, which is an eigenvector with eigenvalue $0$. So form a basis of the whole space by taking the union of a basis of $W$ and a basis of the kernel of $T$, and the matrix of $T$ with respect to that basis is
$$
begin{bmatrix}
1 \ & 1 \ & & 1 \ & & & ddots \ & & & & 1 \ & & & & & 0 \ & & & & & & ddots \ & & & & & & & 0
end{bmatrix}
$$
(and all off-diagonal entries are $0$) where the number of $1$s is the dimension of $W$ and the number of $0$s is the dimension of the kernel of $T$.
answered Oct 19 '11 at 2:56
Michael HardyMichael Hardy
1
$endgroup$
$begingroup$
You are perfectly right. But can you throw more light on how to set up this matrix. That is where I have a problem.
$endgroup$
– smanoos
Oct 19 '11 at 3:03
2
$begingroup$
@smanoos: Using your notation: take a basis $m_1,ldots,m_k$ for $M$, a basis $n_{k+1},ldots,n_r$ for $N$. Show that $m_1,ldots,m_k,n_{k+1},ldots,n_r$ is a basis for $V$, and that the matrix of $T$ relative to that basis is the one that Michael Hardy writes above.
$endgroup$
– Arturo Magidin
Oct 19 '11 at 4:36
add a comment |
$begingroup$
@Michael Hardy's answer is nice and complete. I'd like to write down how I think about this question.
Let $P:{mathbb R}^mto{mathbb R}^m$ be the projection transformation. By rank–nullity theorem,
$$dim(ker(P))+dim(operatorname{range}(P))=m
$$
On the other hand, by definition, $P^2=P$, which implies that the eigenvalues of $P$ are $lambda=0$ or $1$. It's not hard to show that $ker(P)$ is the eigenspace of $lambda=0$ and $operatorname{range}(P)$ is the eigenspace of $lambda=1$. Therefore, we an independent set of $m$ eigenvectors, which implies that $P$ is diagonalizable.
edited Aug 17 '16 at 17:50
Michael Hardy
1
answered Oct 14 '12 at 2:06
JackJack
27.5k1782202
$endgroup$
$begingroup$
Could you explain to me why the range(P) is the eigenspace of $lambda=1$?
$endgroup$
– user74261
Oct 23 '15 at 2:06
$begingroup$
$P(Px)=1cdot Px$.
$endgroup$
– Jack
Oct 23 '15 at 3:04
add a comment |
$begingroup$
If $P$ is a projection and $J$ is the Jordan matrix associated with $P$ (that is, for some nonsingular $Q$, $P=QJQ^{-1}$) then
$$J^2=left(Q^{-1}PQright)left(Q^{-1}PQright)=Q^{=1}P^2Q=Q^{-1}PQ=J$$
This is possible only when $J$ is diagonal.
answered Dec 17 '18 at 6:47
user165536user165536
61
$endgroup$
$begingroup$
This doesn't really add much to the existing answers.
$endgroup$
– Henrik
Dec 17 '18 at 7:14
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $T$ is a projection, that means there's a subspace $W$ onto which it projects. It maps every vector in $W$ to itself. Therefore every vector in $W$ is an eigenvector with eigenvalue $1$. Every vector not in $W$ is mapped to a vector in $W$. Take any vector $v$ and write
$$
v = Tv + (v-Tv),
$$
so the first term $Tv$ is in $W$. It is easy to see that the second term, $v-Tv$, is in the kernel of $T$: the first term is mapped to $Tv$, and the second is mapped to $Tv-T^2v$. But since $Tv$ is in $W$, it must be fixed by $T$, so $T^2v=Tv$; thus $T(v-TV)=0$. In this way, every vector $v$ is written as the sum of a vector in $W$, which is an eigenvector with eigenvalue $1$, and a vector in the kernel of $T$, which is an eigenvector with eigenvalue $0$. So form a basis of the whole space by taking the union of a basis of $W$ and a basis of the kernel of $T$, and the matrix of $T$ with respect to that basis is
$$
begin{bmatrix}
1 \ & 1 \ & & 1 \ & & & ddots \ & & & & 1 \ & & & & & 0 \ & & & & & & ddots \ & & & & & & & 0
end{bmatrix}
$$
(and all off-diagonal entries are $0$) where the number of $1$s is the dimension of $W$ and the number of $0$s is the dimension of the kernel of $T$.
answered Oct 19 '11 at 2:56
Michael HardyMichael Hardy
1
$endgroup$
$begingroup$
You are perfectly right. But can you throw more light on how to set up this matrix. That is where I have a problem.
