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Determining eigenvalues with limited information
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The following question is from a System Theory test with only answers (no solutions). Maybe someone here knows how to tackle it.
Consider the discrete time system $$x(k+1) = Ax(k)$$ with a matrix $$P = beginbmatrix 8 & 2\ 2 & 4 endbmatrix$$ such that $$A^TPA-P=- beginbmatrix 4 & 1 \ 1 & 2 endbmatrix$$ Which of the following statements is correct about the location of the eigenvalues of $A$?
$A$ has only eigenvalues with modulus (absolute value) between $0.8$ and $1$.
$A$ has only eigenvalues with modulus (absolute value) between $0.6$ and $0.8$.
$A$ has only eigenvalues with modulus (absolute value) between $0.4$ and $0.6$.
$A$ has only eigenvalues with modulus (absolute value) smaller than $0.4$.
None of the above.
I thought about moving $P$ to the other side of the equals sign but that doesn't get me any further:
$$A^T beginbmatrix 8 & 2\ 2 & 4 endbmatrixA = beginbmatrix 4 & 1\ 1 & 2 endbmatrix$$
Thanks in advance for the help.
linear-algebra eigenvalues-eigenvectors matrix-equations control-theory linear-control
edited Aug 1 at 9:59
Rodrigo de Azevedo
12.6k41751
asked Jul 30 at 12:53
user463102
1197
add a comment |Â
up vote
1
down vote
favorite
The following question is from a System Theory test with only answers (no solutions). Maybe someone here knows how to tackle it.
Consider the discrete time system $$x(k+1) = Ax(k)$$ with a matrix $$P = beginbmatrix 8 & 2\ 2 & 4 endbmatrix$$ such that $$A^TPA-P=- beginbmatrix 4 & 1 \ 1 & 2 endbmatrix$$ Which of the following statements is correct about the location of the eigenvalues of $A$?
$A$ has only eigenvalues with modulus (absolute value) between $0.8$ and $1$.
$A$ has only eigenvalues with modulus (absolute value) between $0.6$ and $0.8$.
$A$ has only eigenvalues with modulus (absolute value) between $0.4$ and $0.6$.
$A$ has only eigenvalues with modulus (absolute value) smaller than $0.4$.
None of the above.
I thought about moving $P$ to the other side of the equals sign but that doesn't get me any further:
$$A^T beginbmatrix 8 & 2\ 2 & 4 endbmatrixA = beginbmatrix 4 & 1\ 1 & 2 endbmatrix$$
Thanks in advance for the help.
linear-algebra eigenvalues-eigenvectors matrix-equations control-theory linear-control
edited Aug 1 at 9:59
Rodrigo de Azevedo
12.6k41751
asked Jul 30 at 12:53
user463102
1197
We have an equality between two quadratic forms. Apply these forms two an eigenvector of $A$. You will have an equality involving $A$'s corresponding eigenvalue.
– Sasha Kozachinskiy
Jul 30 at 13:33
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The following question is from a System Theory test with only answers (no solutions). Maybe someone here knows how to tackle it.
Consider the discrete time system $$x(k+1) = Ax(k)$$ with a matrix $$P = beginbmatrix 8 & 2\ 2 & 4 endbmatrix$$ such that $$A^TPA-P=- beginbmatrix 4 & 1 \ 1 & 2 endbmatrix$$ Which of the following statements is correct about the location of the eigenvalues of $A$?
$A$ has only eigenvalues with modulus (absolute value) between $0.8$ and $1$.
$A$ has only eigenvalues with modulus (absolute value) between $0.6$ and $0.8$.
$A$ has only eigenvalues with modulus (absolute value) between $0.4$ and $0.6$.
$A$ has only eigenvalues with modulus (absolute value) smaller than $0.4$.
None of the above.
I thought about moving $P$ to the other side of the equals sign but that doesn't get me any further:
$$A^T beginbmatrix 8 & 2\ 2 & 4 endbmatrixA = beginbmatrix 4 & 1\ 1 & 2 endbmatrix$$
Thanks in advance for the help.
linear-algebra eigenvalues-eigenvectors matrix-equations control-theory linear-control
edited Aug 1 at 9:59
Rodrigo de Azevedo
12.6k41751
asked Jul 30 at 12:53
user463102
1197
The following question is from a System Theory test with only answers (no solutions). Maybe someone here knows how to tackle it.
Consider the discrete time system $$x(k+1) = Ax(k)$$ with a matrix $$P = beginbmatrix 8 & 2\ 2 & 4 endbmatrix$$ such that $$A^TPA-P=- beginbmatrix 4 & 1 \ 1 & 2 endbmatrix$$ Which of the following statements is correct about the location of the eigenvalues of $A$?
$A$ has only eigenvalues with modulus (absolute value) between $0.8$ and $1$.
$A$ has only eigenvalues with modulus (absolute value) between $0.6$ and $0.8$.
$A$ has only eigenvalues with modulus (absolute value) between $0.4$ and $0.6$.
$A$ has only eigenvalues with modulus (absolute value) smaller than $0.4$.
