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For a real square matrix $S$, what can be said about $Tr(S^2)$?
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For a real square matrix $S$ with all eigenvalues on the imaginary axis, what can be said about $Tr(S^2)$? Since all eigenvalues of $S$ are on imaginary axis, $Tr(S)=0$ but I was wondering if anything more I can find about $Tr(S^2)$.
I know the following property is associated with the trace of the product of two positive semi-definite matrices $Xgeq0,Ygeq0$,
$Tr(X^TY)leqTr(X)lambda_max(Y)$ where $lambda_max(Y)$ denotes the maximum eigenvalue of $Y$. Since $S$ is not specified whether positive semi-definite or not, I cannot use the property. Is there any other property that I can use to say anything about $Tr(S^2)$?
linear-algebra matrices matrix-decomposition trace symmetric-matrices
asked Aug 2 at 20:55
jbgujgu
1139
add a comment |Â
up vote
0
down vote
favorite
For a real square matrix $S$ with all eigenvalues on the imaginary axis, what can be said about $Tr(S^2)$? Since all eigenvalues of $S$ are on imaginary axis, $Tr(S)=0$ but I was wondering if anything more I can find about $Tr(S^2)$.
I know the following property is associated with the trace of the product of two positive semi-definite matrices $Xgeq0,Ygeq0$,
$Tr(X^TY)leqTr(X)lambda_max(Y)$ where $lambda_max(Y)$ denotes the maximum eigenvalue of $Y$. Since $S$ is not specified whether positive semi-definite or not, I cannot use the property. Is there any other property that I can use to say anything about $Tr(S^2)$?
linear-algebra matrices matrix-decomposition trace symmetric-matrices
asked Aug 2 at 20:55
jbgujgu
1139
Why do you say that $tr(S^2)=0$? Take $S=beginpmatrix 0 & 1 cr -1 & 0endpmatrix$. Then $tr(S^2)=-2neq 0$.
– Dietrich Burde
Aug 2 at 21:00
Sorry it was a typo. I only meant what can be said about $Tr(S^2)$?
– jbgujgu
Aug 2 at 21:03
I think I found an answer and correct me if I am wrong and please let me know if you find anything new. Since $S$ is real and can be put in the real Jordan form $S=M^-1KM$ and therefore $Tr(S^2)=Tr(K^2)$. Since $K$ is a block diagonal matrix with each block being in the form $K=beginbmatrix0 & pmsigma\mpsigma & 0endbmatrix$, $K^2=beginbmatrix-sigma^2 & 0\0 & -sigma^2endbmatrix$ and from which $Tr(K^2)=-2sigma^2<0$.
– jbgujgu
Aug 2 at 21:08
If $ilambda$ is an eigen value of $S$, then $-lambda^2$is an eigen value of $S^2$. The trace being sum of eigen values means $textTr(S^2)=sum_kleft(-lambda_k^2right)<0$. So the trace will be a negative number. Of course $lambda=0$ also lies on the imaginary axis and it will change the result a bit.
– Anurag A
Aug 2 at 21:37
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For a real square matrix $S$ with all eigenvalues on the imaginary axis, what can be said about $Tr(S^2)$? Since all eigenvalues of $S$ are on imaginary axis, $Tr(S)=0$ but I was wondering if anything more I can find about $Tr(S^2)$.
I know the following property is associated with the trace of the product of two positive semi-definite matrices $Xgeq0,Ygeq0$,
$Tr(X^TY)leqTr(X)lambda_max(Y)$ where $lambda_max(Y)$ denotes the maximum eigenvalue of $Y$. Since $S$ is not specified whether positive semi-definite or not, I cannot use the property. Is there any other property that I can use to say anything about $Tr(S^2)$?
linear-algebra matrices matrix-decomposition trace symmetric-matrices
asked Aug 2 at 20:55
jbgujgu
1139
For a real square matrix $S$ with all eigenvalues on the imaginary axis, what can be said about $Tr(S^2)$? Since all eigenvalues of $S$ are on imaginary axis, $Tr(S)=0$ but I was wondering if anything more I can find about $Tr(S^2)$.
I know the following property is associated with the trace of the product of two positive semi-definite matrices $Xgeq0,Ygeq0$,
$Tr(X^TY)leqTr(X)lambda_max(Y)$ where $lambda_max(Y)$ denotes the maximum eigenvalue of $Y$. Since $S$ is not specified whether positive semi-definite or not, I cannot use the property. Is there any other property that I can use to say anything about $Tr(S^2)$?
linear-algebra matrices matrix-decomposition trace symmetric-matrices
asked Aug 2 at 20:55
jbgujgu
1139
edited Aug 2 at 21:02
asked Aug 2 at 20:55
jbgujgu
1139
asked Aug 2 at 20:55
jbgujgu
1139
asked Aug 2 at 20:55
jbgujgu
1139
1139
Why do you say that $tr(S^2)=0$? Take $S=beginpmatrix 0 & 1 cr -1 & 0endpmatrix$. Then $tr(S^2)=-2neq 0$.
– Dietrich Burde
Aug 2 at 21:00
Sorry it was a typo. I only meant what can be said about $Tr(S^2)$?
– jbgujgu
Aug 2 at 21:03
I think I found an answer and correct me if I am wrong and please let me know if you find anything new. Since $S$ is real and can be put in the real Jordan form $S=M^-1KM$ and therefore $Tr(S^2)=Tr(K^2)$. Since $K$ is a block diagonal matrix with each block being in the form $K=beginbmatrix0 & pmsigma\mpsigma & 0endbmatrix$, $K^2=beginbmatrix-sigma^2 & 0\0 & -sigma^2endbmatrix$ and from which $Tr(K^2)=-2sigma^2<0$.