$endgroup$
– smanoos
Oct 19 '11 at 3:03
2
$begingroup$
@smanoos: Using your notation: take a basis $m_1,ldots,m_k$ for $M$, a basis $n_{k+1},ldots,n_r$ for $N$. Show that $m_1,ldots,m_k,n_{k+1},ldots,n_r$ is a basis for $V$, and that the matrix of $T$ relative to that basis is the one that Michael Hardy writes above.
$endgroup$
– Arturo Magidin
Oct 19 '11 at 4:36
add a comment |
$begingroup$
If $T$ is a projection, that means there's a subspace $W$ onto which it projects. It maps every vector in $W$ to itself. Therefore every vector in $W$ is an eigenvector with eigenvalue $1$. Every vector not in $W$ is mapped to a vector in $W$. Take any vector $v$ and write
$$
v = Tv + (v-Tv),
$$
so the first term $Tv$ is in $W$. It is easy to see that the second term, $v-Tv$, is in the kernel of $T$: the first term is mapped to $Tv$, and the second is mapped to $Tv-T^2v$. But since $Tv$ is in $W$, it must be fixed by $T$, so $T^2v=Tv$; thus $T(v-TV)=0$. In this way, every vector $v$ is written as the sum of a vector in $W$, which is an eigenvector with eigenvalue $1$, and a vector in the kernel of $T$, which is an eigenvector with eigenvalue $0$. So form a basis of the whole space by taking the union of a basis of $W$ and a basis of the kernel of $T$, and the matrix of $T$ with respect to that basis is
$$
begin{bmatrix}
1 \ & 1 \ & & 1 \ & & & ddots \ & & & & 1 \ & & & & & 0 \ & & & & & & ddots \ & & & & & & & 0
end{bmatrix}
$$
(and all off-diagonal entries are $0$) where the number of $1$s is the dimension of $W$ and the number of $0$s is the dimension of the kernel of $T$.
answered Oct 19 '11 at 2:56
Michael HardyMichael Hardy
1
$endgroup$
$begingroup$
You are perfectly right. But can you throw more light on how to set up this matrix. That is where I have a problem.
$endgroup$
– smanoos
Oct 19 '11 at 3:03
2
$begingroup$
@smanoos: Using your notation: take a basis $m_1,ldots,m_k$ for $M$, a basis $n_{k+1},ldots,n_r$ for $N$. Show that $m_1,ldots,m_k,n_{k+1},ldots,n_r$ is a basis for $V$, and that the matrix of $T$ relative to that basis is the one that Michael Hardy writes above.
$endgroup$
– Arturo Magidin
Oct 19 '11 at 4:36
add a comment |
$begingroup$
If $T$ is a projection, that means there's a subspace $W$ onto which it projects. It maps every vector in $W$ to itself. Therefore every vector in $W$ is an eigenvector with eigenvalue $1$. Every vector not in $W$ is mapped to a vector in $W$. Take any vector $v$ and write
$$
v = Tv + (v-Tv),
$$
so the first term $Tv$ is in $W$. It is easy to see that the second term, $v-Tv$, is in the kernel of $T$: the first term is mapped to $Tv$, and the second is mapped to $Tv-T^2v$. But since $Tv$ is in $W$, it must be fixed by $T$, so $T^2v=Tv$; thus $T(v-TV)=0$. In this way, every vector $v$ is written as the sum of a vector in $W$, which is an eigenvector with eigenvalue $1$, and a vector in the kernel of $T$, which is an eigenvector with eigenvalue $0$. So form a basis of the whole space by taking the union of a basis of $W$ and a basis of the kernel of $T$, and the matrix of $T$ with respect to that basis is
$$
begin{bmatrix}
1 \ & 1 \ & & 1 \ & & & ddots \ & & & & 1 \ & & & & & 0 \ & & & & & & ddots \ & & & & & & & 0
end{bmatrix}
$$
(and all off-diagonal entries are $0$) where the number of $1$s is the dimension of $W$ and the number of $0$s is the dimension of the kernel of $T$.
