Mixing
quadratic eigenvalue problem reformulation with eigenvalues on unit cycle
$begingroup$
Suppose a quadratic eigenvalue problem is $$(A_2lambda^2+A_1lambda+A_0)x=0$$
where $lambda$ is the eigenvalue and $x$ is the eigenvector. the quadratic form can be rewritten by
$$left[ {begin{array}{*{20}{c}}
0&I\
{{A_0}}&{{A_1}}
end{array}} right]left[ {begin{array}{*{20}{c}}
x\
{lambda x}
end{array}} right] = lambda left[ {begin{array}{*{20}{c}}
I&0\
0&{ - {A_2}}
end{array}} right]left[ {begin{array}{*{20}{c}}
x\
{lambda x}
end{array}} right]$$
Now, suppose $lambda$ is restricted to eigenvalues on unit circles,i.e.,$lambda = frac{{1 + isigma }}{{1 - isigma }}$. I want to further change the equation above as
$$left[ {begin{array}{*{20}{c}}
{ - I}&I\
{{X_1}}&{{X_2}}
end{array}} right]left[ {begin{array}{*{20}{c}}
x\
{lambda x}
end{array}} right] = isigma left[ {begin{array}{*{20}{c}}
I&I\
{{X_3}}&{{X_4}}
end{array}} right]left[ {begin{array}{*{20}{c}}
x\
{lambda x}
end{array}} right]$$
What should $X_{1,2,3,4}$ be?
linear-algebra matrices matrix-calculus
asked Jan 15 at 13:23
user4756766user4756766
357
$endgroup$
add a comment |
$begingroup$
Suppose a quadratic eigenvalue problem is $$(A_2lambda^2+A_1lambda+A_0)x=0$$
where $lambda$ is the eigenvalue and $x$ is the eigenvector. the quadratic form can be rewritten by
$$left[ {begin{array}{*{20}{c}}
0&I\
{{A_0}}&{{A_1}}
end{array}} right]left[ {begin{array}{*{20}{c}}
x\
{lambda x}
end{array}} right] = lambda left[ {begin{array}{*{20}{c}}
I&0\
0&{ - {A_2}}
end{array}} right]left[ {begin{array}{*{20}{c}}
x\
{lambda x}
end{array}} right]$$
Now, suppose $lambda$ is restricted to eigenvalues on unit circles,i.e.,$lambda = frac{{1 + isigma }}{{1 - isigma }}$. I want to further change the equation above as
$$left[ {begin{array}{*{20}{c}}
{ - I}&I\
{{X_1}}&{{X_2}}
end{array}} right]left[ {begin{array}{*{20}{c}}
x\
{lambda x}
end{array}} right] = isigma left[ {begin{array}{*{20}{c}}
I&I\
{{X_3}}&{{X_4}}
end{array}} right]left[ {begin{array}{*{20}{c}}
x\
{lambda x}
end{array}} right]$$
What should $X_{1,2,3,4}$ be?
linear-algebra matrices matrix-calculus
asked Jan 15 at 13:23
user4756766user4756766
357
$endgroup$
$begingroup$
I'm not sure why you call $lambda$ an "eingenvalue" and $x$ and eigenvector
$endgroup$
– leonbloy
Jan 15 at 17:04
$begingroup$
$lambda$ is the nonlinear eigenvalue of the nonlinear eigenproblem, and $x$ is the corresponding nonlinear eigenvector
$endgroup$
– user4756766
Jan 16 at 0:17
$begingroup$
By using the Caley-transformation and using the characteristics: if $Ay = lambda By$, then $left( {A - B} right)y = isigma left( {A + B} right)y$. one has ${X_1} = {A_0}$, ${X_2} = {A_1}+{A_2}$, ${X_3} = {A_0}$, ${X_4} = {A_1}-{A_2}$,
$endgroup$
– user4756766
Jan 18 at 11:22
add a comment |
$begingroup$
Suppose a quadratic eigenvalue problem is $$(A_2lambda^2+A_1lambda+A_0)x=0$$
where $lambda$ is the eigenvalue and $x$ is the eigenvector. the quadratic form can be rewritten by
$$left[ {begin{array}{*{20}{c}}
0&I\
{{A_0}}&{{A_1}}
end{array}} right]left[ {begin{array}{*{20}{c}}
x\
{lambda x}
end{array}} right] = lambda left[ {begin{array}{*{20}{c}}
I&0\
0&{ - {A_2}}
end{array}} right]left[ {begin{array}{*{20}{c}}
x\
{lambda x}
end{array}} right]$$
Now, suppose $lambda$ is restricted to eigenvalues on unit circles,i.e.,$lambda = frac{{1 + isigma }}{{1 - isigma }}$. I want to further change the equation above as
$$left[ {begin{array}{*{20}{c}}
{ - I}&I\
{{X_1}}&{{X_2}}
end{array}} right]left[ {begin{array}{*{20}{c}}
x\
{lambda x}
end{array}} right] = isigma left[ {begin{array}{*{20}{c}}
I&I\
{{X_3}}&{{X_4}}
end{array}} right]left[ {begin{array}{*{20}{c}}
x\
{lambda x}
end{array}} right]$$
What should $X_{1,2,3,4}$ be?
