What is the projector decomposition of a symmetric matrix?
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Hi I am studying linear algebra and I bumped into projector decomposition of symmetric matrices and I just don’t know what it is can anyone of you help please?
linear-algebra matrices
asked Jan 4 at 21:42
Ali DahudAli Dahud
31
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add a comment |
$begingroup$
Hi I am studying linear algebra and I bumped into projector decomposition of symmetric matrices and I just don’t know what it is can anyone of you help please?
linear-algebra matrices
asked Jan 4 at 21:42
Ali DahudAli Dahud
31
$endgroup$
add a comment |
$begingroup$
Hi I am studying linear algebra and I bumped into projector decomposition of symmetric matrices and I just don’t know what it is can anyone of you help please?
linear-algebra matrices
asked Jan 4 at 21:42
Ali DahudAli Dahud
31
$endgroup$
Hi I am studying linear algebra and I bumped into projector decomposition of symmetric matrices and I just don’t know what it is can anyone of you help please?
linear-algebra matrices
linear-algebra matrices
asked Jan 4 at 21:42
Ali DahudAli Dahud
31
asked Jan 4 at 21:42
Ali DahudAli Dahud
31
asked Jan 4 at 21:42
Ali DahudAli Dahud
31
asked Jan 4 at 21:42
Ali DahudAli Dahud
31
asked Jan 4 at 21:42
Ali DahudAli Dahud
31
31
add a comment |
add a comment |
1 Answer
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If $A$ is a symmetric matrix over a real space or a Hermitian matrix over a complex space, then
$$
Ax = sum_{n=1}^{N}lambda_n P_nx
$$
where ${ lambda_1,cdots,lambda_N }$ are the distinct eigenvalues of $A$, and where $P_n$ is the orthogonal projection onto the eigenspace $mathcal{ker}(A-lambda_n I)$. For example, if ${ e_{n,1},e_{n,2},cdots,e_{n,k_n} }$ is an orthonormal basis of $mbox{ker}(A-lambda_n I)$, then the orthogona projection $P_n$ may be written as
$$
P_nx = sum_{j=1}^{k_n}langle x,e_{n,j}rangle e_{n,j}
$$
These projections satisfy:
$$
P_n^2 = P_n, \ P_n^*= P_n; (mbox{or } P_n^{T}=P_n mbox{ for real spaces}) \
P_nP_m = 0 mbox{ for } nne m \
AP_n = lambda_n P_n \
sum_{n=1}^{N} P_n = I,\
A = sum_{n=1}^{N}lambda_n P_n.
$$
answered Jan 4 at 22:15
DisintegratingByPartsDisintegratingByParts
59.8k42681
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $A$ is a symmetric matrix over a real space or a Hermitian matrix over a complex space, then
$$
Ax = sum_{n=1}^{N}lambda_n P_nx
$$
where ${ lambda_1,cdots,lambda_N }$ are the distinct eigenvalues of $A$, and where $P_n$ is the orthogonal projection onto the eigenspace $mathcal{ker}(A-lambda_n I)$. For example, if ${ e_{n,1},e_{n,2},cdots,e_{n,k_n} }$ is an orthonormal basis of $mbox{ker}(A-lambda_n I)$, then the orthogona projection $P_n$ may be written as
$$
P_nx = sum_{j=1}^{k_n}langle x,e_{n,j}rangle e_{n,j}
$$
These projections satisfy:
$$
P_n^2 = P_n, \ P_n^*= P_n; (mbox{or } P_n^{T}=P_n mbox{ for real spaces}) \
P_nP_m = 0 mbox{ for } nne m \
AP_n = lambda_n P_n \
sum_{n=1}^{N} P_n = I,\
A = sum_{n=1}^{N}lambda_n P_n.
