Mixing
Quadratic eigenvalue problem (QEP)
$begingroup$
$Q(lambda)x=0$ and $Q(lambda) = lambda^2 M+lambda C +K$ are defined in this PDF file
The matrices $M$, $C$, $K$ are $ntimes n $ matrices. The thesis said that when $M$, $C$, $K$ are real or hermitian. the eigenvalues are real or come in pairs $(lambda,overline{lambda})$. But the explanation in the thesis is hard to understand. Would you give me a more clear explanation?
matrices eigenvalues-eigenvectors quadratics
edited Jan 21 at 23:07
egreg
183k1486205
asked Jan 21 at 5:50
chen xi chen xi
82
$endgroup$
add a comment |
$begingroup$
$Q(lambda)x=0$ and $Q(lambda) = lambda^2 M+lambda C +K$ are defined in this PDF file
The matrices $M$, $C$, $K$ are $ntimes n $ matrices. The thesis said that when $M$, $C$, $K$ are real or hermitian. the eigenvalues are real or come in pairs $(lambda,overline{lambda})$. But the explanation in the thesis is hard to understand. Would you give me a more clear explanation?
matrices eigenvalues-eigenvectors quadratics
edited Jan 21 at 23:07
egreg
183k1486205
asked Jan 21 at 5:50
chen xi chen xi
82
$endgroup$
add a comment |
$begingroup$
$Q(lambda)x=0$ and $Q(lambda) = lambda^2 M+lambda C +K$ are defined in this PDF file
The matrices $M$, $C$, $K$ are $ntimes n $ matrices. The thesis said that when $M$, $C$, $K$ are real or hermitian. the eigenvalues are real or come in pairs $(lambda,overline{lambda})$. But the explanation in the thesis is hard to understand. Would you give me a more clear explanation?
matrices eigenvalues-eigenvectors quadratics
edited Jan 21 at 23:07
egreg
183k1486205
asked Jan 21 at 5:50
chen xi chen xi
82
$endgroup$
$Q(lambda)x=0$ and $Q(lambda) = lambda^2 M+lambda C +K$ are defined in this PDF file
The matrices $M$, $C$, $K$ are $ntimes n $ matrices. The thesis said that when $M$, $C$, $K$ are real or hermitian. the eigenvalues are real or come in pairs $(lambda,overline{lambda})$. But the explanation in the thesis is hard to understand. Would you give me a more clear explanation?
matrices eigenvalues-eigenvectors quadratics
matrices eigenvalues-eigenvectors quadratics
edited Jan 21 at 23:07
egreg
183k1486205
asked Jan 21 at 5:50
chen xi chen xi
82
edited Jan 21 at 23:07
egreg
183k1486205
asked Jan 21 at 5:50
chen xi chen xi
82
edited Jan 21 at 23:07
egreg
183k1486205
edited Jan 21 at 23:07
egreg
183k1486205
edited Jan 21 at 23:07
egreg
183k1486205
183k1486205
asked Jan 21 at 5:50
chen xi chen xi
82
asked Jan 21 at 5:50
chen xi chen xi
82
asked Jan 21 at 5:50
chen xi chen xi
82
82
add a comment |
add a comment |
1 Answer
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oldest
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If a square matrix has real entries, its characteristic polynomial has real coefficients as well. Therefore its non real roots must come in pairs $lambda$ and $bar{lambda}$, because if $lambdainmathbb{C}$ is a root of a real polynomial $f(x)$, then $f(lambda)=0$ and so
$$
0=bar{0}=overline{f(lambda)}=f(bar{lambda})
$$
and therefore also $bar{lambda}$ is a root of $f$.
For a real symmetric matrix every eigenvalue is real. Suppose $M$ is real symmetric and $lambda$ is an eigenvalue, with eigenvector $vne0$ (possibly with complex entries). Then, denoting by $^H$ the Hermitian transpose,
$$
lambda v^Hv=v^H(lambda v)=v^HMv=v^HM^Hv=(Mv)^Hv=(lambda v)^Hv=bar{lambda}v^Hv
$$
Therefore $(lambda-bar{lambda})(v^Hv)=0$. Since $vne0$, also $v^Hvne0$ and therefore $lambda-bar{lambda}=0$, which means $lambda$ is real.
answered Jan 21 at 23:15
egregegreg
183k1486205
$endgroup$
$begingroup$
What if the matrix is Hermitian, why it is the same with the real matrix?
$endgroup$
– chen xi
Jan 22 at 11:29
$begingroup$
@chenxi The proof for real symmetric goes the same for a Hermitian matrix: I just used that $M=M^H$.
$endgroup$
– egreg
Jan 22 at 11:50
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If a square matrix has real entries, its characteristic polynomial has real coefficients as well. Therefore its non real roots must come in pairs $lambda$ and $bar{lambda}$, because if $lambdainmathbb{C}$ is a root of a real polynomial $f(x)$, then $f(lambda)=0$ and so
$$
0=bar{0}=overline{f(lambda)}=f(bar{lambda})
$$
and therefore also $bar{lambda}$ is a root of $f$.
For a real symmetric matrix every eigenvalue is real. Suppose $M$ is real symmetric and $lambda$ is an eigenvalue, with eigenvector $vne0$ (possibly with complex entries). Then, denoting by $^H$ the Hermitian transpose,
$$
lambda v^Hv=v^H(lambda v)=v^HMv=v^HM^Hv=(Mv)^Hv=(lambda v)^Hv=bar{lambda}v^Hv
$$
Therefore $(lambda-bar{lambda})(v^Hv)=0$. Since $vne0$, also $v^Hvne0$ and therefore $lambda-bar{lambda}=0$, which means $lambda$ is real.
answered Jan 21 at 23:15
egregegreg
183k1486205
$endgroup$
$begingroup$
What if the matrix is Hermitian, why it is the same with the real matrix?
