Mixing
Trace of a matrix by eigendecomposition
Clash Royale CLAN TAG#URR8PPP
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Let $A$ denote matrix based on another matrix $B$:
beginalign
A = (I + lambda B)^-1
endalign
$I$ is the identity matrix and $lambda$ is a coefficient.
Decomposing $B$ as $USU^T$ where $U^TU=I$:
beginalign
A = (I + lambda USU^T)^-1 = U(I + lambda S)^-1U^T quad text(eq. 1)
endalign
and the trace of $A$ is given as:
beginalign
operatornametr(A) = operatornametrleft(U(I + lambda S)^-1U^Tright) = operatornametrleft((I + lambda S)^-1U^TUright) = sum_i=1^n frac11 + lambda s_ii
endalign
where $s_ii$ are the eigenvalues of $B$.
Given diagonal matrix $X$, we define a new matrix $C$:
beginalign
C = (X + lambda B)^-1 X
endalign
Do the identities in eq. (1) hold when there is an arbitrary diagonal matrix in place of the identity matrix such that the trace of $C$ can be written simply as shown below?
beginalign
operatornametr(C) = sum_i=1^n fracx_iix_ii + lambda s_ii
endalign
linear-algebra matrices trace
asked Aug 2 at 11:30
hatmatrix
1758
add a comment |Â
up vote
0
down vote
favorite
Let $A$ denote matrix based on another matrix $B$:
beginalign
A = (I + lambda B)^-1
endalign
$I$ is the identity matrix and $lambda$ is a coefficient.
Decomposing $B$ as $USU^T$ where $U^TU=I$:
beginalign
A = (I + lambda USU^T)^-1 = U(I + lambda S)^-1U^T quad text(eq. 1)
endalign
and the trace of $A$ is given as:
beginalign
operatornametr(A) = operatornametrleft(U(I + lambda S)^-1U^Tright) = operatornametrleft((I + lambda S)^-1U^TUright) = sum_i=1^n frac11 + lambda s_ii
endalign
where $s_ii$ are the eigenvalues of $B$.
Given diagonal matrix $X$, we define a new matrix $C$:
beginalign
C = (X + lambda B)^-1 X
endalign
Do the identities in eq. (1) hold when there is an arbitrary diagonal matrix in place of the identity matrix such that the trace of $C$ can be written simply as shown below?
beginalign
operatornametr(C) = sum_i=1^n fracx_iix_ii + lambda s_ii
endalign
linear-algebra matrices trace
asked Aug 2 at 11:30
hatmatrix
1758
You're right - it's the trace that's equal. I've corrected.
– hatmatrix
Aug 2 at 11:54
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $A$ denote matrix based on another matrix $B$:
beginalign
A = (I + lambda B)^-1
endalign
$I$ is the identity matrix and $lambda$ is a coefficient.
Decomposing $B$ as $USU^T$ where $U^TU=I$:
beginalign
A = (I + lambda USU^T)^-1 = U(I + lambda S)^-1U^T quad text(eq. 1)
endalign
and the trace of $A$ is given as:
beginalign
operatornametr(A) = operatornametrleft(U(I + lambda S)^-1U^Tright) = operatornametrleft((I + lambda S)^-1U^TUright) = sum_i=1^n frac11 + lambda s_ii
endalign
where $s_ii$ are the eigenvalues of $B$.
Given diagonal matrix $X$, we define a new matrix $C$:
beginalign
C = (X + lambda B)^-1 X
endalign
Do the identities in eq. (1) hold when there is an arbitrary diagonal matrix in place of the identity matrix such that the trace of $C$ can be written simply as shown below?
beginalign
operatornametr(C) = sum_i=1^n fracx_iix_ii + lambda s_ii
endalign
linear-algebra matrices trace
asked Aug 2 at 11:30
hatmatrix
1758
Let $A$ denote matrix based on another matrix $B$:
beginalign
A = (I + lambda B)^-1
endalign
$I$ is the identity matrix and $lambda$ is a coefficient.
