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Sum of entries of powers of symmetric matrix related to eigenvalues?
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I wish to prove:
For any symmetric matrix $A$ with nonnegative entries, the grand sum (sum of all entries) of $A^n$ is equal to the sum of the $n$th powers of the eigenvalues of $A$.
Proof sketch:
Diagonalize the matrix. We know we can because it is symmetric and all entries are positive or zero. Say $S$ is the diagonalization matrix (i.e. the combination of all the eigenvectors), so that $B = S^-1 A S$ is a diagonal matrix with the eigenvalues of $A$ on the diagonal. Then $$A^n = (S B S^-1)^n = S B^n S^-1$$
so
$$grandsum(A^n)=grandsum(S B^n S^-1) stackrel?= grandsum(B^n) = trace(B^n) = sum_a in eigvals(A)a^n$$
The only thing I'm missing to complete the proof is to show that the grand sum of $S B^n S^-1$ is the grand sum of $B^n$. My thought is that the diagonalization shouldn't change the magnitude of $A$ but I don't know how to express this.
linear-algebra eigenvalues-eigenvectors
asked Jul 24 at 5:01
Snowbody
1416
add a comment |Â
up vote
4
down vote
favorite
I wish to prove:
For any symmetric matrix $A$ with nonnegative entries, the grand sum (sum of all entries) of $A^n$ is equal to the sum of the $n$th powers of the eigenvalues of $A$.
Proof sketch:
Diagonalize the matrix. We know we can because it is symmetric and all entries are positive or zero. Say $S$ is the diagonalization matrix (i.e. the combination of all the eigenvectors), so that $B = S^-1 A S$ is a diagonal matrix with the eigenvalues of $A$ on the diagonal. Then $$A^n = (S B S^-1)^n = S B^n S^-1$$
so
$$grandsum(A^n)=grandsum(S B^n S^-1) stackrel?= grandsum(B^n) = trace(B^n) = sum_a in eigvals(A)a^n$$
The only thing I'm missing to complete the proof is to show that the grand sum of $S B^n S^-1$ is the grand sum of $B^n$. My thought is that the diagonalization shouldn't change the magnitude of $A$ but I don't know how to express this.
linear-algebra eigenvalues-eigenvectors
asked Jul 24 at 5:01
Snowbody
1416
2
It doesn't seem like it would be true to me. I would suggest you try to find a counter example instead. e.g. if $A = beginbmatrix 4&1\1&4 end bmatrix$ the eigenvalues are $3,5$ and the "grand sum" = 10. But "grand sum" of $A^1 ne 3^1 + 5^1$ And for $A^2$ the results get no better.
– Doug M
Jul 24 at 5:17
1
Sum of squares of the entries of $A^n$ is the sum of $n$-th powers of the squares of the eigenvalues. Try that with the matrices from the numerous counterexamples already given.
– WimC
Jul 24 at 5:28
1
For any $2times2$ symmetric matrix $A$, the sum of the squares of the entries of $A^2$ is identically equal to the sum of the $4$-th powers of the eigenvalues of $A$.
– quasi
Jul 24 at 6:59
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I wish to prove:
For any symmetric matrix $A$ with nonnegative entries, the grand sum (sum of all entries) of $A^n$ is equal to the sum of the $n$th powers of the eigenvalues of $A$.
Proof sketch:
Diagonalize the matrix. We know we can because it is symmetric and all entries are positive or zero. Say $S$ is the diagonalization matrix (i.e. the combination of all the eigenvectors), so that $B = S^-1 A S$ is a diagonal matrix with the eigenvalues of $A$ on the diagonal. Then $$A^n = (S B S^-1)^n = S B^n S^-1$$
so
$$grandsum(A^n)=grandsum(S B^n S^-1) stackrel?= grandsum(B^n) = trace(B^n) = sum_a in eigvals(A)a^n$$
The only thing I'm missing to complete the proof is to show that the grand sum of $S B^n S^-1$ is the grand sum of $B^n$. My thought is that the diagonalization shouldn't change the magnitude of $A$ but I don't know how to express this.
linear-algebra eigenvalues-eigenvectors
asked Jul 24 at 5:01
Snowbody
1416
I wish to prove:
For any symmetric matrix $A$ with nonnegative entries, the grand sum (sum of all entries) of $A^n$ is equal to the sum of the $n$th powers of the eigenvalues of $A$.