None of the above.
I thought about moving $P$ to the other side of the equals sign but that doesn't get me any further:
$$A^T beginbmatrix 8 & 2\ 2 & 4 endbmatrixA = beginbmatrix 4 & 1\ 1 & 2 endbmatrix$$
Thanks in advance for the help.
linear-algebra eigenvalues-eigenvectors matrix-equations control-theory linear-control
edited Aug 1 at 9:59
Rodrigo de Azevedo
12.6k41751
asked Jul 30 at 12:53
user463102
1197
edited Aug 1 at 9:59
Rodrigo de Azevedo
12.6k41751
edited Aug 1 at 9:59
Rodrigo de Azevedo
12.6k41751
edited Aug 1 at 9:59
Rodrigo de Azevedo
12.6k41751
12.6k41751
asked Jul 30 at 12:53
user463102
1197
asked Jul 30 at 12:53
user463102
1197
asked Jul 30 at 12:53
user463102
1197
1197
We have an equality between two quadratic forms. Apply these forms two an eigenvector of $A$. You will have an equality involving $A$'s corresponding eigenvalue.
– Sasha Kozachinskiy
Jul 30 at 13:33
add a comment |Â
We have an equality between two quadratic forms. Apply these forms two an eigenvector of $A$. You will have an equality involving $A$'s corresponding eigenvalue.
– Sasha Kozachinskiy
Jul 30 at 13:33
We have an equality between two quadratic forms. Apply these forms two an eigenvector of $A$. You will have an equality involving $A$'s corresponding eigenvalue.
– Sasha Kozachinskiy
Jul 30 at 13:33
We have an equality between two quadratic forms. Apply these forms two an eigenvector of $A$. You will have an equality involving $A$'s corresponding eigenvalue.
– Sasha Kozachinskiy
Jul 30 at 13:33
add a comment |Â
1 Answer
1
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votes
up vote
3
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accepted
First note that $A^T P A-P=-frac12P$
So, $A^T P A=frac12P$
Let $Av=lambda v$.
Then, we note that $v^T A^T=lambda v^T$.
So in particular,
$v^TA^T P A v=frac12v^TPv$
$lambda^2 v^T P v=frac12 v^T P v$
Note that $v^T P v>0$ for a non-zero vector $v$. ($v^T Pv$ defines a positive definite bilinear form or a inner product)
So $lambda^2=frac12$
answered Jul 30 at 13:50
daruma
736512
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
First note that $A^T P A-P=-frac12P$
So, $A^T P A=frac12P$
Let $Av=lambda v$.
Then, we note that $v^T A^T=lambda v^T$.
So in particular,
$v^TA^T P A v=frac12v^TPv$
$lambda^2 v^T P v=frac12 v^T P v$
Note that $v^T P v>0$ for a non-zero vector $v$. ($v^T Pv$ defines a positive definite bilinear form or a inner product)
So $lambda^2=frac12$
answered Jul 30 at 13:50
daruma
736512
add a comment |Â
up vote
3
down vote
accepted
First note that $A^T P A-P=-frac12P$
So, $A^T P A=frac12P$
Let $Av=lambda v$.
Then, we note that $v^T A^T=lambda v^T$.
So in particular,
$v^TA^T P A v=frac12v^TPv$
$lambda^2 v^T P v=frac12 v^T P v$
Note that $v^T P v>0$ for a non-zero vector $v$. ($v^T Pv$ defines a positive definite bilinear form or a inner product)
So $lambda^2=frac12$
answered Jul 30 at 13:50
daruma
736512
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
First note that $A^T P A-P=-frac12P$
So, $A^T P A=frac12P$
Let $Av=lambda v$.
Then, we note that $v^T A^T=lambda v^T$.
So in particular,
$v^TA^T P A v=frac12v^TPv$
$lambda^2 v^T P v=frac12 v^T P v$
Note that $v^T P v>0$ for a non-zero vector $v$. ($v^T Pv$ defines a positive definite bilinear form or a inner product)
So $lambda^2=frac12$
answered Jul 30 at 13:50
daruma
736512
First note that $A^T P A-P=-frac12P$
So, $A^T P A=frac12P$
Let $Av=lambda v$.
Then, we note that $v^T A^T=lambda v^T$.
So in particular,
$v^TA^T P A v=frac12v^TPv$
$lambda^2 v^T P v=frac12 v^T P v$
Note that $v^T P v>0$ for a non-zero vector $v$. ($v^T Pv$ defines a positive definite bilinear form or a inner product)
So $lambda^2=frac12$
answered Jul 30 at 13:50
daruma
736512
answered Jul 30 at 13:50
daruma
736512
answered Jul 30 at 13:50
daruma
736512
answered Jul 30 at 13:50
daruma
736512
736512
add a comment |Â
add a comment |Â
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We have an equality between two quadratic forms. Apply these forms two an eigenvector of $A$. You will have an equality involving $A$'s corresponding eigenvalue.
– Sasha Kozachinskiy
Jul 30 at 13:33