Lyapunov stability of 4x4 matrix.












0














Consider the following continuous-time state space representation of the form:



$frac{d}{dx}x(t) = Ax(t)+Bu(t), quad y(t)=Cx(t), quad tin mathbb{R}^{+}$



$A=begin{bmatrix}-1&3&0&0\-3&-1&0&0\0&0&0&3\0&0&-3&0 end{bmatrix} quad B = begin{bmatrix}0\1\0\0 end{bmatrix} quad C=begin{bmatrix}1&0&0&1 end{bmatrix}$



The corresponding eigenvalues are: $-1+3i, -1-3i, 0+3i text{and} 0-3i$.



The answer states that this system is Lyaponov stable.
But I'm wondering why.



Is it because the Jordan blocks of the eigenvalues with zero real-part are $1$x$1$. Because this matrix is in Jordan Form?






control-theory linear-control







share|cite|improve this question




















  • 1




    Your $A$ matrix in not in Jordan form, but in its real Jordan form, so you have to look at the size of each full real Jordan block.
    – Kwin van der Veen
    Nov 21 '18 at 14:28










  • I thought that a 4x4 real jordan form had an identity matrix in the upper right corner? like this:$begin{bmatrix}-1&3&1&0\-3&-1&0&1\0&0&0&3\0&0&-3&0 end{bmatrix}$
    – user463102
    Nov 21 '18 at 17:00






  • 1




    Yes, only your example is not a real Jordan block, because then both 2x2 matrices on the diagonal need to be the same.
    – Kwin van der Veen
    Nov 21 '18 at 18:54










  • Thank you. I did not know that a right upper 2x2 identity only appears if both 2x2 matrices on the diagonal are the same.
    – user463102
    Nov 22 '18 at 7:44


















0














Consider the following continuous-time state space representation of the form:



$frac{d}{dx}x(t) = Ax(t)+Bu(t), quad y(t)=Cx(t), quad tin mathbb{R}^{+}$



$A=begin{bmatrix}-1&3&0&0\-3&-1&0&0\0&0&0&3\0&0&-3&0 end{bmatrix} quad B = begin{bmatrix}0\1\0\0 end{bmatrix} quad C=begin{bmatrix}1&0&0&1 end{bmatrix}$



The corresponding eigenvalues are: $-1+3i, -1-3i, 0+3i text{and} 0-3i$.



The answer states that this system is Lyaponov stable.
But I'm wondering why.



Is it because the Jordan blocks of the eigenvalues with zero real-part are $1$x$1$. Because this matrix is in Jordan Form?






control-theory linear-control







share|cite|improve this question




















  • 1




    Your $A$ matrix in not in Jordan form, but in its real Jordan form, so you have to look at the size of each full real Jordan block.
    – Kwin van der Veen
    Nov 21 '18 at 14:28










  • I thought that a 4x4 real jordan form had an identity matrix in the upper right corner? like this:$begin{bmatrix}-1&3&1&0\-3&-1&0&1\0&0&0&3\0&0&-3&0 end{bmatrix}$
    – user463102
    Nov 21 '18 at 17:00






  • 1




    Yes, only your example is not a real Jordan block, because then both 2x2 matrices on the diagonal need to be the same.
    – Kwin van der Veen
    Nov 21 '18 at 18:54










  • Thank you. I did not know that a right upper 2x2 identity only appears if both 2x2 matrices on the diagonal are the same.
    – user463102
    Nov 22 '18 at 7:44
















0












0








0







Consider the following continuous-time state space representation of the form:



$frac{d}{dx}x(t) = Ax(t)+Bu(t), quad y(t)=Cx(t), quad tin mathbb{R}^{+}$



$A=begin{bmatrix}-1&3&0&0\-3&-1&0&0\0&0&0&3\0&0&-3&0 end{bmatrix} quad B = begin{bmatrix}0\1\0\0 end{bmatrix} quad C=begin{bmatrix}1&0&0&1 end{bmatrix}$



The corresponding eigenvalues are: $-1+3i, -1-3i, 0+3i text{and} 0-3i$.



The answer states that this system is Lyaponov stable.
But I'm wondering why.



