Mixing
Eigenvectors in a block diagonal matrix
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In spatial statistics, I am trying to deal with the following topic:
I have a real-valued, symmetric, full-rank matrix $textbf{A}$, say $N times N$. It's a connectivity matrix, i.e. $a_{ij}$ values are either 1 or 0. Eigenvectors of $textbf{A}$ are orthogonal (and real). Now, for an empirical analysis, I need to generalize my $textbf{A}$ matrix into a block-diagonal matrix such as $mathcal{A}= (I_T otimes textbf{A})$ - also real-valued, symmetric and full rank. I assume it also has orthogonal eigenvectors (of a corresponding lenght $NT$.
My question is, are the eigenvectors of $textbf{A}$ and $mathcal{A}$ related? E.g. through some basic transformation or "rule"?
I searched this web for questions on eigenvectors in block diagonal matrices, but did not find an answer for this one... Thank you.
eigenvalues-eigenvectors block-matrices
asked 2 days ago
Tomas
1085
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Tomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
1
down vote
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In spatial statistics, I am trying to deal with the following topic:
I have a real-valued, symmetric, full-rank matrix $textbf{A}$, say $N times N$. It's a connectivity matrix, i.e. $a_{ij}$ values are either 1 or 0. Eigenvectors of $textbf{A}$ are orthogonal (and real). Now, for an empirical analysis, I need to generalize my $textbf{A}$ matrix into a block-diagonal matrix such as $mathcal{A}= (I_T otimes textbf{A})$ - also real-valued, symmetric and full rank. I assume it also has orthogonal eigenvectors (of a corresponding lenght $NT$.
My question is, are the eigenvectors of $textbf{A}$ and $mathcal{A}$ related? E.g. through some basic transformation or "rule"?
I searched this web for questions on eigenvectors in block diagonal matrices, but did not find an answer for this one... Thank you.
eigenvalues-eigenvectors block-matrices
asked 2 days ago
Tomas
1085
New contributor
Tomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Could you clarify what you mean by $mathcal A = (I_T otimes mathbf A)$? Is $otimes$ the Kronecker product? Is $I_T$ an identity matrix?
– Omnomnomnom
2 days ago
Sorry, yes, it's a Kroenecker product and identity matrix.
– Tomas
2 days ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In spatial statistics, I am trying to deal with the following topic:
I have a real-valued, symmetric, full-rank matrix $textbf{A}$, say $N times N$. It's a connectivity matrix, i.e. $a_{ij}$ values are either 1 or 0. Eigenvectors of $textbf{A}$ are orthogonal (and real). Now, for an empirical analysis, I need to generalize my $textbf{A}$ matrix into a block-diagonal matrix such as $mathcal{A}= (I_T otimes textbf{A})$ - also real-valued, symmetric and full rank. I assume it also has orthogonal eigenvectors (of a corresponding lenght $NT$.
My question is, are the eigenvectors of $textbf{A}$ and $mathcal{A}$ related? E.g. through some basic transformation or "rule"?
I searched this web for questions on eigenvectors in block diagonal matrices, but did not find an answer for this one... Thank you.
eigenvalues-eigenvectors block-matrices
asked 2 days ago
Tomas
1085
New contributor
Tomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In spatial statistics, I am trying to deal with the following topic:
I have a real-valued, symmetric, full-rank matrix $textbf{A}$, say $N times N$. It's a connectivity matrix, i.e. $a_{ij}$ values are either 1 or 0. Eigenvectors of $textbf{A}$ are orthogonal (and real). Now, for an empirical analysis, I need to generalize my $textbf{A}$ matrix into a block-diagonal matrix such as $mathcal{A}= (I_T otimes textbf{A})$ - also real-valued, symmetric and full rank. I assume it also has orthogonal eigenvectors (of a corresponding lenght $NT$.
My question is, are the eigenvectors of $textbf{A}$ and $mathcal{A}$ related? E.g. through some basic transformation or "rule"?
I searched this web for questions on eigenvectors in block diagonal matrices, but did not find an answer for this one... Thank you.
eigenvalues-eigenvectors block-matrices
eigenvalues-eigenvectors block-matrices
asked 2 days ago
Tomas
1085
New contributor
Tomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Tomas
1085
New contributor
Tomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Tomas
1085
New contributor
Tomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Tomas
1085
asked 2 days ago
Tomas
1085
1085
New contributor
Tomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Tomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Tomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Could you clarify what you mean by $mathcal A = (I_T otimes mathbf A)$? Is $otimes$ the Kronecker product? Is $I_T$ an identity matrix?
– Omnomnomnom
2 days ago
Sorry, yes, it's a Kroenecker product and identity matrix.
– Tomas
2 days ago
add a comment |
Could you clarify what you mean by $mathcal A = (I_T otimes mathbf A)$? Is $otimes$ the Kronecker product? Is $I_T$ an identity matrix?
– Omnomnomnom
2 days ago
Sorry, yes, it's a Kroenecker product and identity matrix.
– Tomas
2 days ago
Could you clarify what you mean by $mathcal A = (I_T otimes mathbf A)$? Is $otimes$ the Kronecker product? Is $I_T$ an identity matrix?
– Omnomnomnom
2 days ago
Could you clarify what you mean by $mathcal A = (I_T otimes mathbf A)$? Is $otimes$ the Kronecker product? Is $I_T$ an identity matrix?