linear-algebra matrices matrix-calculus
asked Jan 15 at 13:23
user4756766user4756766
357
$endgroup$
Suppose a quadratic eigenvalue problem is $$(A_2lambda^2+A_1lambda+A_0)x=0$$
where $lambda$ is the eigenvalue and $x$ is the eigenvector. the quadratic form can be rewritten by
$$left[ {begin{array}{*{20}{c}}
0&I\
{{A_0}}&{{A_1}}
end{array}} right]left[ {begin{array}{*{20}{c}}
x\
{lambda x}
end{array}} right] = lambda left[ {begin{array}{*{20}{c}}
I&0\
0&{ - {A_2}}
end{array}} right]left[ {begin{array}{*{20}{c}}
x\
{lambda x}
end{array}} right]$$
Now, suppose $lambda$ is restricted to eigenvalues on unit circles,i.e.,$lambda = frac{{1 + isigma }}{{1 - isigma }}$. I want to further change the equation above as
$$left[ {begin{array}{*{20}{c}}
{ - I}&I\
{{X_1}}&{{X_2}}
end{array}} right]left[ {begin{array}{*{20}{c}}
x\
{lambda x}
end{array}} right] = isigma left[ {begin{array}{*{20}{c}}
I&I\
{{X_3}}&{{X_4}}
end{array}} right]left[ {begin{array}{*{20}{c}}
x\
{lambda x}
end{array}} right]$$
What should $X_{1,2,3,4}$ be?
linear-algebra matrices matrix-calculus
linear-algebra matrices matrix-calculus
asked Jan 15 at 13:23
user4756766user4756766
357
asked Jan 15 at 13:23
user4756766user4756766
357
asked Jan 15 at 13:23
user4756766user4756766
357
asked Jan 15 at 13:23
user4756766user4756766
357
asked Jan 15 at 13:23
user4756766user4756766
357
357
$begingroup$
I'm not sure why you call $lambda$ an "eingenvalue" and $x$ and eigenvector
$endgroup$
– leonbloy
Jan 15 at 17:04
$begingroup$
$lambda$ is the nonlinear eigenvalue of the nonlinear eigenproblem, and $x$ is the corresponding nonlinear eigenvector
$endgroup$
– user4756766
Jan 16 at 0:17
$begingroup$
By using the Caley-transformation and using the characteristics: if $Ay = lambda By$, then $left( {A - B} right)y = isigma left( {A + B} right)y$. one has ${X_1} = {A_0}$, ${X_2} = {A_1}+{A_2}$, ${X_3} = {A_0}$, ${X_4} = {A_1}-{A_2}$,
$endgroup$
– user4756766
Jan 18 at 11:22
add a comment |
$begingroup$
I'm not sure why you call $lambda$ an "eingenvalue" and $x$ and eigenvector
$endgroup$
– leonbloy
Jan 15 at 17:04
$begingroup$
$lambda$ is the nonlinear eigenvalue of the nonlinear eigenproblem, and $x$ is the corresponding nonlinear eigenvector
$endgroup$
– user4756766
Jan 16 at 0:17
$begingroup$
By using the Caley-transformation and using the characteristics: if $Ay = lambda By$, then $left( {A - B} right)y = isigma left( {A + B} right)y$. one has ${X_1} = {A_0}$, ${X_2} = {A_1}+{A_2}$, ${X_3} = {A_0}$, ${X_4} = {A_1}-{A_2}$,
$endgroup$
– user4756766
Jan 18 at 11:22
$begingroup$
I'm not sure why you call $lambda$ an "eingenvalue" and $x$ and eigenvector
$endgroup$
– leonbloy
Jan 15 at 17:04
$begingroup$
I'm not sure why you call $lambda$ an "eingenvalue" and $x$ and eigenvector
$endgroup$
– leonbloy
Jan 15 at 17:04
$begingroup$
$lambda$ is the nonlinear eigenvalue of the nonlinear eigenproblem, and $x$ is the corresponding nonlinear eigenvector
$endgroup$
– user4756766
Jan 16 at 0:17
$begingroup$
$lambda$ is the nonlinear eigenvalue of the nonlinear eigenproblem, and $x$ is the corresponding nonlinear eigenvector
$endgroup$
– user4756766
Jan 16 at 0:17
$begingroup$
By using the Caley-transformation and using the characteristics: if $Ay = lambda By$, then $left( {A - B} right)y = isigma left( {A + B} right)y$. one has ${X_1} = {A_0}$, ${X_2} = {A_1}+{A_2}$, ${X_3} = {A_0}$, ${X_4} = {A_1}-{A_2}$,
$endgroup$
– user4756766
Jan 18 at 11:22
$begingroup$
By using the Caley-transformation and using the characteristics: if $Ay = lambda By$, then $left( {A - B} right)y = isigma left( {A + B} right)y$. one has ${X_1} = {A_0}$, ${X_2} = {A_1}+{A_2}$, ${X_3} = {A_0}$, ${X_4} = {A_1}-{A_2}$,
$endgroup$
– user4756766
Jan 18 at 11:22
add a comment |
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$begingroup$
I'm not sure why you call $lambda$ an "eingenvalue" and $x$ and eigenvector
$endgroup$
– leonbloy
Jan 15 at 17:04
$begingroup$
$lambda$ is the nonlinear eigenvalue of the nonlinear eigenproblem, and $x$ is the corresponding nonlinear eigenvector
$endgroup$
– user4756766
Jan 16 at 0:17
$begingroup$
By using the Caley-transformation and using the characteristics: if $Ay = lambda By$, then $left( {A - B} right)y = isigma left( {A + B} right)y$. one has ${X_1} = {A_0}$, ${X_2} = {A_1}+{A_2}$, ${X_3} = {A_0}$, ${X_4} = {A_1}-{A_2}$,
$endgroup$
– user4756766
Jan 18 at 11:22