Is it because the Jordan blocks of the eigenvalues with zero real-part are $1$x$1$. Because this matrix is in Jordan Form?






control-theory linear-control







share|cite|improve this question















Consider the following continuous-time state space representation of the form:



$frac{d}{dx}x(t) = Ax(t)+Bu(t), quad y(t)=Cx(t), quad tin mathbb{R}^{+}$



$A=begin{bmatrix}-1&3&0&0\-3&-1&0&0\0&0&0&3\0&0&-3&0 end{bmatrix} quad B = begin{bmatrix}0\1\0\0 end{bmatrix} quad C=begin{bmatrix}1&0&0&1 end{bmatrix}$



The corresponding eigenvalues are: $-1+3i, -1-3i, 0+3i text{and} 0-3i$.



The answer states that this system is Lyaponov stable.
But I'm wondering why.



Is it because the Jordan blocks of the eigenvalues with zero real-part are $1$x$1$. Because this matrix is in Jordan Form?






control-theory linear-control




control-theory linear-control






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 '18 at 13:11

























asked Nov 21 '18 at 13:00









user463102

14213




14213








  • 1




    Your $A$ matrix in not in Jordan form, but in its real Jordan form, so you have to look at the size of each full real Jordan block.
    – Kwin van der Veen
    Nov 21 '18 at 14:28










  • I thought that a 4x4 real jordan form had an identity matrix in the upper right corner? like this:$begin{bmatrix}-1&3&1&0\-3&-1&0&1\0&0&0&3\0&0&-3&0 end{bmatrix}$
    – user463102
    Nov 21 '18 at 17:00






  • 1




    Yes, only your example is not a real Jordan block, because then both 2x2 matrices on the diagonal need to be the same.
    – Kwin van der Veen
    Nov 21 '18 at 18:54










  • Thank you. I did not know that a right upper 2x2 identity only appears if both 2x2 matrices on the diagonal are the same.
    – user463102
    Nov 22 '18 at 7:44
















  • 1




    Your $A$ matrix in not in Jordan form, but in its real Jordan form, so you have to look at the size of each full real Jordan block.
    – Kwin van der Veen
    Nov 21 '18 at 14:28










  • I thought that a 4x4 real jordan form had an identity matrix in the upper right corner? like this:$begin{bmatrix}-1&3&1&0\-3&-1&0&1\0&0&0&3\0&0&-3&0 end{bmatrix}$
    – user463102
    Nov 21 '18 at 17:00






  • 1




    Yes, only your example is not a real Jordan block, because then both 2x2 matrices on the diagonal need to be the same.
    – Kwin van der Veen
    Nov 21 '18 at 18:54










  • Thank you. I did not know that a right upper 2x2 identity only appears if both 2x2 matrices on the diagonal are the same.
    – user463102
    Nov 22 '18 at 7:44










1




1




Your $A$ matrix in not in Jordan form, but in its real Jordan form, so you have to look at the size of each full real Jordan block.
– Kwin van der Veen
Nov 21 '18 at 14:28




Your $A$ matrix in not in Jordan form, but in its real Jordan form, so you have to look at the size of each full real Jordan block.
– Kwin van der Veen
Nov 21 '18 at 14:28












I thought that a 4x4 real jordan form had an identity matrix in the upper right corner? like this:$begin{bmatrix}-1&3&1&0\-3&-1&0&1\0&0&0&3\0&0&-3&0 end{bmatrix}$
– user463102
Nov 21 '18 at 17:00




I thought that a 4x4 real jordan form had an identity matrix in the upper right corner? like this:$begin{bmatrix}-1&3&1&0\-3&-1&0&1\0&0&0&3\0&0&-3&0 end{bmatrix}$
– user463102
Nov 21 '18 at 17:00




1




1




Yes, only your example is not a real Jordan block, because then both 2x2 matrices on the diagonal need to be the same.
– Kwin van der Veen
Nov 21 '18 at 18:54




Yes, only your example is not a real Jordan block, because then both 2x2 matrices on the diagonal need to be the same.
– Kwin van der Veen
Nov 21 '18 at 18:54












Thank you. I did not know that a right upper 2x2 identity only appears if both 2x2 matrices on the diagonal are the same.
– user463102
Nov 22 '18 at 7:44






Thank you. I did not know that a right upper 2x2 identity only appears if both 2x2 matrices on the diagonal are the same.
– user463102
Nov 22 '18 at 7:44












1 Answer
1






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oldest

votes


















1














Write $I_n$ for a $n times n$ identity matrix.



