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Eigenvalues of $A+A^T$
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Knowing the eigenvalues of $A$, what we can say about the eigenvalues of $A+A^T$?
I know that $A$ and $A^T$ have the same spectrum, but also I think that it is not true that $lambda_i,(A+A^T) = 2lambda_i,A$ where $lambda_i,A$ is the $i$-th eigenvalue of $A$.
For example if $A$ is a orthogonal matrix we have: $A+A^T=A+A^-1$ and thus $lambda_i,(A+A^T) = lambda_i,A + frac1lambda_i,A$.
linear-algebra matrices eigenvalues-eigenvectors transpose
asked Jul 31 at 9:22
Alex
1729
 |Â
show 7 more comments
up vote
1
down vote
favorite
Knowing the eigenvalues of $A$, what we can say about the eigenvalues of $A+A^T$?
I know that $A$ and $A^T$ have the same spectrum, but also I think that it is not true that $lambda_i,(A+A^T) = 2lambda_i,A$ where $lambda_i,A$ is the $i$-th eigenvalue of $A$.
For example if $A$ is a orthogonal matrix we have: $A+A^T=A+A^-1$ and thus $lambda_i,(A+A^T) = lambda_i,A + frac1lambda_i,A$.
linear-algebra matrices eigenvalues-eigenvectors transpose
asked Jul 31 at 9:22
Alex
1729
1
How can you write $lambda_i,(A+A^T) = lambda_i,A + frac1lambda_i,A$? Other than the fact that it's not true that an eigenvalue of the sum of two matrices is the sum of two eigenvalues respectively, this expression still doesn't make any sense if I start relabeling eigenvalues of one of the matrices.
– Mathematician 42
Jul 31 at 9:26
It is not true? The $i$-th engevalue of $A^-1$ is equal to the inverse of the $i$-th engevalue of $A$
– Alex
Jul 31 at 9:31
Considering that the eigenvalues are evaluated by a determinant it's not always true that $$det(A+B)=det(A)+det(B)$$ so you cannot say that
– Davide Morgante
Jul 31 at 9:34
Ok, I understand that, thanks. So we cannot say anything.
– Alex
Jul 31 at 9:41
1
@Mathematician42 If you still don't believe the statement for the eigenvalues of $A+A^-1$ then could you tell me, which part of my explanation is erroneous? (Also, I believe now that the statement for the eigenvalues of $A+A^-1$ holds for arbitrary matrices -- since both matrices can be simultaneously put in upper triangular form (over $Bbb C$).)
– Claudius
Jul 31 at 10:08
 |Â
show 7 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Knowing the eigenvalues of $A$, what we can say about the eigenvalues of $A+A^T$?
I know that $A$ and $A^T$ have the same spectrum, but also I think that it is not true that $lambda_i,(A+A^T) = 2lambda_i,A$ where $lambda_i,A$ is the $i$-th eigenvalue of $A$.
For example if $A$ is a orthogonal matrix we have: $A+A^T=A+A^-1$ and thus $lambda_i,(A+A^T) = lambda_i,A + frac1lambda_i,A$.
linear-algebra matrices eigenvalues-eigenvectors transpose
asked Jul 31 at 9:22
Alex
1729
Knowing the eigenvalues of $A$, what we can say about the eigenvalues of $A+A^T$?
I know that $A$ and $A^T$ have the same spectrum, but also I think that it is not true that $lambda_i,(A+A^T) = 2lambda_i,A$ where $lambda_i,A$ is the $i$-th eigenvalue of $A$.
For example if $A$ is a orthogonal matrix we have: $A+A^T=A+A^-1$ and thus $lambda_i,(A+A^T) = lambda_i,A + frac1lambda_i,A$.
linear-algebra matrices eigenvalues-eigenvectors transpose
asked Jul 31 at 9:22
Alex
1729
asked Jul 31 at 9:22
Alex
1729
asked Jul 31 at 9:22
Alex
1729
asked Jul 31 at 9:22
Alex
1729
1729
1
How can you write $lambda_i,(A+A^T) = lambda_i,A + frac1lambda_i,A$? Other than the fact that it's not true that an eigenvalue of the sum of two matrices is the sum of two eigenvalues respectively, this expression still doesn't make any sense if I start relabeling eigenvalues of one of the matrices.