Decomposing $B$ as $USU^T$ where $U^TU=I$:
beginalign
A = (I + lambda USU^T)^-1 = U(I + lambda S)^-1U^T quad text(eq. 1)
endalign
and the trace of $A$ is given as:
beginalign
operatornametr(A) = operatornametrleft(U(I + lambda S)^-1U^Tright) = operatornametrleft((I + lambda S)^-1U^TUright) = sum_i=1^n frac11 + lambda s_ii
endalign
where $s_ii$ are the eigenvalues of $B$.
Given diagonal matrix $X$, we define a new matrix $C$:
beginalign
C = (X + lambda B)^-1 X
endalign
Do the identities in eq. (1) hold when there is an arbitrary diagonal matrix in place of the identity matrix such that the trace of $C$ can be written simply as shown below?
beginalign
operatornametr(C) = sum_i=1^n fracx_iix_ii + lambda s_ii
endalign
linear-algebra matrices trace
asked Aug 2 at 11:30
hatmatrix
1758
edited Aug 2 at 11:53
asked Aug 2 at 11:30
hatmatrix
1758
asked Aug 2 at 11:30
hatmatrix
1758
asked Aug 2 at 11:30
hatmatrix
1758
1758
You're right - it's the trace that's equal. I've corrected.
– hatmatrix
Aug 2 at 11:54
add a comment |Â
You're right - it's the trace that's equal. I've corrected.
– hatmatrix
Aug 2 at 11:54
You're right - it's the trace that's equal. I've corrected.
– hatmatrix
Aug 2 at 11:54
You're right - it's the trace that's equal. I've corrected.
– hatmatrix
Aug 2 at 11:54
add a comment |Â
1 Answer
1
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1
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If $X$ commutes with $B$, and (possibly not necessarily) if $X$ or $B$ is invertible, then the claim is true for $lambda$ such that $X+lambda ,B$ is invertible, provided that the ordering of $left(x_i,iright)_i=1,2,ldots,n$ and the ordering of $left(s_i,iright)_i=1,2,ldots,n$ are compatible. If not, then the claim can fail. Here is a counterexample.
Let $X:=beginbmatrix1&0\0&2endbmatrix$ and $B:=beginbmatrix0&1\1&0endbmatrix$. Then, $x_1,1=1$, $x_2,2=2$, and $bigs_1,1,s_2,2big=-1,+1$. Now, for $lambda:=1$, we get
$$C=(X+lambda,B)^-1,X=beginbmatrix1&1\1&2endbmatrix^-1,beginbmatrix1&0\0&2endbmatrix=beginbmatrix2&-1\-1&1endbmatrix,beginbmatrix1&0\0&2endbmatrix=beginbmatrix2&-2\-1&2endbmatrix,.$$
Thus, $$texttrace(C)=4neq sumlimits_i=1^2,fracx_i,ix_i,i+lambda,s_i,i,,$$
as the only way the right-hand side is defined is when $s_1,1=+1$ and $s_2,2=-1$, which makes
$$ sumlimits_i=1^2,fracx_i,ix_i,i+lambda,s_i,i=frac11+1+frac22-1=frac52,.$$
answered Aug 2 at 12:01
Batominovski
22.7k22776
Thanks for the example. Is it conceivable that there is an expression as simple as $sum_i (1/1+lambda s_i)$ when the matrix is not the identity matrix?
– hatmatrix
Aug 2 at 12:35
1
Not that I know of, and I don't expect that there is anything simple. Inverting a matrix often result in something very nasty.
– Batominovski
Aug 2 at 13:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If $X$ commutes with $B$, and (possibly not necessarily) if $X$ or $B$ is invertible, then the claim is true for $lambda$ such that $X+lambda ,B$ is invertible, provided that the ordering of $left(x_i,iright)_i=1,2,ldots,n$ and the ordering of $left(s_i,iright)_i=1,2,ldots,n$ are compatible. If not, then the claim can fail. Here is a counterexample.