answered Oct 19 '11 at 2:56
Michael HardyMichael Hardy
1
$endgroup$
If $T$ is a projection, that means there's a subspace $W$ onto which it projects. It maps every vector in $W$ to itself. Therefore every vector in $W$ is an eigenvector with eigenvalue $1$. Every vector not in $W$ is mapped to a vector in $W$. Take any vector $v$ and write
$$
v = Tv + (v-Tv),
$$
so the first term $Tv$ is in $W$. It is easy to see that the second term, $v-Tv$, is in the kernel of $T$: the first term is mapped to $Tv$, and the second is mapped to $Tv-T^2v$. But since $Tv$ is in $W$, it must be fixed by $T$, so $T^2v=Tv$; thus $T(v-TV)=0$. In this way, every vector $v$ is written as the sum of a vector in $W$, which is an eigenvector with eigenvalue $1$, and a vector in the kernel of $T$, which is an eigenvector with eigenvalue $0$. So form a basis of the whole space by taking the union of a basis of $W$ and a basis of the kernel of $T$, and the matrix of $T$ with respect to that basis is
$$
begin{bmatrix}
1 \ & 1 \ & & 1 \ & & & ddots \ & & & & 1 \ & & & & & 0 \ & & & & & & ddots \ & & & & & & & 0
end{bmatrix}
$$
(and all off-diagonal entries are $0$) where the number of $1$s is the dimension of $W$ and the number of $0$s is the dimension of the kernel of $T$.
answered Oct 19 '11 at 2:56
Michael HardyMichael Hardy
1
answered Oct 19 '11 at 2:56
Michael HardyMichael Hardy
1
answered Oct 19 '11 at 2:56
Michael HardyMichael Hardy
1
answered Oct 19 '11 at 2:56
Michael HardyMichael Hardy
1
1
$begingroup$
You are perfectly right. But can you throw more light on how to set up this matrix. That is where I have a problem.
$endgroup$
– smanoos
Oct 19 '11 at 3:03
2
$begingroup$
@smanoos: Using your notation: take a basis $m_1,ldots,m_k$ for $M$, a basis $n_{k+1},ldots,n_r$ for $N$. Show that $m_1,ldots,m_k,n_{k+1},ldots,n_r$ is a basis for $V$, and that the matrix of $T$ relative to that basis is the one that Michael Hardy writes above.
$endgroup$
– Arturo Magidin
Oct 19 '11 at 4:36
add a comment |
$begingroup$
You are perfectly right. But can you throw more light on how to set up this matrix. That is where I have a problem.
$endgroup$
– smanoos
Oct 19 '11 at 3:03
2
$begingroup$
@smanoos: Using your notation: take a basis $m_1,ldots,m_k$ for $M$, a basis $n_{k+1},ldots,n_r$ for $N$. Show that $m_1,ldots,m_k,n_{k+1},ldots,n_r$ is a basis for $V$, and that the matrix of $T$ relative to that basis is the one that Michael Hardy writes above.
$endgroup$
– Arturo Magidin
Oct 19 '11 at 4:36
$begingroup$
You are perfectly right. But can you throw more light on how to set up this matrix. That is where I have a problem.
$endgroup$
– smanoos
Oct 19 '11 at 3:03
$begingroup$
You are perfectly right. But can you throw more light on how to set up this matrix. That is where I have a problem.
$endgroup$
– smanoos
Oct 19 '11 at 3:03
2
2
$begingroup$
@smanoos: Using your notation: take a basis $m_1,ldots,m_k$ for $M$, a basis $n_{k+1},ldots,n_r$ for $N$. Show that $m_1,ldots,m_k,n_{k+1},ldots,n_r$ is a basis for $V$, and that the matrix of $T$ relative to that basis is the one that Michael Hardy writes above.
$endgroup$
– Arturo Magidin
Oct 19 '11 at 4:36
$begingroup$
@smanoos: Using your notation: take a basis $m_1,ldots,m_k$ for $M$, a basis $n_{k+1},ldots,n_r$ for $N$. Show that $m_1,ldots,m_k,n_{k+1},ldots,n_r$ is a basis for $V$, and that the matrix of $T$ relative to that basis is the one that Michael Hardy writes above.
$endgroup$
– Arturo Magidin
Oct 19 '11 at 4:36
add a comment |
$begingroup$
@Michael Hardy's answer is nice and complete. I'd like to write down how I think about this question.