Take $P = (1/2)I_4$ and $V(x) = x^T P x$, which is clearly positive definite. Now calculate the directional derivative:



$$
begin{align}
dot{V}(x) &= dot{x}^T P x + x^T P dot{x} \
&= x^T A^T P x + x^T P A x \
&= x^T(A^T P + P A) x \
&= x^T Q x,
end{align}
$$



insert $A$ and $P$ and derive $Q$:



$$
begin{align}
Q = A^T P + P A &=
begin{bmatrix}
-1 & -3 & 0 & 0 \
3 & -1 & 0 & 0 \
0 & 0 & 0 & -3 \
0 & 0 & 3 & 0
end{bmatrix} frac{1}{2} I_4 + frac{1}{2} I_4
begin{bmatrix}
-1 & 3 & 0 & 0 \
-3 & -1 & 0 & 0 \
0 & 0 & 0 & 3 \
0 & 0 & -3 & 0
end{bmatrix} \ &=
begin{bmatrix}
-1 & 0 & 0 & 0 \
0 & -1 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 end{bmatrix} .
end{align}
$$



So your directional derivative is $dot{V}(x) = -x_1^2 - x_2^2$, which is negative semi-definite. Therefore, the system is Lyapunov stable (but not asymptotically Lyapunov stable).






share|cite|improve this answer

















  • 1




    In this case it might be even easier to just look at the eigenvalues, because they all have an algebraic multiplicity of one. Only if the algebraic multiplicity of an eigenvalue with zero real part is higher than one, you also need to check its geometric multiplicity.
    – Kwin van der Veen
    Nov 21 '18 at 14:37











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1














Write $I_n$ for a $n times n$ identity matrix.



Take $P = (1/2)I_4$ and $V(x) = x^T P x$, which is clearly positive definite. Now calculate the directional derivative:



$$
begin{align}
dot{V}(x) &= dot{x}^T P x + x^T P dot{x} \
&= x^T A^T P x + x^T P A x \
&= x^T(A^T P + P A) x \
&= x^T Q x,
end{align}
$$



insert $A$ and $P$ and derive $Q$:



$$
begin{align}
Q = A^T P + P A &=
begin{bmatrix}
-1 & -3 & 0 & 0 \
3 & -1 & 0 & 0 \
0 & 0 & 0 & -3 \
0 & 0 & 3 & 0
end{bmatrix} frac{1}{2} I_4 + frac{1}{2} I_4
begin{bmatrix}
-1 & 3 & 0 & 0 \
-3 & -1 & 0 & 0 \
0 & 0 & 0 & 3 \
0 & 0 & -3 & 0
end{bmatrix} \ &=
begin{bmatrix}
-1 & 0 & 0 & 0 \
0 & -1 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 end{bmatrix} .
end{align}
$$



So your directional derivative is $dot{V}(x) = -x_1^2 - x_2^2$, which is negative semi-definite. Therefore, the system is Lyapunov stable (but not asymptotically Lyapunov stable).






share|cite|improve this answer

















  • 1




    In this case it might be even easier to just look at the eigenvalues, because they all have an algebraic multiplicity of one. Only if the algebraic multiplicity of an eigenvalue with zero real part is higher than one, you also need to check its geometric multiplicity.
    – Kwin van der Veen
    Nov 21 '18 at 14:37
















1














Write $I_n$ for a $n times n$ identity matrix.



Take $P = (1/2)I_4$ and $V(x) = x^T P x$, which is clearly positive definite. Now calculate the directional derivative:



$$
begin{align}
dot{V}(x) &= dot{x}^T P x + x^T P dot{x} \
&= x^T A^T P x + x^T P A x \
&= x^T(A^T P + P A) x \
&= x^T Q x,
end{align}
$$



insert $A$ and $P$ and derive $Q$:



$$
begin{align}
Q = A^T P + P A &=
begin{bmatrix}
-1 & -3 & 0 & 0 \
3 & -1 & 0 & 0 \
0 & 0 & 0 & -3 \
0 & 0 & 3 & 0
end{bmatrix} frac{1}{2} I_4 + frac{1}{2} I_4
begin{bmatrix}
-1 & 3 & 0 & 0 \
-3 & -1 & 0 & 0 \
0 & 0 & 0 & 3 \
0 & 0 & -3 & 0
end{bmatrix} \ &=
begin{bmatrix}
-1 & 0 & 0 & 0 \
0 & -1 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 end{bmatrix} .
end{align}
$$



So your directional derivative is $dot{V}(x) = -x_1^2 - x_2^2$, which is negative semi-definite. Therefore, the system is Lyapunov stable (but not asymptotically Lyapunov stable).






share|cite|improve this answer

















  • 1




    In this case it might be even easier to just look at the eigenvalues, because they all have an algebraic multiplicity of one. Only if the algebraic multiplicity of an eigenvalue with zero real part is higher than one, you also need to check its geometric multiplicity.
    – Kwin van der Veen
    Nov 21 '18 at 14:37














1












1








1






Write $I_n$ for a $n times n$ identity matrix.



