What can we say about the eigenvalues of a matrix with a complex trace and determinant?
up vote
1
down vote
favorite
In one of my classes, we showed that, given a 2x2 matrix $mathbf{A}$ with eigenvalues $lambda_pm$,
$Re[lambdapm]<0 iff text{Tr}[mathbf{A}]<0$ and $det[mathbf{A}]>0$
by computing the eigenvectors as
$lambda_pm = text{Tr}[mathbf{A}]pmsqrt{(text{Tr}[mathbf{A}])^2-4det[mathbf{A}]}$
It seemed to me that from the derivation, we were assuming the trace and determinant were both Real numbers; however, this isn't the general case.
Does this statement hold for $text{Tr}[mathbf{A}],det[mathbf{A}]inmathbb{C}$?
matrices complex-numbers eigenvalues-eigenvectors
asked Nov 13 at 20:36
Nathaniel D. Hoffman
356
add a comment |
up vote
1
down vote
favorite
In one of my classes, we showed that, given a 2x2 matrix $mathbf{A}$ with eigenvalues $lambda_pm$,
$Re[lambdapm]<0 iff text{Tr}[mathbf{A}]<0$ and $det[mathbf{A}]>0$
by computing the eigenvectors as
$lambda_pm = text{Tr}[mathbf{A}]pmsqrt{(text{Tr}[mathbf{A}])^2-4det[mathbf{A}]}$
It seemed to me that from the derivation, we were assuming the trace and determinant were both Real numbers; however, this isn't the general case.
Does this statement hold for $text{Tr}[mathbf{A}],det[mathbf{A}]inmathbb{C}$?
matrices complex-numbers eigenvalues-eigenvectors
asked Nov 13 at 20:36
Nathaniel D. Hoffman
356
It's not meaningful to say "trace less than zero" if the trace is complex and not real.
– vadim123
Nov 13 at 20:39
@Nathaniel D. Hoffman Eigenvalues here ($2times 2$ matrix) are either real or complex conjugates. So trace is always real for that matrix.
– AnyAD
Nov 13 at 20:56
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In one of my classes, we showed that, given a 2x2 matrix $mathbf{A}$ with eigenvalues $lambda_pm$,
$Re[lambdapm]<0 iff text{Tr}[mathbf{A}]<0$ and $det[mathbf{A}]>0$
by computing the eigenvectors as
$lambda_pm = text{Tr}[mathbf{A}]pmsqrt{(text{Tr}[mathbf{A}])^2-4det[mathbf{A}]}$
It seemed to me that from the derivation, we were assuming the trace and determinant were both Real numbers; however, this isn't the general case.
Does this statement hold for $text{Tr}[mathbf{A}],det[mathbf{A}]inmathbb{C}$?
matrices complex-numbers eigenvalues-eigenvectors
asked Nov 13 at 20:36
Nathaniel D. Hoffman
356
In one of my classes, we showed that, given a 2x2 matrix $mathbf{A}$ with eigenvalues $lambda_pm$,
$Re[lambdapm]<0 iff text{Tr}[mathbf{A}]<0$ and $det[mathbf{A}]>0$
by computing the eigenvectors as
$lambda_pm = text{Tr}[mathbf{A}]pmsqrt{(text{Tr}[mathbf{A}])^2-4det[mathbf{A}]}$
It seemed to me that from the derivation, we were assuming the trace and determinant were both Real numbers; however, this isn't the general case.
Does this statement hold for $text{Tr}[mathbf{A}],det[mathbf{A}]inmathbb{C}$?
matrices complex-numbers eigenvalues-eigenvectors
matrices complex-numbers eigenvalues-eigenvectors
asked Nov 13 at 20:36
Nathaniel D. Hoffman
356
asked Nov 13 at 20:36
Nathaniel D. Hoffman
356
asked Nov 13 at 20:36
Nathaniel D. Hoffman
356
asked Nov 13 at 20:36
Nathaniel D. Hoffman
356
asked Nov 13 at 20:36
Nathaniel D. Hoffman
356
356
It's not meaningful to say "trace less than zero" if the trace is complex and not real.
– vadim123
Nov 13 at 20:39
@Nathaniel D. Hoffman Eigenvalues here ($2times 2$ matrix) are either real or complex conjugates. So trace is always real for that matrix.
– AnyAD
Nov 13 at 20:56
add a comment |
It's not meaningful to say "trace less than zero" if the trace is complex and not real.
– vadim123
Nov 13 at 20:39
@Nathaniel D. Hoffman Eigenvalues here ($2times 2$ matrix) are either real or complex conjugates. So trace is always real for that matrix.