Let $P:{mathbb R}^mto{mathbb R}^m$ be the projection transformation. By rank–nullity theorem,
$$dim(ker(P))+dim(operatorname{range}(P))=m
$$
On the other hand, by definition, $P^2=P$, which implies that the eigenvalues of $P$ are $lambda=0$ or $1$. It's not hard to show that $ker(P)$ is the eigenspace of $lambda=0$ and $operatorname{range}(P)$ is the eigenspace of $lambda=1$. Therefore, we an independent set of $m$ eigenvectors, which implies that $P$ is diagonalizable.
edited Aug 17 '16 at 17:50
Michael Hardy
1
answered Oct 14 '12 at 2:06
JackJack
27.5k1782202
$endgroup$
$begingroup$
Could you explain to me why the range(P) is the eigenspace of $lambda=1$?
$endgroup$
– user74261
Oct 23 '15 at 2:06
$begingroup$
$P(Px)=1cdot Px$.
$endgroup$
– Jack
Oct 23 '15 at 3:04
add a comment |
$begingroup$
@Michael Hardy's answer is nice and complete. I'd like to write down how I think about this question.
Let $P:{mathbb R}^mto{mathbb R}^m$ be the projection transformation. By rank–nullity theorem,
$$dim(ker(P))+dim(operatorname{range}(P))=m
$$
On the other hand, by definition, $P^2=P$, which implies that the eigenvalues of $P$ are $lambda=0$ or $1$. It's not hard to show that $ker(P)$ is the eigenspace of $lambda=0$ and $operatorname{range}(P)$ is the eigenspace of $lambda=1$. Therefore, we an independent set of $m$ eigenvectors, which implies that $P$ is diagonalizable.
edited Aug 17 '16 at 17:50
Michael Hardy
1
answered Oct 14 '12 at 2:06
JackJack
27.5k1782202
$endgroup$
$begingroup$
Could you explain to me why the range(P) is the eigenspace of $lambda=1$?
$endgroup$
– user74261
Oct 23 '15 at 2:06
$begingroup$
$P(Px)=1cdot Px$.
$endgroup$
– Jack
Oct 23 '15 at 3:04
add a comment |
$begingroup$
@Michael Hardy's answer is nice and complete. I'd like to write down how I think about this question.
Let $P:{mathbb R}^mto{mathbb R}^m$ be the projection transformation. By rank–nullity theorem,
$$dim(ker(P))+dim(operatorname{range}(P))=m
$$
On the other hand, by definition, $P^2=P$, which implies that the eigenvalues of $P$ are $lambda=0$ or $1$. It's not hard to show that $ker(P)$ is the eigenspace of $lambda=0$ and $operatorname{range}(P)$ is the eigenspace of $lambda=1$. Therefore, we an independent set of $m$ eigenvectors, which implies that $P$ is diagonalizable.
edited Aug 17 '16 at 17:50
Michael Hardy
1
answered Oct 14 '12 at 2:06
JackJack
27.5k1782202
$endgroup$
@Michael Hardy's answer is nice and complete. I'd like to write down how I think about this question.
Let $P:{mathbb R}^mto{mathbb R}^m$ be the projection transformation. By rank–nullity theorem,
$$dim(ker(P))+dim(operatorname{range}(P))=m
$$
On the other hand, by definition, $P^2=P$, which implies that the eigenvalues of $P$ are $lambda=0$ or $1$. It's not hard to show that $ker(P)$ is the eigenspace of $lambda=0$ and $operatorname{range}(P)$ is the eigenspace of $lambda=1$. Therefore, we an independent set of $m$ eigenvectors, which implies that $P$ is diagonalizable.
edited Aug 17 '16 at 17:50
Michael Hardy
1
answered Oct 14 '12 at 2:06
JackJack
27.5k1782202
edited Aug 17 '16 at 17:50
Michael Hardy
1
edited Aug 17 '16 at 17:50
Michael Hardy
1
edited Aug 17 '16 at 17:50
Michael Hardy
1
1
answered Oct 14 '12 at 2:06
JackJack
27.5k1782202
answered Oct 14 '12 at 2:06
JackJack
27.5k1782202
answered Oct 14 '12 at 2:06
JackJack
27.5k1782202
27.5k1782202
$begingroup$
Could you explain to me why the range(P) is the eigenspace of $lambda=1$?
$endgroup$
– user74261
Oct 23 '15 at 2:06
$begingroup$
$P(Px)=1cdot Px$.
$endgroup$
– Jack
Oct 23 '15 at 3:04
add a comment |
$begingroup$
Could you explain to me why the range(P) is the eigenspace of $lambda=1$?
$endgroup$
– user74261
Oct 23 '15 at 2:06
$begingroup$
$P(Px)=1cdot Px$.