Take $P = (1/2)I_4$ and $V(x) = x^T P x$, which is clearly positive definite. Now calculate the directional derivative:



$$
begin{align}
dot{V}(x) &= dot{x}^T P x + x^T P dot{x} \
&= x^T A^T P x + x^T P A x \
&= x^T(A^T P + P A) x \
&= x^T Q x,
end{align}
$$



insert $A$ and $P$ and derive $Q$:



$$
begin{align}
Q = A^T P + P A &=
begin{bmatrix}
-1 & -3 & 0 & 0 \
3 & -1 & 0 & 0 \
0 & 0 & 0 & -3 \
0 & 0 & 3 & 0
end{bmatrix} frac{1}{2} I_4 + frac{1}{2} I_4
begin{bmatrix}
-1 & 3 & 0 & 0 \
-3 & -1 & 0 & 0 \
0 & 0 & 0 & 3 \
0 & 0 & -3 & 0
end{bmatrix} \ &=
begin{bmatrix}
-1 & 0 & 0 & 0 \
0 & -1 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 end{bmatrix} .
end{align}
$$



So your directional derivative is $dot{V}(x) = -x_1^2 - x_2^2$, which is negative semi-definite. Therefore, the system is Lyapunov stable (but not asymptotically Lyapunov stable).






share|cite|improve this answer












Write $I_n$ for a $n times n$ identity matrix.



Take $P = (1/2)I_4$ and $V(x) = x^T P x$, which is clearly positive definite. Now calculate the directional derivative:



$$
begin{align}
dot{V}(x) &= dot{x}^T P x + x^T P dot{x} \
&= x^T A^T P x + x^T P A x \
&= x^T(A^T P + P A) x \
&= x^T Q x,
end{align}
$$



insert $A$ and $P$ and derive $Q$:



$$
begin{align}
Q = A^T P + P A &=
begin{bmatrix}
-1 & -3 & 0 & 0 \
3 & -1 & 0 & 0 \
0 & 0 & 0 & -3 \
0 & 0 & 3 & 0
end{bmatrix} frac{1}{2} I_4 + frac{1}{2} I_4
begin{bmatrix}
-1 & 3 & 0 & 0 \
-3 & -1 & 0 & 0 \
0 & 0 & 0 & 3 \
0 & 0 & -3 & 0
end{bmatrix} \ &=
begin{bmatrix}
-1 & 0 & 0 & 0 \
0 & -1 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 end{bmatrix} .
end{align}
$$



So your directional derivative is $dot{V}(x) = -x_1^2 - x_2^2$, which is negative semi-definite. Therefore, the system is Lyapunov stable (but not asymptotically Lyapunov stable).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 21 '18 at 13:57









SampleTime

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  • 1




    In this case it might be even easier to just look at the eigenvalues, because they all have an algebraic multiplicity of one. Only if the algebraic multiplicity of an eigenvalue with zero real part is higher than one, you also need to check its geometric multiplicity.
    – Kwin van der Veen
    Nov 21 '18 at 14:37














  • 1




    In this case it might be even easier to just look at the eigenvalues, because they all have an algebraic multiplicity of one. Only if the algebraic multiplicity of an eigenvalue with zero real part is higher than one, you also need to check its geometric multiplicity.
    – Kwin van der Veen
    Nov 21 '18 at 14:37








1




1




In this case it might be even easier to just look at the eigenvalues, because they all have an algebraic multiplicity of one. Only if the algebraic multiplicity of an eigenvalue with zero real part is higher than one, you also need to check its geometric multiplicity.
– Kwin van der Veen
Nov 21 '18 at 14:37




In this case it might be even easier to just look at the eigenvalues, because they all have an algebraic multiplicity of one. Only if the algebraic multiplicity of an eigenvalue with zero real part is higher than one, you also need to check its geometric multiplicity.
– Kwin van der Veen
Nov 21 '18 at 14:37


















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