– Omnomnomnom
2 days ago
Sorry, yes, it's a Kroenecker product and identity matrix.
– Tomas
2 days ago
Sorry, yes, it's a Kroenecker product and identity matrix.
– Tomas
2 days ago
add a comment |
1 Answer
1
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oldest
votes
up vote
2
down vote
accepted
In general: if $P$ and $Q$ are diagonalizable, then whenever $Px = lambda x$ and $Qy = mu y$, we will have $(P otimes Q)(x otimes y) = mu lambda (x otimes y)$. Moreover, we can form an eigenbasis out of the Kronecker products $x otimes y$.
For your particular example: $I_T otimes A$ will have the same eigenvalues as $A$, but each will have its multiplicity multiplied by $T$. For any eigenvector $v$ of $A$, the vectors $e_1 otimes v, dots, e_n otimes v$ will be eigenvectors of $I_T otimes A$ (here $e_1,dots,e_n$ is the canonical basis; so $e_1 = (1,0,dots,0)^T$).
Another general point: the eigenvalues of any block-diagonal matrix is simply the union (with multiplicity) of the eigenvalues of each individual block.
answered 2 days ago
Omnomnomnom
124k788176
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
In general: if $P$ and $Q$ are diagonalizable, then whenever $Px = lambda x$ and $Qy = mu y$, we will have $(P otimes Q)(x otimes y) = mu lambda (x otimes y)$. Moreover, we can form an eigenbasis out of the Kronecker products $x otimes y$.
For your particular example: $I_T otimes A$ will have the same eigenvalues as $A$, but each will have its multiplicity multiplied by $T$. For any eigenvector $v$ of $A$, the vectors $e_1 otimes v, dots, e_n otimes v$ will be eigenvectors of $I_T otimes A$ (here $e_1,dots,e_n$ is the canonical basis; so $e_1 = (1,0,dots,0)^T$).
Another general point: the eigenvalues of any block-diagonal matrix is simply the union (with multiplicity) of the eigenvalues of each individual block.
answered 2 days ago
Omnomnomnom
124k788176
add a comment |
up vote
2
down vote
accepted
In general: if $P$ and $Q$ are diagonalizable, then whenever $Px = lambda x$ and $Qy = mu y$, we will have $(P otimes Q)(x otimes y) = mu lambda (x otimes y)$. Moreover, we can form an eigenbasis out of the Kronecker products $x otimes y$.
For your particular example: $I_T otimes A$ will have the same eigenvalues as $A$, but each will have its multiplicity multiplied by $T$. For any eigenvector $v$ of $A$, the vectors $e_1 otimes v, dots, e_n otimes v$ will be eigenvectors of $I_T otimes A$ (here $e_1,dots,e_n$ is the canonical basis; so $e_1 = (1,0,dots,0)^T$).
Another general point: the eigenvalues of any block-diagonal matrix is simply the union (with multiplicity) of the eigenvalues of each individual block.
answered 2 days ago
Omnomnomnom
124k788176
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
In general: if $P$ and $Q$ are diagonalizable, then whenever $Px = lambda x$ and $Qy = mu y$, we will have $(P otimes Q)(x otimes y) = mu lambda (x otimes y)$. Moreover, we can form an eigenbasis out of the Kronecker products $x otimes y$.
For your particular example: $I_T otimes A$ will have the same eigenvalues as $A$, but each will have its multiplicity multiplied by $T$. For any eigenvector $v$ of $A$, the vectors $e_1 otimes v, dots, e_n otimes v$ will be eigenvectors of $I_T otimes A$ (here $e_1,dots,e_n$ is the canonical basis; so $e_1 = (1,0,dots,0)^T$).
Another general point: the eigenvalues of any block-diagonal matrix is simply the union (with multiplicity) of the eigenvalues of each individual block.
answered 2 days ago
Omnomnomnom
124k788176
In general: if $P$ and $Q$ are diagonalizable, then whenever $Px = lambda x$ and $Qy = mu y$, we will have $(P otimes Q)(x otimes y) = mu lambda (x otimes y)$. Moreover, we can form an eigenbasis out of the Kronecker products $x otimes y$.
For your particular example: $I_T otimes A$ will have the same eigenvalues as $A$, but each will have its multiplicity multiplied by $T$. For any eigenvector $v$ of $A$, the vectors $e_1 otimes v, dots, e_n otimes v$ will be eigenvectors of $I_T otimes A$ (here $e_1,dots,e_n$ is the canonical basis; so $e_1 = (1,0,dots,0)^T$).
Another general point: the eigenvalues of any block-diagonal matrix is simply the union (with multiplicity) of the eigenvalues of each individual block.
answered 2 days ago
Omnomnomnom
124k788176
answered 2 days ago
Omnomnomnom
124k788176
answered 2 days ago
Omnomnomnom
124k788176
answered 2 days ago
Omnomnomnom
124k788176
124k788176
add a comment |
add a comment |
Tomas is a new contributor. Be nice, and check out our Code of Conduct.
Tomas is a new contributor. Be nice, and check out our Code of Conduct.
Tomas is a new contributor. Be nice, and check out our Code of Conduct.
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Could you clarify what you mean by $mathcal A = (I_T otimes mathbf A)$? Is $otimes$ the Kronecker product? Is $I_T$ an identity matrix?
– Omnomnomnom
2 days ago
Sorry, yes, it's a Kroenecker product and identity matrix.
– Tomas
2 days ago