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







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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















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







share|cite|improve this question







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













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







share|cite|improve this question







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






share|cite|improve this question







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.











share|cite|improve this question







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.









share|cite|improve this question




share|cite|improve this question






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asked 2 days ago









Tomas

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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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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.






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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


















  • 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










1 Answer
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2
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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.






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    up vote
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    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.






    share|cite|improve this answer

























      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.






      share|cite|improve this answer























        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Omnomnomnom

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