Proof sketch:
Diagonalize the matrix. We know we can because it is symmetric and all entries are positive or zero. Say $S$ is the diagonalization matrix (i.e. the combination of all the eigenvectors), so that $B = S^-1 A S$ is a diagonal matrix with the eigenvalues of $A$ on the diagonal. Then $$A^n = (S B S^-1)^n = S B^n S^-1$$
so
$$grandsum(A^n)=grandsum(S B^n S^-1) stackrel?= grandsum(B^n) = trace(B^n) = sum_a in eigvals(A)a^n$$
The only thing I'm missing to complete the proof is to show that the grand sum of $S B^n S^-1$ is the grand sum of $B^n$. My thought is that the diagonalization shouldn't change the magnitude of $A$ but I don't know how to express this.
linear-algebra eigenvalues-eigenvectors
asked Jul 24 at 5:01
Snowbody
1416
asked Jul 24 at 5:01
Snowbody
1416
asked Jul 24 at 5:01
Snowbody
1416
asked Jul 24 at 5:01
Snowbody
1416
1416
2
It doesn't seem like it would be true to me. I would suggest you try to find a counter example instead. e.g. if $A = beginbmatrix 4&1\1&4 end bmatrix$ the eigenvalues are $3,5$ and the "grand sum" = 10. But "grand sum" of $A^1 ne 3^1 + 5^1$ And for $A^2$ the results get no better.
– Doug M
Jul 24 at 5:17
1
Sum of squares of the entries of $A^n$ is the sum of $n$-th powers of the squares of the eigenvalues. Try that with the matrices from the numerous counterexamples already given.
– WimC
Jul 24 at 5:28
1
For any $2times2$ symmetric matrix $A$, the sum of the squares of the entries of $A^2$ is identically equal to the sum of the $4$-th powers of the eigenvalues of $A$.
– quasi
Jul 24 at 6:59
add a comment |Â
2
It doesn't seem like it would be true to me. I would suggest you try to find a counter example instead. e.g. if $A = beginbmatrix 4&1\1&4 end bmatrix$ the eigenvalues are $3,5$ and the "grand sum" = 10. But "grand sum" of $A^1 ne 3^1 + 5^1$ And for $A^2$ the results get no better.
– Doug M
Jul 24 at 5:17
1
Sum of squares of the entries of $A^n$ is the sum of $n$-th powers of the squares of the eigenvalues. Try that with the matrices from the numerous counterexamples already given.
– WimC
Jul 24 at 5:28
1
For any $2times2$ symmetric matrix $A$, the sum of the squares of the entries of $A^2$ is identically equal to the sum of the $4$-th powers of the eigenvalues of $A$.
– quasi
Jul 24 at 6:59
2
2
It doesn't seem like it would be true to me. I would suggest you try to find a counter example instead. e.g. if $A = beginbmatrix 4&1\1&4 end bmatrix$ the eigenvalues are $3,5$ and the "grand sum" = 10. But "grand sum" of $A^1 ne 3^1 + 5^1$ And for $A^2$ the results get no better.
– Doug M
Jul 24 at 5:17
It doesn't seem like it would be true to me. I would suggest you try to find a counter example instead. e.g. if $A = beginbmatrix 4&1\1&4 end bmatrix$ the eigenvalues are $3,5$ and the "grand sum" = 10. But "grand sum" of $A^1 ne 3^1 + 5^1$ And for $A^2$ the results get no better.
– Doug M
Jul 24 at 5:17
1
1
Sum of squares of the entries of $A^n$ is the sum of $n$-th powers of the squares of the eigenvalues. Try that with the matrices from the numerous counterexamples already given.
– WimC
Jul 24 at 5:28
Sum of squares of the entries of $A^n$ is the sum of $n$-th powers of the squares of the eigenvalues. Try that with the matrices from the numerous counterexamples already given.
– WimC
Jul 24 at 5:28
1
1
For any $2times2$ symmetric matrix $A$, the sum of the squares of the entries of $A^2$ is identically equal to the sum of the $4$-th powers of the eigenvalues of $A$.
– quasi
Jul 24 at 6:59
For any $2times2$ symmetric matrix $A$, the sum of the squares of the entries of $A^2$ is identically equal to the sum of the $4$-th powers of the eigenvalues of $A$.
– quasi
Jul 24 at 6:59
add a comment |Â
2 Answers
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votes
up vote
5
down vote
This is wrong already for $n=1$, where, for instance, $pmatrix1&1\1&1$ has eigenvalues $0$ and $2$ but the sum of the entries is $4$.
answered Jul 24 at 5:16
joriki
164k10180328
add a comment |Â
up vote
3
down vote
The claim is false. For starters, check it for $n=1$ and almost any matrix $A$, say $A=beginbmatrix1&2\2&1endbmatrix$ (or any non-diagonal symmetric matrix).