Let $X:=beginbmatrix1&0\0&2endbmatrix$ and $B:=beginbmatrix0&1\1&0endbmatrix$. Then, $x_1,1=1$, $x_2,2=2$, and $bigs_1,1,s_2,2big=-1,+1$. Now, for $lambda:=1$, we get
$$C=(X+lambda,B)^-1,X=beginbmatrix1&1\1&2endbmatrix^-1,beginbmatrix1&0\0&2endbmatrix=beginbmatrix2&-1\-1&1endbmatrix,beginbmatrix1&0\0&2endbmatrix=beginbmatrix2&-2\-1&2endbmatrix,.$$
Thus, $$texttrace(C)=4neq sumlimits_i=1^2,fracx_i,ix_i,i+lambda,s_i,i,,$$
as the only way the right-hand side is defined is when $s_1,1=+1$ and $s_2,2=-1$, which makes
$$ sumlimits_i=1^2,fracx_i,ix_i,i+lambda,s_i,i=frac11+1+frac22-1=frac52,.$$
answered Aug 2 at 12:01
Batominovski
22.7k22776
Thanks for the example. Is it conceivable that there is an expression as simple as $sum_i (1/1+lambda s_i)$ when the matrix is not the identity matrix?
– hatmatrix
Aug 2 at 12:35
1
Not that I know of, and I don't expect that there is anything simple. Inverting a matrix often result in something very nasty.
– Batominovski
Aug 2 at 13:07
add a comment |Â
up vote
1
down vote
accepted
If $X$ commutes with $B$, and (possibly not necessarily) if $X$ or $B$ is invertible, then the claim is true for $lambda$ such that $X+lambda ,B$ is invertible, provided that the ordering of $left(x_i,iright)_i=1,2,ldots,n$ and the ordering of $left(s_i,iright)_i=1,2,ldots,n$ are compatible. If not, then the claim can fail. Here is a counterexample.
Let $X:=beginbmatrix1&0\0&2endbmatrix$ and $B:=beginbmatrix0&1\1&0endbmatrix$. Then, $x_1,1=1$, $x_2,2=2$, and $bigs_1,1,s_2,2big=-1,+1$. Now, for $lambda:=1$, we get
$$C=(X+lambda,B)^-1,X=beginbmatrix1&1\1&2endbmatrix^-1,beginbmatrix1&0\0&2endbmatrix=beginbmatrix2&-1\-1&1endbmatrix,beginbmatrix1&0\0&2endbmatrix=beginbmatrix2&-2\-1&2endbmatrix,.$$
Thus, $$texttrace(C)=4neq sumlimits_i=1^2,fracx_i,ix_i,i+lambda,s_i,i,,$$
as the only way the right-hand side is defined is when $s_1,1=+1$ and $s_2,2=-1$, which makes
$$ sumlimits_i=1^2,fracx_i,ix_i,i+lambda,s_i,i=frac11+1+frac22-1=frac52,.$$
answered Aug 2 at 12:01
Batominovski
22.7k22776
Thanks for the example. Is it conceivable that there is an expression as simple as $sum_i (1/1+lambda s_i)$ when the matrix is not the identity matrix?
– hatmatrix
Aug 2 at 12:35
1
Not that I know of, and I don't expect that there is anything simple. Inverting a matrix often result in something very nasty.
– Batominovski
Aug 2 at 13:07
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If $X$ commutes with $B$, and (possibly not necessarily) if $X$ or $B$ is invertible, then the claim is true for $lambda$ such that $X+lambda ,B$ is invertible, provided that the ordering of $left(x_i,iright)_i=1,2,ldots,n$ and the ordering of $left(s_i,iright)_i=1,2,ldots,n$ are compatible. If not, then the claim can fail. Here is a counterexample.
