Mixing
Matrix Exponential Jordan Form Linear System
$begingroup$
Given the Linear System $dot{x}(t)=A x(t)$ with $x_0=(x_{01},x_{02})$ as initial state and $A=begin{pmatrix} 0 & 1 \ -k/M & -h/M end{pmatrix}$, when $h^2=4Mk$ the matrix A has a single eigenvalue $s_0=s_{1,2}=-frac{h}{2M}$ with algebraic multiplicity 2. In this case A is not diagonalizable and its Jordan Form is $J=begin{pmatrix} s_0 & 1 \ 0 & s_0 end{pmatrix}$. My objective is to find the exponential matrix so that $x(t)=e^{At}x_0$ using the Jordan normal form. I see that $e^{Jt}=e^{s_0t}begin{pmatrix} 1 & t \ 0 & 1 end{pmatrix}$ So how can I find a matrix $T$ such that $J=TAT^{-1}$? Why the book suggests to use $T=begin{pmatrix} 0 & -1/s_0^2 \ 1 & -1/s_0 end{pmatrix}$?
dynamical-systems jordan-normal-form matrix-exponential
asked Jan 7 at 18:36
TimothyTimothy
83
$endgroup$
add a comment |
$begingroup$
Given the Linear System $dot{x}(t)=A x(t)$ with $x_0=(x_{01},x_{02})$ as initial state and $A=begin{pmatrix} 0 & 1 \ -k/M & -h/M end{pmatrix}$, when $h^2=4Mk$ the matrix A has a single eigenvalue $s_0=s_{1,2}=-frac{h}{2M}$ with algebraic multiplicity 2. In this case A is not diagonalizable and its Jordan Form is $J=begin{pmatrix} s_0 & 1 \ 0 & s_0 end{pmatrix}$. My objective is to find the exponential matrix so that $x(t)=e^{At}x_0$ using the Jordan normal form. I see that $e^{Jt}=e^{s_0t}begin{pmatrix} 1 & t \ 0 & 1 end{pmatrix}$ So how can I find a matrix $T$ such that $J=TAT^{-1}$? Why the book suggests to use $T=begin{pmatrix} 0 & -1/s_0^2 \ 1 & -1/s_0 end{pmatrix}$?
dynamical-systems jordan-normal-form matrix-exponential
asked Jan 7 at 18:36
TimothyTimothy
83
$endgroup$
add a comment |
$begingroup$
Given the Linear System $dot{x}(t)=A x(t)$ with $x_0=(x_{01},x_{02})$ as initial state and $A=begin{pmatrix} 0 & 1 \ -k/M & -h/M end{pmatrix}$, when $h^2=4Mk$ the matrix A has a single eigenvalue $s_0=s_{1,2}=-frac{h}{2M}$ with algebraic multiplicity 2. In this case A is not diagonalizable and its Jordan Form is $J=begin{pmatrix} s_0 & 1 \ 0 & s_0 end{pmatrix}$. My objective is to find the exponential matrix so that $x(t)=e^{At}x_0$ using the Jordan normal form. I see that $e^{Jt}=e^{s_0t}begin{pmatrix} 1 & t \ 0 & 1 end{pmatrix}$ So how can I find a matrix $T$ such that $J=TAT^{-1}$? Why the book suggests to use $T=begin{pmatrix} 0 & -1/s_0^2 \ 1 & -1/s_0 end{pmatrix}$?
dynamical-systems jordan-normal-form matrix-exponential
asked Jan 7 at 18:36
TimothyTimothy
83
$endgroup$
Given the Linear System $dot{x}(t)=A x(t)$ with $x_0=(x_{01},x_{02})$ as initial state and $A=begin{pmatrix} 0 & 1 \ -k/M & -h/M end{pmatrix}$, when $h^2=4Mk$ the matrix A has a single eigenvalue $s_0=s_{1,2}=-frac{h}{2M}$ with algebraic multiplicity 2. In this case A is not diagonalizable and its Jordan Form is $J=begin{pmatrix} s_0 & 1 \ 0 & s_0 end{pmatrix}$. My objective is to find the exponential matrix so that $x(t)=e^{At}x_0$ using the Jordan normal form. I see that $e^{Jt}=e^{s_0t}begin{pmatrix} 1 & t \ 0 & 1 end{pmatrix}$ So how can I find a matrix $T$ such that $J=TAT^{-1}$? Why the book suggests to use $T=begin{pmatrix} 0 & -1/s_0^2 \ 1 & -1/s_0 end{pmatrix}$?