$endgroup$
– chen xi
Jan 22 at 11:29
$begingroup$
@chenxi The proof for real symmetric goes the same for a Hermitian matrix: I just used that $M=M^H$.
$endgroup$
– egreg
Jan 22 at 11:50
add a comment |
$begingroup$
If a square matrix has real entries, its characteristic polynomial has real coefficients as well. Therefore its non real roots must come in pairs $lambda$ and $bar{lambda}$, because if $lambdainmathbb{C}$ is a root of a real polynomial $f(x)$, then $f(lambda)=0$ and so
$$
0=bar{0}=overline{f(lambda)}=f(bar{lambda})
$$
and therefore also $bar{lambda}$ is a root of $f$.
For a real symmetric matrix every eigenvalue is real. Suppose $M$ is real symmetric and $lambda$ is an eigenvalue, with eigenvector $vne0$ (possibly with complex entries). Then, denoting by $^H$ the Hermitian transpose,
$$
lambda v^Hv=v^H(lambda v)=v^HMv=v^HM^Hv=(Mv)^Hv=(lambda v)^Hv=bar{lambda}v^Hv
$$
Therefore $(lambda-bar{lambda})(v^Hv)=0$. Since $vne0$, also $v^Hvne0$ and therefore $lambda-bar{lambda}=0$, which means $lambda$ is real.
answered Jan 21 at 23:15
egregegreg
183k1486205
$endgroup$
$begingroup$
What if the matrix is Hermitian, why it is the same with the real matrix?
$endgroup$
– chen xi
Jan 22 at 11:29
$begingroup$
@chenxi The proof for real symmetric goes the same for a Hermitian matrix: I just used that $M=M^H$.
$endgroup$
– egreg
Jan 22 at 11:50
add a comment |
$begingroup$
If a square matrix has real entries, its characteristic polynomial has real coefficients as well. Therefore its non real roots must come in pairs $lambda$ and $bar{lambda}$, because if $lambdainmathbb{C}$ is a root of a real polynomial $f(x)$, then $f(lambda)=0$ and so
$$
0=bar{0}=overline{f(lambda)}=f(bar{lambda})
$$
and therefore also $bar{lambda}$ is a root of $f$.
For a real symmetric matrix every eigenvalue is real. Suppose $M$ is real symmetric and $lambda$ is an eigenvalue, with eigenvector $vne0$ (possibly with complex entries). Then, denoting by $^H$ the Hermitian transpose,
$$
lambda v^Hv=v^H(lambda v)=v^HMv=v^HM^Hv=(Mv)^Hv=(lambda v)^Hv=bar{lambda}v^Hv
$$
Therefore $(lambda-bar{lambda})(v^Hv)=0$. Since $vne0$, also $v^Hvne0$ and therefore $lambda-bar{lambda}=0$, which means $lambda$ is real.
answered Jan 21 at 23:15
egregegreg
183k1486205
$endgroup$
If a square matrix has real entries, its characteristic polynomial has real coefficients as well. Therefore its non real roots must come in pairs $lambda$ and $bar{lambda}$, because if $lambdainmathbb{C}$ is a root of a real polynomial $f(x)$, then $f(lambda)=0$ and so
$$
0=bar{0}=overline{f(lambda)}=f(bar{lambda})
$$
and therefore also $bar{lambda}$ is a root of $f$.
For a real symmetric matrix every eigenvalue is real. Suppose $M$ is real symmetric and $lambda$ is an eigenvalue, with eigenvector $vne0$ (possibly with complex entries). Then, denoting by $^H$ the Hermitian transpose,
$$
lambda v^Hv=v^H(lambda v)=v^HMv=v^HM^Hv=(Mv)^Hv=(lambda v)^Hv=bar{lambda}v^Hv
$$
Therefore $(lambda-bar{lambda})(v^Hv)=0$. Since $vne0$, also $v^Hvne0$ and therefore $lambda-bar{lambda}=0$, which means $lambda$ is real.
answered Jan 21 at 23:15
egregegreg
183k1486205
answered Jan 21 at 23:15
egregegreg
183k1486205
answered Jan 21 at 23:15
egregegreg
183k1486205
answered Jan 21 at 23:15
egregegreg
183k1486205
183k1486205
$begingroup$
What if the matrix is Hermitian, why it is the same with the real matrix?
$endgroup$
– chen xi
Jan 22 at 11:29
$begingroup$
@chenxi The proof for real symmetric goes the same for a Hermitian matrix: I just used that $M=M^H$.
$endgroup$
– egreg
Jan 22 at 11:50
add a comment |
$begingroup$
What if the matrix is Hermitian, why it is the same with the real matrix?
$endgroup$
– chen xi
Jan 22 at 11:29
$begingroup$
@chenxi The proof for real symmetric goes the same for a Hermitian matrix: I just used that $M=M^H$.
$endgroup$
– egreg
Jan 22 at 11:50
$begingroup$
What if the matrix is Hermitian, why it is the same with the real matrix?
$endgroup$
– chen xi
Jan 22 at 11:29
$begingroup$
What if the matrix is Hermitian, why it is the same with the real matrix?
$endgroup$
– chen xi
Jan 22 at 11:29
$begingroup$
@chenxi The proof for real symmetric goes the same for a Hermitian matrix: I just used that $M=M^H$.
$endgroup$
– egreg
Jan 22 at 11:50
$begingroup$
@chenxi The proof for real symmetric goes the same for a Hermitian matrix: I just used that $M=M^H$.
$endgroup$
– egreg
Jan 22 at 11:50
add a comment |
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