Mixing
Trace of a the inverse of a complex matrix
$begingroup$
I require the following
$${rm Tr}({bf H} + omega{bf I})^{-1}$$
where $bf H$ is Hermitian, $omega$ is a complex number with a small imaginary component $(Im omega approx 1times 10^{-5})$ and $bf I$ is the unit matrix. $rm Tr$ indicates a trace is required
Unfortunately these matrices are $45times 45$ and are to be numerically integrated ($bf H$ is a function of two variables) thus I was wondering if there was a cheaper way to obtain the trace, i.e. without needing to do a full inversion.
inverse trace
asked Jan 25 at 9:12
AlexDAlexD
32
$endgroup$
add a comment |
$begingroup$
I require the following
$${rm Tr}({bf H} + omega{bf I})^{-1}$$
where $bf H$ is Hermitian, $omega$ is a complex number with a small imaginary component $(Im omega approx 1times 10^{-5})$ and $bf I$ is the unit matrix. $rm Tr$ indicates a trace is required
Unfortunately these matrices are $45times 45$ and are to be numerically integrated ($bf H$ is a function of two variables) thus I was wondering if there was a cheaper way to obtain the trace, i.e. without needing to do a full inversion.
inverse trace
asked Jan 25 at 9:12
AlexDAlexD
32
$endgroup$
add a comment |
$begingroup$
I require the following
$${rm Tr}({bf H} + omega{bf I})^{-1}$$
where $bf H$ is Hermitian, $omega$ is a complex number with a small imaginary component $(Im omega approx 1times 10^{-5})$ and $bf I$ is the unit matrix. $rm Tr$ indicates a trace is required
Unfortunately these matrices are $45times 45$ and are to be numerically integrated ($bf H$ is a function of two variables) thus I was wondering if there was a cheaper way to obtain the trace, i.e. without needing to do a full inversion.
inverse trace
asked Jan 25 at 9:12
AlexDAlexD
32
$endgroup$
I require the following
$${rm Tr}({bf H} + omega{bf I})^{-1}$$
where $bf H$ is Hermitian, $omega$ is a complex number with a small imaginary component $(Im omega approx 1times 10^{-5})$ and $bf I$ is the unit matrix. $rm Tr$ indicates a trace is required
Unfortunately these matrices are $45times 45$ and are to be numerically integrated ($bf H$ is a function of two variables) thus I was wondering if there was a cheaper way to obtain the trace, i.e. without needing to do a full inversion.
inverse trace
inverse trace
asked Jan 25 at 9:12
AlexDAlexD
32
asked Jan 25 at 9:12
AlexDAlexD
32
asked Jan 25 at 9:12
AlexDAlexD
32
asked Jan 25 at 9:12
AlexDAlexD
32
asked Jan 25 at 9:12
AlexDAlexD
32
32
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Let $n=45$ and let $lambda_1,...., lambda_n$ the (real) eigenvalues of $H$. Since $H$ is diagonalisable, we have
$H=P^{-1}DP$, where $D=diag(lambda_1,...., lambda_n).$
Hence
$H+ omega I=P^{-1}D_{omega}P$, where $D_{omega}=diag(lambda_1+omega,...., lambda_n+omega).$
Therefore $(H+ omega I)^{-1}=P^{-1}D_{omega}^{-1}P$, hence
${rm Tr}((H+ omega I)^{-1})={rm Tr}(D_{omega}^{-1})= sum_{k=1}^nfrac{1}{lambda_k+omega}.$
answered Jan 25 at 10:12
FredFred
48.5k11849
$endgroup$
$begingroup$
Hi Fred, thanks for your answer, I just wanted to get some clarification: I am using the Tux Eigen package and it quotes that the computational cost to obtain the eigenvalues only, is $4n^3/3$. I can't find information as to the cost of inversion. I believe in this case it would use LU decomposition with partial pivoting. Are you able to indicate a performance gain, in $mathcal{O} (n)$ notation? Many thanks
$endgroup$
– AlexD
Jan 25 at 11:32
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $n=45$ and let $lambda_1,...., lambda_n$ the (real) eigenvalues of $H$. Since $H$ is diagonalisable, we have
$H=P^{-1}DP$, where $D=diag(lambda_1,...., lambda_n).$
Hence
$H+ omega I=P^{-1}D_{omega}P$, where $D_{omega}=diag(lambda_1+omega,...., lambda_n+omega).$
Therefore $(H+ omega I)^{-1}=P^{-1}D_{omega}^{-1}P$, hence
${rm Tr}((H+ omega I)^{-1})={rm Tr}(D_{omega}^{-1})= sum_{k=1}^nfrac{1}{lambda_k+omega}.$
answered Jan 25 at 10:12
FredFred
48.5k11849
$endgroup$
$begingroup$
Hi Fred, thanks for your answer, I just wanted to get some clarification: I am using the Tux Eigen package and it quotes that the computational cost to obtain the eigenvalues only, is $4n^3/3$. I can't find information as to the cost of inversion. I believe in this case it would use LU decomposition with partial pivoting. Are you able to indicate a performance gain, in $mathcal{O} (n)$ notation? Many thanks