dynamical-systems jordan-normal-form matrix-exponential
dynamical-systems jordan-normal-form matrix-exponential
asked Jan 7 at 18:36
TimothyTimothy
83
asked Jan 7 at 18:36
TimothyTimothy
83
edited Jan 7 at 21:46
Timothy
asked Jan 7 at 18:36
TimothyTimothy
83
asked Jan 7 at 18:36
TimothyTimothy
83
asked Jan 7 at 18:36
TimothyTimothy
83
83
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
using the letter $c,$
$$
left(
begin{array}{rr}
1&0 \
c&1
end{array}
right)
left(
begin{array}{rr}
0&1 \
-c^2&-2c
end{array}
right)
left(
begin{array}{rr}
1&0 \
-c&1
end{array}
right) =
left(
begin{array}{rr}
-c&1 \
0&-c
end{array}
right)
$$
answered Jan 7 at 18:56
Will JagyWill Jagy
103k5101200
$endgroup$
1
$begingroup$
c is for Cookie and it's good enough for me.
$endgroup$
– James S. Cook
Jan 7 at 19:42
$begingroup$
Sorry, I cannot understand, how did you find T?
$endgroup$
– Timothy
Jan 7 at 21:08
$begingroup$
@Timothy let me call my matrix $G,$ I am going to construct $P$ so that $P^{-1} GP = J$ is the Jordan form. Since $(G+cI)^2 = 0,$ I chose whatever I want for the right column $v$ of $P,$ I chose $v = left( begin{array}{rr} 0 \ 1 end{array} right)$ Then the rule just says the left column of $P$ is $u = (G+cI) v$
$endgroup$
– Will Jagy
Jan 7 at 22:38
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
using the letter $c,$
$$
left(
begin{array}{rr}
1&0 \
c&1
end{array}
right)
left(
begin{array}{rr}
0&1 \
-c^2&-2c
end{array}
right)
left(
begin{array}{rr}
1&0 \
-c&1
end{array}
right) =
left(
begin{array}{rr}
-c&1 \
0&-c
end{array}
right)
$$
answered Jan 7 at 18:56
Will JagyWill Jagy
103k5101200
$endgroup$
1
$begingroup$
c is for Cookie and it's good enough for me.
$endgroup$
– James S. Cook
Jan 7 at 19:42
$begingroup$
Sorry, I cannot understand, how did you find T?
$endgroup$
– Timothy
Jan 7 at 21:08
$begingroup$
@Timothy let me call my matrix $G,$ I am going to construct $P$ so that $P^{-1} GP = J$ is the Jordan form. Since $(G+cI)^2 = 0,$ I chose whatever I want for the right column $v$ of $P,$ I chose $v = left( begin{array}{rr} 0 \ 1 end{array} right)$ Then the rule just says the left column of $P$ is $u = (G+cI) v$
$endgroup$
– Will Jagy
Jan 7 at 22:38
add a comment |
$begingroup$
using the letter $c,$
$$
left(
begin{array}{rr}
1&0 \
c&1
end{array}
right)
left(
begin{array}{rr}
0&1 \
-c^2&-2c
end{array}
right)
left(
begin{array}{rr}
1&0 \
-c&1
end{array}
right) =
left(
begin{array}{rr}
-c&1 \
0&-c
end{array}
right)
$$
answered Jan 7 at 18:56
Will JagyWill Jagy
103k5101200
$endgroup$
1
$begingroup$
c is for Cookie and it's good enough for me.
$endgroup$
– James S. Cook
Jan 7 at 19:42
$begingroup$
Sorry, I cannot understand, how did you find T?
