Mixing
Eigenvalues of a block off-diagonal matrix
$begingroup$
Let $A_1,A_2 in mathbb{R}^{n times n}$. Construct the block matrix $A$ as follows:
$$A: = left[ {begin{array}{*{20}{c}}
0&{{A_1}}\
{{A_2}}&0
end{array}} right]$$
My observation is that matrix $A$ cannot have all its eigenvalues in the open left-half plane. In other worlds, if all eigenvalues of $A$ have non-positive real parts, then all of them are imaginary.
I greatly appreciate it if someone can give me some idea/intuition why this happens.
Thanks
linear-algebra matrices eigenvalues-eigenvectors matrix-calculus
asked Jan 21 at 0:29
ArthurArthur
51412
$endgroup$
add a comment |
$begingroup$
Let $A_1,A_2 in mathbb{R}^{n times n}$. Construct the block matrix $A$ as follows:
$$A: = left[ {begin{array}{*{20}{c}}
0&{{A_1}}\
{{A_2}}&0
end{array}} right]$$
My observation is that matrix $A$ cannot have all its eigenvalues in the open left-half plane. In other worlds, if all eigenvalues of $A$ have non-positive real parts, then all of them are imaginary.
I greatly appreciate it if someone can give me some idea/intuition why this happens.
Thanks
linear-algebra matrices eigenvalues-eigenvectors matrix-calculus
asked Jan 21 at 0:29
ArthurArthur
51412
$endgroup$
add a comment |
$begingroup$
Let $A_1,A_2 in mathbb{R}^{n times n}$. Construct the block matrix $A$ as follows:
$$A: = left[ {begin{array}{*{20}{c}}
0&{{A_1}}\
{{A_2}}&0
end{array}} right]$$
My observation is that matrix $A$ cannot have all its eigenvalues in the open left-half plane. In other worlds, if all eigenvalues of $A$ have non-positive real parts, then all of them are imaginary.
I greatly appreciate it if someone can give me some idea/intuition why this happens.
Thanks
linear-algebra matrices eigenvalues-eigenvectors matrix-calculus
asked Jan 21 at 0:29
ArthurArthur
51412
$endgroup$
Let $A_1,A_2 in mathbb{R}^{n times n}$. Construct the block matrix $A$ as follows:
$$A: = left[ {begin{array}{*{20}{c}}
0&{{A_1}}\
{{A_2}}&0
end{array}} right]$$
My observation is that matrix $A$ cannot have all its eigenvalues in the open left-half plane. In other worlds, if all eigenvalues of $A$ have non-positive real parts, then all of them are imaginary.
I greatly appreciate it if someone can give me some idea/intuition why this happens.
Thanks
linear-algebra matrices eigenvalues-eigenvectors matrix-calculus
linear-algebra matrices eigenvalues-eigenvectors matrix-calculus
asked Jan 21 at 0:29
ArthurArthur
51412
asked Jan 21 at 0:29
ArthurArthur
51412
asked Jan 21 at 0:29
ArthurArthur
51412
asked Jan 21 at 0:29
ArthurArthur
51412
asked Jan 21 at 0:29
ArthurArthur
51412
51412
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The characteristic polynomial of this matrix is of the form $p(lambda) = q(lambda^2)$, where $q$ is the characteristic polynomial of $A_1A_2$. Therefore all roots come in pairs.
answered Jan 21 at 0:42
KlausKlaus
2,12711
$endgroup$
$begingroup$
But, why can't we have complex conjugate roots (which are pairs) with negative real parts?
$endgroup$
– Arthur
Jan 21 at 0:45
$begingroup$
Sorry, I wasn't clear enough there. I meant $lambda$ is a root if and only if $-lambda$ is a root.
$endgroup$
– Klaus
Jan 21 at 0:47
$begingroup$
I see. Thanks Klaus!