$$
answered Jan 4 at 22:15
DisintegratingByPartsDisintegratingByParts
59.8k42681
$endgroup$
add a comment |
$begingroup$
If $A$ is a symmetric matrix over a real space or a Hermitian matrix over a complex space, then
$$
Ax = sum_{n=1}^{N}lambda_n P_nx
$$
where ${ lambda_1,cdots,lambda_N }$ are the distinct eigenvalues of $A$, and where $P_n$ is the orthogonal projection onto the eigenspace $mathcal{ker}(A-lambda_n I)$. For example, if ${ e_{n,1},e_{n,2},cdots,e_{n,k_n} }$ is an orthonormal basis of $mbox{ker}(A-lambda_n I)$, then the orthogona projection $P_n$ may be written as
$$
P_nx = sum_{j=1}^{k_n}langle x,e_{n,j}rangle e_{n,j}
$$
These projections satisfy:
$$
P_n^2 = P_n, \ P_n^*= P_n; (mbox{or } P_n^{T}=P_n mbox{ for real spaces}) \
P_nP_m = 0 mbox{ for } nne m \
AP_n = lambda_n P_n \
sum_{n=1}^{N} P_n = I,\
A = sum_{n=1}^{N}lambda_n P_n.
$$
answered Jan 4 at 22:15
DisintegratingByPartsDisintegratingByParts
59.8k42681
$endgroup$
add a comment |
$begingroup$
If $A$ is a symmetric matrix over a real space or a Hermitian matrix over a complex space, then
$$
Ax = sum_{n=1}^{N}lambda_n P_nx
$$
where ${ lambda_1,cdots,lambda_N }$ are the distinct eigenvalues of $A$, and where $P_n$ is the orthogonal projection onto the eigenspace $mathcal{ker}(A-lambda_n I)$. For example, if ${ e_{n,1},e_{n,2},cdots,e_{n,k_n} }$ is an orthonormal basis of $mbox{ker}(A-lambda_n I)$, then the orthogona projection $P_n$ may be written as
$$
P_nx = sum_{j=1}^{k_n}langle x,e_{n,j}rangle e_{n,j}
$$
These projections satisfy:
$$
P_n^2 = P_n, \ P_n^*= P_n; (mbox{or } P_n^{T}=P_n mbox{ for real spaces}) \
P_nP_m = 0 mbox{ for } nne m \
AP_n = lambda_n P_n \
sum_{n=1}^{N} P_n = I,\
A = sum_{n=1}^{N}lambda_n P_n.
$$
answered Jan 4 at 22:15
DisintegratingByPartsDisintegratingByParts
59.8k42681
$endgroup$
If $A$ is a symmetric matrix over a real space or a Hermitian matrix over a complex space, then
$$
Ax = sum_{n=1}^{N}lambda_n P_nx
$$
where ${ lambda_1,cdots,lambda_N }$ are the distinct eigenvalues of $A$, and where $P_n$ is the orthogonal projection onto the eigenspace $mathcal{ker}(A-lambda_n I)$. For example, if ${ e_{n,1},e_{n,2},cdots,e_{n,k_n} }$ is an orthonormal basis of $mbox{ker}(A-lambda_n I)$, then the orthogona projection $P_n$ may be written as
$$
P_nx = sum_{j=1}^{k_n}langle x,e_{n,j}rangle e_{n,j}
$$
These projections satisfy:
$$
P_n^2 = P_n, \ P_n^*= P_n; (mbox{or } P_n^{T}=P_n mbox{ for real spaces}) \
P_nP_m = 0 mbox{ for } nne m \
AP_n = lambda_n P_n \
sum_{n=1}^{N} P_n = I,\
A = sum_{n=1}^{N}lambda_n P_n.
$$
answered Jan 4 at 22:15
DisintegratingByPartsDisintegratingByParts
59.8k42681
answered Jan 4 at 22:15
DisintegratingByPartsDisintegratingByParts
59.8k42681
answered Jan 4 at 22:15
DisintegratingByPartsDisintegratingByParts
59.8k42681
answered Jan 4 at 22:15
DisintegratingByPartsDisintegratingByParts
59.8k42681
59.8k42681
add a comment |
add a comment |
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