– AnyAD
Nov 13 at 20:56
It's not meaningful to say "trace less than zero" if the trace is complex and not real.
– vadim123
Nov 13 at 20:39
It's not meaningful to say "trace less than zero" if the trace is complex and not real.
– vadim123
Nov 13 at 20:39
@Nathaniel D. Hoffman Eigenvalues here ($2times 2$ matrix) are either real or complex conjugates. So trace is always real for that matrix.
– AnyAD
Nov 13 at 20:56
@Nathaniel D. Hoffman Eigenvalues here ($2times 2$ matrix) are either real or complex conjugates. So trace is always real for that matrix.
– AnyAD
Nov 13 at 20:56
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You can show the first result by noting that trace is real in the case of a $2times 2$ matrix since the characteristic polynomial has either real roots or 2 complex conjugate roots.
The determinant is the product of the eigenvalues so again will be positive in any case (you can check) if real parts of the eigenvalues are positive.
Your question then maybe should be if the statement holds for a matrix of size $ntimes n $, or for which $n $?
answered Nov 13 at 21:02
AnyAD
1,796811
That makes sense, I wasn't even thinking of it that way, but I guess one thing that still kind of bugs me is, isn't the trace also defined as the sum of the diagonal elements? Couldn't I have one element be like "4" and another be "3+5i" and then have a complex trace? Or is that a less general way to calculate the trace?
– Nathaniel D. Hoffman
Nov 14 at 14:52
@Nathaniel D.Hoffman You said the result is fir $2times 2$ matrices. If a $2times 2$ matruc has a complex eigenvalue $lambda$ then the conjugate of $lambda $ is also an eigenvalue and there are no others. Add these two eigenvalues now and obviously the sum is $2*Re (lambda) $.
– AnyAD
Nov 14 at 21:21
Yeah, that makes sense, thank you!
– Nathaniel D. Hoffman
Nov 16 at 16:52
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You can show the first result by noting that trace is real in the case of a $2times 2$ matrix since the characteristic polynomial has either real roots or 2 complex conjugate roots.
The determinant is the product of the eigenvalues so again will be positive in any case (you can check) if real parts of the eigenvalues are positive.
Your question then maybe should be if the statement holds for a matrix of size $ntimes n $, or for which $n $?
answered Nov 13 at 21:02
AnyAD
1,796811
That makes sense, I wasn't even thinking of it that way, but I guess one thing that still kind of bugs me is, isn't the trace also defined as the sum of the diagonal elements? Couldn't I have one element be like "4" and another be "3+5i" and then have a complex trace? Or is that a less general way to calculate the trace?
– Nathaniel D. Hoffman
Nov 14 at 14:52
@Nathaniel D.Hoffman You said the result is fir $2times 2$ matrices. If a $2times 2$ matruc has a complex eigenvalue $lambda$ then the conjugate of $lambda $ is also an eigenvalue and there are no others. Add these two eigenvalues now and obviously the sum is $2*Re (lambda) $.
– AnyAD
Nov 14 at 21:21
Yeah, that makes sense, thank you!
– Nathaniel D. Hoffman
Nov 16 at 16:52
add a comment |
up vote
1
down vote
accepted
You can show the first result by noting that trace is real in the case of a $2times 2$ matrix since the characteristic polynomial has either real roots or 2 complex conjugate roots.
The determinant is the product of the eigenvalues so again will be positive in any case (you can check) if real parts of the eigenvalues are positive.
Your question then maybe should be if the statement holds for a matrix of size $ntimes n $, or for which $n $?
answered Nov 13 at 21:02
AnyAD
1,796811
That makes sense, I wasn't even thinking of it that way, but I guess one thing that still kind of bugs me is, isn't the trace also defined as the sum of the diagonal elements? Couldn't I have one element be like "4" and another be "3+5i" and then have a complex trace? Or is that a less general way to calculate the trace?
– Nathaniel D. Hoffman
Nov 14 at 14:52
@Nathaniel D.Hoffman You said the result is fir $2times 2$ matrices. If a $2times 2$ matruc has a complex eigenvalue $lambda$ then the conjugate of $lambda $ is also an eigenvalue and there are no others. Add these two eigenvalues now and obviously the sum is $2*Re (lambda) $.
– AnyAD
Nov 14 at 21:21
Yeah, that makes sense, thank you!
– Nathaniel D. Hoffman
Nov 16 at 16:52
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You can show the first result by noting that trace is real in the case of a $2times 2$ matrix since the characteristic polynomial has either real roots or 2 complex conjugate roots.