$endgroup$
– Arthur
Jan 21 at 0:49
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
The characteristic polynomial of this matrix is of the form $p(lambda) = q(lambda^2)$, where $q$ is the characteristic polynomial of $A_1A_2$. Therefore all roots come in pairs.
answered Jan 21 at 0:42
KlausKlaus
2,12711
$endgroup$
$begingroup$
But, why can't we have complex conjugate roots (which are pairs) with negative real parts?
$endgroup$
– Arthur
Jan 21 at 0:45
$begingroup$
Sorry, I wasn't clear enough there. I meant $lambda$ is a root if and only if $-lambda$ is a root.
$endgroup$
– Klaus
Jan 21 at 0:47
$begingroup$
I see. Thanks Klaus!
$endgroup$
– Arthur
Jan 21 at 0:49
add a comment |
$begingroup$
The characteristic polynomial of this matrix is of the form $p(lambda) = q(lambda^2)$, where $q$ is the characteristic polynomial of $A_1A_2$. Therefore all roots come in pairs.
answered Jan 21 at 0:42
KlausKlaus
2,12711
$endgroup$
$begingroup$
But, why can't we have complex conjugate roots (which are pairs) with negative real parts?
$endgroup$
– Arthur
Jan 21 at 0:45
$begingroup$
Sorry, I wasn't clear enough there. I meant $lambda$ is a root if and only if $-lambda$ is a root.
$endgroup$
– Klaus
Jan 21 at 0:47
$begingroup$
I see. Thanks Klaus!
$endgroup$
– Arthur
Jan 21 at 0:49
add a comment |
$begingroup$
The characteristic polynomial of this matrix is of the form $p(lambda) = q(lambda^2)$, where $q$ is the characteristic polynomial of $A_1A_2$. Therefore all roots come in pairs.
answered Jan 21 at 0:42
KlausKlaus
2,12711
$endgroup$
The characteristic polynomial of this matrix is of the form $p(lambda) = q(lambda^2)$, where $q$ is the characteristic polynomial of $A_1A_2$. Therefore all roots come in pairs.
answered Jan 21 at 0:42
KlausKlaus
2,12711
answered Jan 21 at 0:42
KlausKlaus
2,12711
answered Jan 21 at 0:42
KlausKlaus
2,12711
answered Jan 21 at 0:42
KlausKlaus
2,12711
2,12711
$begingroup$
But, why can't we have complex conjugate roots (which are pairs) with negative real parts?
$endgroup$
– Arthur
Jan 21 at 0:45
$begingroup$
Sorry, I wasn't clear enough there. I meant $lambda$ is a root if and only if $-lambda$ is a root.
$endgroup$
– Klaus
Jan 21 at 0:47
$begingroup$
I see. Thanks Klaus!
$endgroup$
– Arthur
Jan 21 at 0:49
add a comment |
$begingroup$
But, why can't we have complex conjugate roots (which are pairs) with negative real parts?
$endgroup$
– Arthur
Jan 21 at 0:45
$begingroup$
Sorry, I wasn't clear enough there. I meant $lambda$ is a root if and only if $-lambda$ is a root.
$endgroup$
– Klaus
Jan 21 at 0:47
$begingroup$
I see. Thanks Klaus!
$endgroup$
– Arthur
Jan 21 at 0:49
$begingroup$
But, why can't we have complex conjugate roots (which are pairs) with negative real parts?
$endgroup$
– Arthur
Jan 21 at 0:45
$begingroup$
But, why can't we have complex conjugate roots (which are pairs) with negative real parts?
$endgroup$
– Arthur
Jan 21 at 0:45
$begingroup$
Sorry, I wasn't clear enough there. I meant $lambda$ is a root if and only if $-lambda$ is a root.
$endgroup$
– Klaus
Jan 21 at 0:47
$begingroup$
Sorry, I wasn't clear enough there. I meant $lambda$ is a root if and only if $-lambda$ is a root.
$endgroup$
– Klaus
Jan 21 at 0:47
$begingroup$
I see. Thanks Klaus!
$endgroup$
– Arthur
Jan 21 at 0:49
$begingroup$
I see. Thanks Klaus!
$endgroup$
– Arthur
Jan 21 at 0:49
add a comment |
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