Solving a complicated inequality
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I have got the following inequality
$|(a_{12}+a_{21}) x_1 - i (a_{12}-a_{21}) x_2 + (a_{11}-a_{22}) x_3 + (a_{11}+a_{21}) X| le sqrt{a_{11}^2 + |a_{21}|^2} + sqrt{a_{22}^2 + |a_{12}|^2}$,
where $a_{12}=a_{21}^*$ are complex conjugates; and $a_{11}$ and $a_{22}$ are real coefficients. Also, $|vec{x}| + |X|le 1$. Here $vec{x}=(x_1,x_2,x_3)$.
Problem: I need to solve for $x_1,x_2,x_3$ and $X$.
This is my first question on StackExchange. So please don't mind if you find something wrong in the way I am asking.
Edit: I have rewritten my question in a more simplified form which I just got.
linear-algebra inequality absolute-value
asked Nov 20 at 17:30
Zilch
63
add a comment |
up vote
0
down vote
favorite
I have got the following inequality
$|(a_{12}+a_{21}) x_1 - i (a_{12}-a_{21}) x_2 + (a_{11}-a_{22}) x_3 + (a_{11}+a_{21}) X| le sqrt{a_{11}^2 + |a_{21}|^2} + sqrt{a_{22}^2 + |a_{12}|^2}$,
where $a_{12}=a_{21}^*$ are complex conjugates; and $a_{11}$ and $a_{22}$ are real coefficients. Also, $|vec{x}| + |X|le 1$. Here $vec{x}=(x_1,x_2,x_3)$.
Problem: I need to solve for $x_1,x_2,x_3$ and $X$.
This is my first question on StackExchange. So please don't mind if you find something wrong in the way I am asking.
Edit: I have rewritten my question in a more simplified form which I just got.
linear-algebra inequality absolute-value
asked Nov 20 at 17:30
Zilch
63
$vec{x}=X=0$ is always a solution. If $C_4>0$, then any sufficiently small $vec{x}$ and $X$ would constitute a solution.
– user1551
Nov 20 at 17:34
Welcome to MathStackExchange! As @user1551 pointed out, there are trivial solutions to your inequality. Maybe you need to add some kind of restrictions or requirements to the solution space. Note that you can edit your question to clarify it.
– Mefitico
Nov 20 at 20:23
Are $x_1, x_2, x_3, X$ real, or are they allowed to be complex?
– Paul Sinclair
Nov 21 at 1:26
$x_1,x_2,x_3$ and $X$ are real.
– Zilch
Nov 21 at 6:55
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have got the following inequality
$|(a_{12}+a_{21}) x_1 - i (a_{12}-a_{21}) x_2 + (a_{11}-a_{22}) x_3 + (a_{11}+a_{21}) X| le sqrt{a_{11}^2 + |a_{21}|^2} + sqrt{a_{22}^2 + |a_{12}|^2}$,
where $a_{12}=a_{21}^*$ are complex conjugates; and $a_{11}$ and $a_{22}$ are real coefficients. Also, $|vec{x}| + |X|le 1$. Here $vec{x}=(x_1,x_2,x_3)$.
Problem: I need to solve for $x_1,x_2,x_3$ and $X$.
This is my first question on StackExchange. So please don't mind if you find something wrong in the way I am asking.
Edit: I have rewritten my question in a more simplified form which I just got.
linear-algebra inequality absolute-value
asked Nov 20 at 17:30
Zilch
63
I have got the following inequality
$|(a_{12}+a_{21}) x_1 - i (a_{12}-a_{21}) x_2 + (a_{11}-a_{22}) x_3 + (a_{11}+a_{21}) X| le sqrt{a_{11}^2 + |a_{21}|^2} + sqrt{a_{22}^2 + |a_{12}|^2}$,
where $a_{12}=a_{21}^*$ are complex conjugates; and $a_{11}$ and $a_{22}$ are real coefficients. Also, $|vec{x}| + |X|le 1$. Here $vec{x}=(x_1,x_2,x_3)$.