For a matrix $A$, the sum of its eigenvalues is equal to its trace $operatornametr(A)=a_11+a_22+cdots+a_nn$. As long as you have nonzero elements outside of the main diagonal, i.e. if your matrix is not diagonal, the claim doesn't hold true.
answered Jul 24 at 5:20
zipirovich
10k11630
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
This is wrong already for $n=1$, where, for instance, $pmatrix1&1\1&1$ has eigenvalues $0$ and $2$ but the sum of the entries is $4$.
answered Jul 24 at 5:16
joriki
164k10180328
add a comment |Â
up vote
5
down vote
This is wrong already for $n=1$, where, for instance, $pmatrix1&1\1&1$ has eigenvalues $0$ and $2$ but the sum of the entries is $4$.
answered Jul 24 at 5:16
joriki
164k10180328
add a comment |Â
up vote
5
down vote
up vote
5
down vote
This is wrong already for $n=1$, where, for instance, $pmatrix1&1\1&1$ has eigenvalues $0$ and $2$ but the sum of the entries is $4$.
answered Jul 24 at 5:16
joriki
164k10180328
This is wrong already for $n=1$, where, for instance, $pmatrix1&1\1&1$ has eigenvalues $0$ and $2$ but the sum of the entries is $4$.
answered Jul 24 at 5:16
joriki
164k10180328
answered Jul 24 at 5:16
joriki
164k10180328
answered Jul 24 at 5:16
joriki
164k10180328
answered Jul 24 at 5:16
joriki
164k10180328
164k10180328
add a comment |Â
add a comment |Â
up vote
3
down vote
The claim is false. For starters, check it for $n=1$ and almost any matrix $A$, say $A=beginbmatrix1&2\2&1endbmatrix$ (or any non-diagonal symmetric matrix).
For a matrix $A$, the sum of its eigenvalues is equal to its trace $operatornametr(A)=a_11+a_22+cdots+a_nn$. As long as you have nonzero elements outside of the main diagonal, i.e. if your matrix is not diagonal, the claim doesn't hold true.
answered Jul 24 at 5:20
zipirovich
10k11630
add a comment |Â
up vote
3
down vote
The claim is false. For starters, check it for $n=1$ and almost any matrix $A$, say $A=beginbmatrix1&2\2&1endbmatrix$ (or any non-diagonal symmetric matrix).
For a matrix $A$, the sum of its eigenvalues is equal to its trace $operatornametr(A)=a_11+a_22+cdots+a_nn$. As long as you have nonzero elements outside of the main diagonal, i.e. if your matrix is not diagonal, the claim doesn't hold true.
answered Jul 24 at 5:20
zipirovich
10k11630
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The claim is false. For starters, check it for $n=1$ and almost any matrix $A$, say $A=beginbmatrix1&2\2&1endbmatrix$ (or any non-diagonal symmetric matrix).
For a matrix $A$, the sum of its eigenvalues is equal to its trace $operatornametr(A)=a_11+a_22+cdots+a_nn$. As long as you have nonzero elements outside of the main diagonal, i.e. if your matrix is not diagonal, the claim doesn't hold true.
answered Jul 24 at 5:20
zipirovich
10k11630
The claim is false. For starters, check it for $n=1$ and almost any matrix $A$, say $A=beginbmatrix1&2\2&1endbmatrix$ (or any non-diagonal symmetric matrix).
For a matrix $A$, the sum of its eigenvalues is equal to its trace $operatornametr(A)=a_11+a_22+cdots+a_nn$. As long as you have nonzero elements outside of the main diagonal, i.e. if your matrix is not diagonal, the claim doesn't hold true.
answered Jul 24 at 5:20
zipirovich
10k11630
answered Jul 24 at 5:20
zipirovich
10k11630
answered Jul 24 at 5:20
zipirovich
10k11630
answered Jul 24 at 5:20
zipirovich
10k11630
10k11630
add a comment |Â
add a comment |Â
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2
It doesn't seem like it would be true to me. I would suggest you try to find a counter example instead. e.g. if $A = beginbmatrix 4&1\1&4 end bmatrix$ the eigenvalues are $3,5$ and the "grand sum" = 10. But "grand sum" of $A^1 ne 3^1 + 5^1$ And for $A^2$ the results get no better.
– Doug M
Jul 24 at 5:17
1
Sum of squares of the entries of $A^n$ is the sum of $n$-th powers of the squares of the eigenvalues. Try that with the matrices from the numerous counterexamples already given.
– WimC
Jul 24 at 5:28
1
For any $2times2$ symmetric matrix $A$, the sum of the squares of the entries of $A^2$ is identically equal to the sum of the $4$-th powers of the eigenvalues of $A$.
– quasi
Jul 24 at 6:59