$endgroup$
– Timothy
Jan 7 at 21:08
$begingroup$
@Timothy let me call my matrix $G,$ I am going to construct $P$ so that $P^{-1} GP = J$ is the Jordan form. Since $(G+cI)^2 = 0,$ I chose whatever I want for the right column $v$ of $P,$ I chose $v = left( begin{array}{rr} 0 \ 1 end{array} right)$ Then the rule just says the left column of $P$ is $u = (G+cI) v$
$endgroup$
– Will Jagy
Jan 7 at 22:38
add a comment |
$begingroup$
using the letter $c,$
$$
left(
begin{array}{rr}
1&0 \
c&1
end{array}
right)
left(
begin{array}{rr}
0&1 \
-c^2&-2c
end{array}
right)
left(
begin{array}{rr}
1&0 \
-c&1
end{array}
right) =
left(
begin{array}{rr}
-c&1 \
0&-c
end{array}
right)
$$
answered Jan 7 at 18:56
Will JagyWill Jagy
103k5101200
$endgroup$
using the letter $c,$
$$
left(
begin{array}{rr}
1&0 \
c&1
end{array}
right)
left(
begin{array}{rr}
0&1 \
-c^2&-2c
end{array}
right)
left(
begin{array}{rr}
1&0 \
-c&1
end{array}
right) =
left(
begin{array}{rr}
-c&1 \
0&-c
end{array}
right)
$$
answered Jan 7 at 18:56
Will JagyWill Jagy
103k5101200
edited Jan 7 at 19:40
answered Jan 7 at 18:56
Will JagyWill Jagy
103k5101200
answered Jan 7 at 18:56
Will JagyWill Jagy
103k5101200
answered Jan 7 at 18:56
Will JagyWill Jagy
103k5101200
103k5101200
1
$begingroup$
c is for Cookie and it's good enough for me.
$endgroup$
– James S. Cook
Jan 7 at 19:42
$begingroup$
Sorry, I cannot understand, how did you find T?
$endgroup$
– Timothy
Jan 7 at 21:08
$begingroup$
@Timothy let me call my matrix $G,$ I am going to construct $P$ so that $P^{-1} GP = J$ is the Jordan form. Since $(G+cI)^2 = 0,$ I chose whatever I want for the right column $v$ of $P,$ I chose $v = left( begin{array}{rr} 0 \ 1 end{array} right)$ Then the rule just says the left column of $P$ is $u = (G+cI) v$
$endgroup$
– Will Jagy
Jan 7 at 22:38
add a comment |
1
$begingroup$
c is for Cookie and it's good enough for me.
$endgroup$
– James S. Cook
Jan 7 at 19:42
$begingroup$
Sorry, I cannot understand, how did you find T?
$endgroup$
– Timothy
Jan 7 at 21:08
$begingroup$
@Timothy let me call my matrix $G,$ I am going to construct $P$ so that $P^{-1} GP = J$ is the Jordan form. Since $(G+cI)^2 = 0,$ I chose whatever I want for the right column $v$ of $P,$ I chose $v = left( begin{array}{rr} 0 \ 1 end{array} right)$ Then the rule just says the left column of $P$ is $u = (G+cI) v$
$endgroup$
– Will Jagy
Jan 7 at 22:38
1
1
$begingroup$
c is for Cookie and it's good enough for me.
$endgroup$
– James S. Cook
Jan 7 at 19:42
$begingroup$
c is for Cookie and it's good enough for me.
$endgroup$
– James S. Cook
Jan 7 at 19:42
$begingroup$
Sorry, I cannot understand, how did you find T?
$endgroup$
– Timothy
Jan 7 at 21:08
$begingroup$
Sorry, I cannot understand, how did you find T?
$endgroup$
– Timothy
Jan 7 at 21:08
$begingroup$
@Timothy let me call my matrix $G,$ I am going to construct $P$ so that $P^{-1} GP = J$ is the Jordan form. Since $(G+cI)^2 = 0,$ I chose whatever I want for the right column $v$ of $P,$ I chose $v = left( begin{array}{rr} 0 \ 1 end{array} right)$ Then the rule just says the left column of $P$ is $u = (G+cI) v$
$endgroup$
– Will Jagy
Jan 7 at 22:38
$begingroup$
@Timothy let me call my matrix $G,$ I am going to construct $P$ so that $P^{-1} GP = J$ is the Jordan form. Since $(G+cI)^2 = 0,$ I chose whatever I want for the right column $v$ of $P,$ I chose $v = left( begin{array}{rr} 0 \ 1 end{array} right)$ Then the rule just says the left column of $P$ is $u = (G+cI) v$
$endgroup$
– Will Jagy
Jan 7 at 22:38
add a comment |
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