Let $X:=beginbmatrix1&0\0&2endbmatrix$ and $B:=beginbmatrix0&1\1&0endbmatrix$. Then, $x_1,1=1$, $x_2,2=2$, and $bigs_1,1,s_2,2big=-1,+1$. Now, for $lambda:=1$, we get
$$C=(X+lambda,B)^-1,X=beginbmatrix1&1\1&2endbmatrix^-1,beginbmatrix1&0\0&2endbmatrix=beginbmatrix2&-1\-1&1endbmatrix,beginbmatrix1&0\0&2endbmatrix=beginbmatrix2&-2\-1&2endbmatrix,.$$
Thus, $$texttrace(C)=4neq sumlimits_i=1^2,fracx_i,ix_i,i+lambda,s_i,i,,$$
as the only way the right-hand side is defined is when $s_1,1=+1$ and $s_2,2=-1$, which makes
$$ sumlimits_i=1^2,fracx_i,ix_i,i+lambda,s_i,i=frac11+1+frac22-1=frac52,.$$
answered Aug 2 at 12:01
Batominovski
22.7k22776
If $X$ commutes with $B$, and (possibly not necessarily) if $X$ or $B$ is invertible, then the claim is true for $lambda$ such that $X+lambda ,B$ is invertible, provided that the ordering of $left(x_i,iright)_i=1,2,ldots,n$ and the ordering of $left(s_i,iright)_i=1,2,ldots,n$ are compatible. If not, then the claim can fail. Here is a counterexample.
Let $X:=beginbmatrix1&0\0&2endbmatrix$ and $B:=beginbmatrix0&1\1&0endbmatrix$. Then, $x_1,1=1$, $x_2,2=2$, and $bigs_1,1,s_2,2big=-1,+1$. Now, for $lambda:=1$, we get
$$C=(X+lambda,B)^-1,X=beginbmatrix1&1\1&2endbmatrix^-1,beginbmatrix1&0\0&2endbmatrix=beginbmatrix2&-1\-1&1endbmatrix,beginbmatrix1&0\0&2endbmatrix=beginbmatrix2&-2\-1&2endbmatrix,.$$
Thus, $$texttrace(C)=4neq sumlimits_i=1^2,fracx_i,ix_i,i+lambda,s_i,i,,$$
as the only way the right-hand side is defined is when $s_1,1=+1$ and $s_2,2=-1$, which makes
$$ sumlimits_i=1^2,fracx_i,ix_i,i+lambda,s_i,i=frac11+1+frac22-1=frac52,.$$
answered Aug 2 at 12:01
Batominovski
22.7k22776
edited Aug 2 at 12:07
answered Aug 2 at 12:01
Batominovski
22.7k22776
answered Aug 2 at 12:01
Batominovski
22.7k22776
answered Aug 2 at 12:01
Batominovski
22.7k22776
22.7k22776
Thanks for the example. Is it conceivable that there is an expression as simple as $sum_i (1/1+lambda s_i)$ when the matrix is not the identity matrix?
– hatmatrix
Aug 2 at 12:35
1
Not that I know of, and I don't expect that there is anything simple. Inverting a matrix often result in something very nasty.
– Batominovski
Aug 2 at 13:07
add a comment |Â
Thanks for the example. Is it conceivable that there is an expression as simple as $sum_i (1/1+lambda s_i)$ when the matrix is not the identity matrix?
– hatmatrix
Aug 2 at 12:35
1
Not that I know of, and I don't expect that there is anything simple. Inverting a matrix often result in something very nasty.
– Batominovski
Aug 2 at 13:07
Thanks for the example. Is it conceivable that there is an expression as simple as $sum_i (1/1+lambda s_i)$ when the matrix is not the identity matrix?
– hatmatrix
Aug 2 at 12:35
Thanks for the example. Is it conceivable that there is an expression as simple as $sum_i (1/1+lambda s_i)$ when the matrix is not the identity matrix?
– hatmatrix
Aug 2 at 12:35
1
1
Not that I know of, and I don't expect that there is anything simple. Inverting a matrix often result in something very nasty.
– Batominovski
Aug 2 at 13:07
Not that I know of, and I don't expect that there is anything simple. Inverting a matrix often result in something very nasty.
– Batominovski
Aug 2 at 13:07
add a comment |Â
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You're right - it's the trace that's equal. I've corrected.
– hatmatrix
Aug 2 at 11:54