– Mathematician 42
Jul 31 at 9:26
It is not true? The $i$-th engevalue of $A^-1$ is equal to the inverse of the $i$-th engevalue of $A$
– Alex
Jul 31 at 9:31
Considering that the eigenvalues are evaluated by a determinant it's not always true that $$det(A+B)=det(A)+det(B)$$ so you cannot say that
– Davide Morgante
Jul 31 at 9:34
Ok, I understand that, thanks. So we cannot say anything.
– Alex
Jul 31 at 9:41
1
@Mathematician42 If you still don't believe the statement for the eigenvalues of $A+A^-1$ then could you tell me, which part of my explanation is erroneous? (Also, I believe now that the statement for the eigenvalues of $A+A^-1$ holds for arbitrary matrices -- since both matrices can be simultaneously put in upper triangular form (over $Bbb C$).)
– Claudius
Jul 31 at 10:08
 |Â
show 7 more comments
1
How can you write $lambda_i,(A+A^T) = lambda_i,A + frac1lambda_i,A$? Other than the fact that it's not true that an eigenvalue of the sum of two matrices is the sum of two eigenvalues respectively, this expression still doesn't make any sense if I start relabeling eigenvalues of one of the matrices.
– Mathematician 42
Jul 31 at 9:26
It is not true? The $i$-th engevalue of $A^-1$ is equal to the inverse of the $i$-th engevalue of $A$
– Alex
Jul 31 at 9:31
Considering that the eigenvalues are evaluated by a determinant it's not always true that $$det(A+B)=det(A)+det(B)$$ so you cannot say that
– Davide Morgante
Jul 31 at 9:34
Ok, I understand that, thanks. So we cannot say anything.
– Alex
Jul 31 at 9:41
1
@Mathematician42 If you still don't believe the statement for the eigenvalues of $A+A^-1$ then could you tell me, which part of my explanation is erroneous? (Also, I believe now that the statement for the eigenvalues of $A+A^-1$ holds for arbitrary matrices -- since both matrices can be simultaneously put in upper triangular form (over $Bbb C$).)
– Claudius
Jul 31 at 10:08
1
1
How can you write $lambda_i,(A+A^T) = lambda_i,A + frac1lambda_i,A$? Other than the fact that it's not true that an eigenvalue of the sum of two matrices is the sum of two eigenvalues respectively, this expression still doesn't make any sense if I start relabeling eigenvalues of one of the matrices.
– Mathematician 42
Jul 31 at 9:26
How can you write $lambda_i,(A+A^T) = lambda_i,A + frac1lambda_i,A$? Other than the fact that it's not true that an eigenvalue of the sum of two matrices is the sum of two eigenvalues respectively, this expression still doesn't make any sense if I start relabeling eigenvalues of one of the matrices.
– Mathematician 42
Jul 31 at 9:26
It is not true? The $i$-th engevalue of $A^-1$ is equal to the inverse of the $i$-th engevalue of $A$
– Alex
Jul 31 at 9:31
It is not true? The $i$-th engevalue of $A^-1$ is equal to the inverse of the $i$-th engevalue of $A$
– Alex
Jul 31 at 9:31
Considering that the eigenvalues are evaluated by a determinant it's not always true that $$det(A+B)=det(A)+det(B)$$ so you cannot say that
– Davide Morgante
Jul 31 at 9:34
Considering that the eigenvalues are evaluated by a determinant it's not always true that $$det(A+B)=det(A)+det(B)$$ so you cannot say that
– Davide Morgante
Jul 31 at 9:34
Ok, I understand that, thanks. So we cannot say anything.
– Alex
Jul 31 at 9:41
Ok, I understand that, thanks. So we cannot say anything.
– Alex
Jul 31 at 9:41
1
1
@Mathematician42 If you still don't believe the statement for the eigenvalues of $A+A^-1$ then could you tell me, which part of my explanation is erroneous? (Also, I believe now that the statement for the eigenvalues of $A+A^-1$ holds for arbitrary matrices -- since both matrices can be simultaneously put in upper triangular form (over $Bbb C$).)