– jbgujgu
Aug 2 at 21:08
If $ilambda$ is an eigen value of $S$, then $-lambda^2$is an eigen value of $S^2$. The trace being sum of eigen values means $textTr(S^2)=sum_kleft(-lambda_k^2right)<0$. So the trace will be a negative number. Of course $lambda=0$ also lies on the imaginary axis and it will change the result a bit.
– Anurag A
Aug 2 at 21:37
add a comment |Â
Why do you say that $tr(S^2)=0$? Take $S=beginpmatrix 0 & 1 cr -1 & 0endpmatrix$. Then $tr(S^2)=-2neq 0$.
– Dietrich Burde
Aug 2 at 21:00
Sorry it was a typo. I only meant what can be said about $Tr(S^2)$?
– jbgujgu
Aug 2 at 21:03
I think I found an answer and correct me if I am wrong and please let me know if you find anything new. Since $S$ is real and can be put in the real Jordan form $S=M^-1KM$ and therefore $Tr(S^2)=Tr(K^2)$. Since $K$ is a block diagonal matrix with each block being in the form $K=beginbmatrix0 & pmsigma\mpsigma & 0endbmatrix$, $K^2=beginbmatrix-sigma^2 & 0\0 & -sigma^2endbmatrix$ and from which $Tr(K^2)=-2sigma^2<0$.
– jbgujgu
Aug 2 at 21:08
If $ilambda$ is an eigen value of $S$, then $-lambda^2$is an eigen value of $S^2$. The trace being sum of eigen values means $textTr(S^2)=sum_kleft(-lambda_k^2right)<0$. So the trace will be a negative number. Of course $lambda=0$ also lies on the imaginary axis and it will change the result a bit.
– Anurag A
Aug 2 at 21:37
Why do you say that $tr(S^2)=0$? Take $S=beginpmatrix 0 & 1 cr -1 & 0endpmatrix$. Then $tr(S^2)=-2neq 0$.
– Dietrich Burde
Aug 2 at 21:00
Why do you say that $tr(S^2)=0$? Take $S=beginpmatrix 0 & 1 cr -1 & 0endpmatrix$. Then $tr(S^2)=-2neq 0$.
– Dietrich Burde
Aug 2 at 21:00
Sorry it was a typo. I only meant what can be said about $Tr(S^2)$?
– jbgujgu
Aug 2 at 21:03
Sorry it was a typo. I only meant what can be said about $Tr(S^2)$?
– jbgujgu
Aug 2 at 21:03
I think I found an answer and correct me if I am wrong and please let me know if you find anything new. Since $S$ is real and can be put in the real Jordan form $S=M^-1KM$ and therefore $Tr(S^2)=Tr(K^2)$. Since $K$ is a block diagonal matrix with each block being in the form $K=beginbmatrix0 & pmsigma\mpsigma & 0endbmatrix$, $K^2=beginbmatrix-sigma^2 & 0\0 & -sigma^2endbmatrix$ and from which $Tr(K^2)=-2sigma^2<0$.
– jbgujgu
Aug 2 at 21:08
I think I found an answer and correct me if I am wrong and please let me know if you find anything new. Since $S$ is real and can be put in the real Jordan form $S=M^-1KM$ and therefore $Tr(S^2)=Tr(K^2)$. Since $K$ is a block diagonal matrix with each block being in the form $K=beginbmatrix0 & pmsigma\mpsigma & 0endbmatrix$, $K^2=beginbmatrix-sigma^2 & 0\0 & -sigma^2endbmatrix$ and from which $Tr(K^2)=-2sigma^2<0$.
– jbgujgu
Aug 2 at 21:08
If $ilambda$ is an eigen value of $S$, then $-lambda^2$is an eigen value of $S^2$. The trace being sum of eigen values means $textTr(S^2)=sum_kleft(-lambda_k^2right)<0$. So the trace will be a negative number. Of course $lambda=0$ also lies on the imaginary axis and it will change the result a bit.
– Anurag A
Aug 2 at 21:37
If $ilambda$ is an eigen value of $S$, then $-lambda^2$is an eigen value of $S^2$. The trace being sum of eigen values means $textTr(S^2)=sum_kleft(-lambda_k^2right)<0$. So the trace will be a negative number. Of course $lambda=0$ also lies on the imaginary axis and it will change the result a bit.
– Anurag A
Aug 2 at 21:37
add a comment |Â
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Why do you say that $tr(S^2)=0$? Take $S=beginpmatrix 0 & 1 cr -1 & 0endpmatrix$. Then $tr(S^2)=-2neq 0$.
– Dietrich Burde
Aug 2 at 21:00
Sorry it was a typo. I only meant what can be said about $Tr(S^2)$?
– jbgujgu
Aug 2 at 21:03
I think I found an answer and correct me if I am wrong and please let me know if you find anything new. Since $S$ is real and can be put in the real Jordan form $S=M^-1KM$ and therefore $Tr(S^2)=Tr(K^2)$. Since $K$ is a block diagonal matrix with each block being in the form $K=beginbmatrix0 & pmsigma\mpsigma & 0endbmatrix$, $K^2=beginbmatrix-sigma^2 & 0\0 & -sigma^2endbmatrix$ and from which $Tr(K^2)=-2sigma^2<0$.
– jbgujgu
Aug 2 at 21:08
If $ilambda$ is an eigen value of $S$, then $-lambda^2$is an eigen value of $S^2$. The trace being sum of eigen values means $textTr(S^2)=sum_kleft(-lambda_k^2right)<0$. So the trace will be a negative number. Of course $lambda=0$ also lies on the imaginary axis and it will change the result a bit.
– Anurag A
Aug 2 at 21:37