$endgroup$
– Jack
Oct 23 '15 at 3:04
$begingroup$
Could you explain to me why the range(P) is the eigenspace of $lambda=1$?
$endgroup$
– user74261
Oct 23 '15 at 2:06
$begingroup$
Could you explain to me why the range(P) is the eigenspace of $lambda=1$?
$endgroup$
– user74261
Oct 23 '15 at 2:06
$begingroup$
$P(Px)=1cdot Px$.
$endgroup$
– Jack
Oct 23 '15 at 3:04
$begingroup$
$P(Px)=1cdot Px$.
$endgroup$
– Jack
Oct 23 '15 at 3:04
add a comment |
$begingroup$
If $P$ is a projection and $J$ is the Jordan matrix associated with $P$ (that is, for some nonsingular $Q$, $P=QJQ^{-1}$) then
$$J^2=left(Q^{-1}PQright)left(Q^{-1}PQright)=Q^{=1}P^2Q=Q^{-1}PQ=J$$
This is possible only when $J$ is diagonal.
answered Dec 17 '18 at 6:47
user165536user165536
61
$endgroup$
$begingroup$
This doesn't really add much to the existing answers.
$endgroup$
– Henrik
Dec 17 '18 at 7:14
add a comment |
$begingroup$
If $P$ is a projection and $J$ is the Jordan matrix associated with $P$ (that is, for some nonsingular $Q$, $P=QJQ^{-1}$) then
$$J^2=left(Q^{-1}PQright)left(Q^{-1}PQright)=Q^{=1}P^2Q=Q^{-1}PQ=J$$
This is possible only when $J$ is diagonal.
answered Dec 17 '18 at 6:47
user165536user165536
61
$endgroup$
$begingroup$
This doesn't really add much to the existing answers.
$endgroup$
– Henrik
Dec 17 '18 at 7:14
add a comment |
$begingroup$
If $P$ is a projection and $J$ is the Jordan matrix associated with $P$ (that is, for some nonsingular $Q$, $P=QJQ^{-1}$) then
$$J^2=left(Q^{-1}PQright)left(Q^{-1}PQright)=Q^{=1}P^2Q=Q^{-1}PQ=J$$
This is possible only when $J$ is diagonal.
answered Dec 17 '18 at 6:47
user165536user165536
61
$endgroup$
If $P$ is a projection and $J$ is the Jordan matrix associated with $P$ (that is, for some nonsingular $Q$, $P=QJQ^{-1}$) then
$$J^2=left(Q^{-1}PQright)left(Q^{-1}PQright)=Q^{=1}P^2Q=Q^{-1}PQ=J$$
This is possible only when $J$ is diagonal.
answered Dec 17 '18 at 6:47
user165536user165536
61
answered Dec 17 '18 at 6:47
user165536user165536
61
answered Dec 17 '18 at 6:47
user165536user165536
61
answered Dec 17 '18 at 6:47
user165536user165536
61
61
$begingroup$
This doesn't really add much to the existing answers.
$endgroup$
– Henrik
Dec 17 '18 at 7:14
add a comment |
$begingroup$
This doesn't really add much to the existing answers.
$endgroup$
– Henrik
Dec 17 '18 at 7:14
$begingroup$
This doesn't really add much to the existing answers.
$endgroup$
– Henrik
Dec 17 '18 at 7:14
$begingroup$
This doesn't really add much to the existing answers.
$endgroup$
– Henrik
Dec 17 '18 at 7:14
add a comment |
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$begingroup$
Have you determined the eigenvectors and eigenvalues of T?
$endgroup$
– Adam Saltz
Oct 19 '11 at 2:33
$begingroup$
No. As a matter of fact I do not know the matrix for T.
$endgroup$
– smanoos
Oct 19 '11 at 2:35
$begingroup$
Ok. What is your definition of projection?
$endgroup$
– Adam Saltz
Oct 19 '11 at 2:48
$begingroup$
If $V$ is the direct sum of $M$ and $N$, so that every $z$ in $V$ may be written uniquely in the form $z=x+y$ with $x$ in $M$ and $y$ in $N$, the projection on $M$ along $N$ is the transformation $T$ defined by $Tz=x$. That is the definition of projection.
$endgroup$
– smanoos
Oct 19 '11 at 2:54
$begingroup$
Sometimes the definition is taken to be that $T$ is a projection precisely if $T^2=T$, i.e. $T$ is idempotent. That's actually exactly equivalent.
$endgroup$
– Michael Hardy
Oct 19 '11 at 21:26