– Claudius
Jul 31 at 10:08
@Mathematician42 If you still don't believe the statement for the eigenvalues of $A+A^-1$ then could you tell me, which part of my explanation is erroneous? (Also, I believe now that the statement for the eigenvalues of $A+A^-1$ holds for arbitrary matrices -- since both matrices can be simultaneously put in upper triangular form (over $Bbb C$).)
– Claudius
Jul 31 at 10:08
 |Â
show 7 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
On this forum post on MathOverflow the user Denis Serre gave some interesting insights. Probably is the most you can say about your problem
answered Jul 31 at 9:29
Davide Morgante
1,679220
add a comment |Â
up vote
2
down vote
$A^T$ can be equal to $-A$ so there is hardly anything you can say about eigen values of $A+A^T$.
answered Jul 31 at 9:30
Kavi Rama Murthy
19.5k2829
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
On this forum post on MathOverflow the user Denis Serre gave some interesting insights. Probably is the most you can say about your problem
answered Jul 31 at 9:29
Davide Morgante
1,679220
add a comment |Â
up vote
2
down vote
On this forum post on MathOverflow the user Denis Serre gave some interesting insights. Probably is the most you can say about your problem
answered Jul 31 at 9:29
Davide Morgante
1,679220
add a comment |Â
up vote
2
down vote
up vote
2
down vote
On this forum post on MathOverflow the user Denis Serre gave some interesting insights. Probably is the most you can say about your problem
answered Jul 31 at 9:29
Davide Morgante
1,679220
On this forum post on MathOverflow the user Denis Serre gave some interesting insights. Probably is the most you can say about your problem
answered Jul 31 at 9:29
Davide Morgante
1,679220
answered Jul 31 at 9:29
Davide Morgante
1,679220
answered Jul 31 at 9:29
Davide Morgante
1,679220
answered Jul 31 at 9:29
Davide Morgante
1,679220
1,679220
add a comment |Â
add a comment |Â
up vote
2
down vote
$A^T$ can be equal to $-A$ so there is hardly anything you can say about eigen values of $A+A^T$.
answered Jul 31 at 9:30
Kavi Rama Murthy
19.5k2829
add a comment |Â
up vote
2
down vote
$A^T$ can be equal to $-A$ so there is hardly anything you can say about eigen values of $A+A^T$.
answered Jul 31 at 9:30
Kavi Rama Murthy
19.5k2829
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$A^T$ can be equal to $-A$ so there is hardly anything you can say about eigen values of $A+A^T$.
answered Jul 31 at 9:30
Kavi Rama Murthy
19.5k2829
$A^T$ can be equal to $-A$ so there is hardly anything you can say about eigen values of $A+A^T$.
answered Jul 31 at 9:30
Kavi Rama Murthy
19.5k2829
answered Jul 31 at 9:30
Kavi Rama Murthy
19.5k2829
answered Jul 31 at 9:30
Kavi Rama Murthy
19.5k2829
answered Jul 31 at 9:30
Kavi Rama Murthy
19.5k2829
19.5k2829
add a comment |Â
add a comment |Â
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1
How can you write $lambda_i,(A+A^T) = lambda_i,A + frac1lambda_i,A$? Other than the fact that it's not true that an eigenvalue of the sum of two matrices is the sum of two eigenvalues respectively, this expression still doesn't make any sense if I start relabeling eigenvalues of one of the matrices.
– Mathematician 42
Jul 31 at 9:26
It is not true? The $i$-th engevalue of $A^-1$ is equal to the inverse of the $i$-th engevalue of $A$
– Alex
Jul 31 at 9:31
Considering that the eigenvalues are evaluated by a determinant it's not always true that $$det(A+B)=det(A)+det(B)$$ so you cannot say that
– Davide Morgante
Jul 31 at 9:34
Ok, I understand that, thanks. So we cannot say anything.
– Alex
Jul 31 at 9:41
1
@Mathematician42 If you still don't believe the statement for the eigenvalues of $A+A^-1$ then could you tell me, which part of my explanation is erroneous? (Also, I believe now that the statement for the eigenvalues of $A+A^-1$ holds for arbitrary matrices -- since both matrices can be simultaneously put in upper triangular form (over $Bbb C$).)
– Claudius
Jul 31 at 10:08