Problem: I need to solve for $x_1,x_2,x_3$ and $X$.
This is my first question on StackExchange. So please don't mind if you find something wrong in the way I am asking.
Edit: I have rewritten my question in a more simplified form which I just got.
linear-algebra inequality absolute-value
linear-algebra inequality absolute-value
asked Nov 20 at 17:30
Zilch
63
asked Nov 20 at 17:30
Zilch
63
edited Nov 21 at 13:07
asked Nov 20 at 17:30
Zilch
63
asked Nov 20 at 17:30
Zilch
63
asked Nov 20 at 17:30
Zilch
63
63
$vec{x}=X=0$ is always a solution. If $C_4>0$, then any sufficiently small $vec{x}$ and $X$ would constitute a solution.
– user1551
Nov 20 at 17:34
Welcome to MathStackExchange! As @user1551 pointed out, there are trivial solutions to your inequality. Maybe you need to add some kind of restrictions or requirements to the solution space. Note that you can edit your question to clarify it.
– Mefitico
Nov 20 at 20:23
Are $x_1, x_2, x_3, X$ real, or are they allowed to be complex?
– Paul Sinclair
Nov 21 at 1:26
$x_1,x_2,x_3$ and $X$ are real.
– Zilch
Nov 21 at 6:55
add a comment |
$vec{x}=X=0$ is always a solution. If $C_4>0$, then any sufficiently small $vec{x}$ and $X$ would constitute a solution.
– user1551
Nov 20 at 17:34
Welcome to MathStackExchange! As @user1551 pointed out, there are trivial solutions to your inequality. Maybe you need to add some kind of restrictions or requirements to the solution space. Note that you can edit your question to clarify it.
– Mefitico
Nov 20 at 20:23
Are $x_1, x_2, x_3, X$ real, or are they allowed to be complex?
– Paul Sinclair
Nov 21 at 1:26
$x_1,x_2,x_3$ and $X$ are real.
– Zilch
Nov 21 at 6:55
$vec{x}=X=0$ is always a solution. If $C_4>0$, then any sufficiently small $vec{x}$ and $X$ would constitute a solution.
– user1551
Nov 20 at 17:34
$vec{x}=X=0$ is always a solution. If $C_4>0$, then any sufficiently small $vec{x}$ and $X$ would constitute a solution.
– user1551
Nov 20 at 17:34
Welcome to MathStackExchange! As @user1551 pointed out, there are trivial solutions to your inequality. Maybe you need to add some kind of restrictions or requirements to the solution space. Note that you can edit your question to clarify it.
– Mefitico
Nov 20 at 20:23
Welcome to MathStackExchange! As @user1551 pointed out, there are trivial solutions to your inequality. Maybe you need to add some kind of restrictions or requirements to the solution space. Note that you can edit your question to clarify it.
– Mefitico
Nov 20 at 20:23
Are $x_1, x_2, x_3, X$ real, or are they allowed to be complex?
– Paul Sinclair
Nov 21 at 1:26
Are $x_1, x_2, x_3, X$ real, or are they allowed to be complex?
– Paul Sinclair
Nov 21 at 1:26
$x_1,x_2,x_3$ and $X$ are real.
– Zilch
Nov 21 at 6:55
$x_1,x_2,x_3$ and $X$ are real.
– Zilch
Nov 21 at 6:55
add a comment |
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$vec{x}=X=0$ is always a solution. If $C_4>0$, then any sufficiently small $vec{x}$ and $X$ would constitute a solution.
– user1551
Nov 20 at 17:34
Welcome to MathStackExchange! As @user1551 pointed out, there are trivial solutions to your inequality. Maybe you need to add some kind of restrictions or requirements to the solution space. Note that you can edit your question to clarify it.
– Mefitico
Nov 20 at 20:23
Are $x_1, x_2, x_3, X$ real, or are they allowed to be complex?
– Paul Sinclair
Nov 21 at 1:26
$x_1,x_2,x_3$ and $X$ are real.
– Zilch
Nov 21 at 6:55