$endgroup$
– AlexD
Jan 25 at 11:32
add a comment |
$begingroup$
Let $n=45$ and let $lambda_1,...., lambda_n$ the (real) eigenvalues of $H$. Since $H$ is diagonalisable, we have
$H=P^{-1}DP$, where $D=diag(lambda_1,...., lambda_n).$
Hence
$H+ omega I=P^{-1}D_{omega}P$, where $D_{omega}=diag(lambda_1+omega,...., lambda_n+omega).$
Therefore $(H+ omega I)^{-1}=P^{-1}D_{omega}^{-1}P$, hence
${rm Tr}((H+ omega I)^{-1})={rm Tr}(D_{omega}^{-1})= sum_{k=1}^nfrac{1}{lambda_k+omega}.$
answered Jan 25 at 10:12
FredFred
48.5k11849
$endgroup$
$begingroup$
Hi Fred, thanks for your answer, I just wanted to get some clarification: I am using the Tux Eigen package and it quotes that the computational cost to obtain the eigenvalues only, is $4n^3/3$. I can't find information as to the cost of inversion. I believe in this case it would use LU decomposition with partial pivoting. Are you able to indicate a performance gain, in $mathcal{O} (n)$ notation? Many thanks
$endgroup$
– AlexD
Jan 25 at 11:32
add a comment |
$begingroup$
Let $n=45$ and let $lambda_1,...., lambda_n$ the (real) eigenvalues of $H$. Since $H$ is diagonalisable, we have
$H=P^{-1}DP$, where $D=diag(lambda_1,...., lambda_n).$
Hence
$H+ omega I=P^{-1}D_{omega}P$, where $D_{omega}=diag(lambda_1+omega,...., lambda_n+omega).$
Therefore $(H+ omega I)^{-1}=P^{-1}D_{omega}^{-1}P$, hence
${rm Tr}((H+ omega I)^{-1})={rm Tr}(D_{omega}^{-1})= sum_{k=1}^nfrac{1}{lambda_k+omega}.$
answered Jan 25 at 10:12
FredFred
48.5k11849
$endgroup$
Let $n=45$ and let $lambda_1,...., lambda_n$ the (real) eigenvalues of $H$. Since $H$ is diagonalisable, we have
$H=P^{-1}DP$, where $D=diag(lambda_1,...., lambda_n).$
Hence
$H+ omega I=P^{-1}D_{omega}P$, where $D_{omega}=diag(lambda_1+omega,...., lambda_n+omega).$
Therefore $(H+ omega I)^{-1}=P^{-1}D_{omega}^{-1}P$, hence
${rm Tr}((H+ omega I)^{-1})={rm Tr}(D_{omega}^{-1})= sum_{k=1}^nfrac{1}{lambda_k+omega}.$
answered Jan 25 at 10:12
FredFred
48.5k11849
answered Jan 25 at 10:12
FredFred
48.5k11849
answered Jan 25 at 10:12
FredFred
48.5k11849
answered Jan 25 at 10:12
FredFred
48.5k11849
48.5k11849
$begingroup$
Hi Fred, thanks for your answer, I just wanted to get some clarification: I am using the Tux Eigen package and it quotes that the computational cost to obtain the eigenvalues only, is $4n^3/3$. I can't find information as to the cost of inversion. I believe in this case it would use LU decomposition with partial pivoting. Are you able to indicate a performance gain, in $mathcal{O} (n)$ notation? Many thanks
$endgroup$
– AlexD
Jan 25 at 11:32
add a comment |
$begingroup$
Hi Fred, thanks for your answer, I just wanted to get some clarification: I am using the Tux Eigen package and it quotes that the computational cost to obtain the eigenvalues only, is $4n^3/3$. I can't find information as to the cost of inversion. I believe in this case it would use LU decomposition with partial pivoting. Are you able to indicate a performance gain, in $mathcal{O} (n)$ notation? Many thanks
$endgroup$
– AlexD
Jan 25 at 11:32
$begingroup$
Hi Fred, thanks for your answer, I just wanted to get some clarification: I am using the Tux Eigen package and it quotes that the computational cost to obtain the eigenvalues only, is $4n^3/3$. I can't find information as to the cost of inversion. I believe in this case it would use LU decomposition with partial pivoting. Are you able to indicate a performance gain, in $mathcal{O} (n)$ notation? Many thanks
$endgroup$
– AlexD
Jan 25 at 11:32
$begingroup$
Hi Fred, thanks for your answer, I just wanted to get some clarification: I am using the Tux Eigen package and it quotes that the computational cost to obtain the eigenvalues only, is $4n^3/3$. I can't find information as to the cost of inversion. I believe in this case it would use LU decomposition with partial pivoting. Are you able to indicate a performance gain, in $mathcal{O} (n)$ notation? Many thanks
$endgroup$
– AlexD
Jan 25 at 11:32
add a comment |
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