The determinant is the product of the eigenvalues so again will be positive in any case (you can check) if real parts of the eigenvalues are positive.
Your question then maybe should be if the statement holds for a matrix of size $ntimes n $, or for which $n $?
answered Nov 13 at 21:02
AnyAD
1,796811
You can show the first result by noting that trace is real in the case of a $2times 2$ matrix since the characteristic polynomial has either real roots or 2 complex conjugate roots.
The determinant is the product of the eigenvalues so again will be positive in any case (you can check) if real parts of the eigenvalues are positive.
Your question then maybe should be if the statement holds for a matrix of size $ntimes n $, or for which $n $?
answered Nov 13 at 21:02
AnyAD
1,796811
answered Nov 13 at 21:02
AnyAD
1,796811
answered Nov 13 at 21:02
AnyAD
1,796811
answered Nov 13 at 21:02
AnyAD
1,796811
1,796811
That makes sense, I wasn't even thinking of it that way, but I guess one thing that still kind of bugs me is, isn't the trace also defined as the sum of the diagonal elements? Couldn't I have one element be like "4" and another be "3+5i" and then have a complex trace? Or is that a less general way to calculate the trace?
– Nathaniel D. Hoffman
Nov 14 at 14:52
@Nathaniel D.Hoffman You said the result is fir $2times 2$ matrices. If a $2times 2$ matruc has a complex eigenvalue $lambda$ then the conjugate of $lambda $ is also an eigenvalue and there are no others. Add these two eigenvalues now and obviously the sum is $2*Re (lambda) $.
– AnyAD
Nov 14 at 21:21
Yeah, that makes sense, thank you!
– Nathaniel D. Hoffman
Nov 16 at 16:52
add a comment |
That makes sense, I wasn't even thinking of it that way, but I guess one thing that still kind of bugs me is, isn't the trace also defined as the sum of the diagonal elements? Couldn't I have one element be like "4" and another be "3+5i" and then have a complex trace? Or is that a less general way to calculate the trace?
– Nathaniel D. Hoffman
Nov 14 at 14:52
@Nathaniel D.Hoffman You said the result is fir $2times 2$ matrices. If a $2times 2$ matruc has a complex eigenvalue $lambda$ then the conjugate of $lambda $ is also an eigenvalue and there are no others. Add these two eigenvalues now and obviously the sum is $2*Re (lambda) $.
– AnyAD
Nov 14 at 21:21
Yeah, that makes sense, thank you!
– Nathaniel D. Hoffman
Nov 16 at 16:52
That makes sense, I wasn't even thinking of it that way, but I guess one thing that still kind of bugs me is, isn't the trace also defined as the sum of the diagonal elements? Couldn't I have one element be like "4" and another be "3+5i" and then have a complex trace? Or is that a less general way to calculate the trace?
– Nathaniel D. Hoffman
Nov 14 at 14:52
That makes sense, I wasn't even thinking of it that way, but I guess one thing that still kind of bugs me is, isn't the trace also defined as the sum of the diagonal elements? Couldn't I have one element be like "4" and another be "3+5i" and then have a complex trace? Or is that a less general way to calculate the trace?
– Nathaniel D. Hoffman
Nov 14 at 14:52
@Nathaniel D.Hoffman You said the result is fir $2times 2$ matrices. If a $2times 2$ matruc has a complex eigenvalue $lambda$ then the conjugate of $lambda $ is also an eigenvalue and there are no others. Add these two eigenvalues now and obviously the sum is $2*Re (lambda) $.
– AnyAD
Nov 14 at 21:21
@Nathaniel D.Hoffman You said the result is fir $2times 2$ matrices. If a $2times 2$ matruc has a complex eigenvalue $lambda$ then the conjugate of $lambda $ is also an eigenvalue and there are no others. Add these two eigenvalues now and obviously the sum is $2*Re (lambda) $.
– AnyAD
Nov 14 at 21:21
Yeah, that makes sense, thank you!
– Nathaniel D. Hoffman
Nov 16 at 16:52
Yeah, that makes sense, thank you!
– Nathaniel D. Hoffman
Nov 16 at 16:52
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997304%2fwhat-can-we-say-about-the-eigenvalues-of-a-matrix-with-a-complex-trace-and-deter%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
It's not meaningful to say "trace less than zero" if the trace is complex and not real.
– vadim123
Nov 13 at 20:39
@Nathaniel D. Hoffman Eigenvalues here ($2times 2$ matrix) are either real or complex conjugates. So trace is always real for that matrix.
– AnyAD
Nov 13 at 20:56