Why the set S(t) is the subspace of the null space of A?
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The picture below is on the page 5 of the paper On linear differential-algebraic equations and linearizations.
What I have learnt from the pic is as follows:
$N(t):=ker A(t)subset mathbb{R}^m$ is that $N(t)$ is the solution space of $A(t)x=0$.
$B(t)z in mathbb{im} : A(t)$ is that $forall z,exists y, B(t)z=A(t)y$
So why we can get $S(t)$ is the subspace of the solution space of $A(t)x'(t)+B(t)x(t)=0, : tin J$
And what book do I need read next?
Thanks
linear-algebra differential-equations
edited Nov 16 at 13:21
Rahul
32.9k466161
asked Nov 16 at 12:52
Hewie Ding
13
add a comment |
up vote
0
down vote
favorite
The picture below is on the page 5 of the paper On linear differential-algebraic equations and linearizations.
What I have learnt from the pic is as follows:
$N(t):=ker A(t)subset mathbb{R}^m$ is that $N(t)$ is the solution space of $A(t)x=0$.
$B(t)z in mathbb{im} : A(t)$ is that $forall z,exists y, B(t)z=A(t)y$
So why we can get $S(t)$ is the subspace of the solution space of $A(t)x'(t)+B(t)x(t)=0, : tin J$
And what book do I need read next?
Thanks
linear-algebra differential-equations
edited Nov 16 at 13:21
Rahul
32.9k466161
asked Nov 16 at 12:52
Hewie Ding
13
What they mean by "$S(t)$ is the subspace where the homogeneous solutions proceed" is that for the homogeneous equation, $x(t)$ has to lie in $S(t)$. This is because otherwise it would not be possible to satisfy the homogeneous equation for any value of $x'(t)$.
– Rahul
Nov 16 at 13:26
@Rahul Sorry, i don't know,That "$S(t)$ is the subspace of the homogeneous equation solutions" is that $forall z in S(t)$ is the solution of $A(t)x'(t)+B(t)x(t)=0$?? And $A(t)z'+B(t)z=0, B(t)z=A(t)y$
– Hewie Ding
Nov 16 at 14:07
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The picture below is on the page 5 of the paper On linear differential-algebraic equations and linearizations.
What I have learnt from the pic is as follows:
$N(t):=ker A(t)subset mathbb{R}^m$ is that $N(t)$ is the solution space of $A(t)x=0$.
$B(t)z in mathbb{im} : A(t)$ is that $forall z,exists y, B(t)z=A(t)y$
So why we can get $S(t)$ is the subspace of the solution space of $A(t)x'(t)+B(t)x(t)=0, : tin J$
And what book do I need read next?
Thanks
linear-algebra differential-equations
edited Nov 16 at 13:21
Rahul
32.9k466161
asked Nov 16 at 12:52
Hewie Ding
13
The picture below is on the page 5 of the paper On linear differential-algebraic equations and linearizations.
What I have learnt from the pic is as follows:
$N(t):=ker A(t)subset mathbb{R}^m$ is that $N(t)$ is the solution space of $A(t)x=0$.
$B(t)z in mathbb{im} : A(t)$ is that $forall z,exists y, B(t)z=A(t)y$
So why we can get $S(t)$ is the subspace of the solution space of $A(t)x'(t)+B(t)x(t)=0, : tin J$
And what book do I need read next?
Thanks
linear-algebra differential-equations
linear-algebra differential-equations
edited Nov 16 at 13:21
Rahul
32.9k466161
asked Nov 16 at 12:52
Hewie Ding
13
edited Nov 16 at 13:21
Rahul
32.9k466161
asked Nov 16 at 12:52
Hewie Ding
13
edited Nov 16 at 13:21
Rahul
32.9k466161
edited Nov 16 at 13:21
Rahul
32.9k466161
edited Nov 16 at 13:21
Rahul
32.9k466161
32.9k466161
asked Nov 16 at 12:52
Hewie Ding
13
asked Nov 16 at 12:52
Hewie Ding
13
asked Nov 16 at 12:52
Hewie Ding
13
13
What they mean by "$S(t)$ is the subspace where the homogeneous solutions proceed" is that for the homogeneous equation, $x(t)$ has to lie in $S(t)$. This is because otherwise it would not be possible to satisfy the homogeneous equation for any value of $x'(t)$.
– Rahul
Nov 16 at 13:26
@Rahul Sorry, i don't know,That "$S(t)$ is the subspace of the homogeneous equation solutions" is that $forall z in S(t)$ is the solution of $A(t)x'(t)+B(t)x(t)=0$?? And $A(t)z'+B(t)z=0, B(t)z=A(t)y$
– Hewie Ding
Nov 16 at 14:07
add a comment |
What they mean by "$S(t)$ is the subspace where the homogeneous solutions proceed" is that for the homogeneous equation, $x(t)$ has to lie in $S(t)$. This is because otherwise it would not be possible to satisfy the homogeneous equation for any value of $x'(t)$.
– Rahul
Nov 16 at 13:26
@Rahul Sorry, i don't know,That "$S(t)$ is the subspace of the homogeneous equation solutions" is that $forall z in S(t)$ is the solution of $A(t)x'(t)+B(t)x(t)=0$?? And $A(t)z'+B(t)z=0, B(t)z=A(t)y$
– Hewie Ding
Nov 16 at 14:07
What they mean by "$S(t)$ is the subspace where the homogeneous solutions proceed" is that for the homogeneous equation, $x(t)$ has to lie in $S(t)$. This is because otherwise it would not be possible to satisfy the homogeneous equation for any value of $x'(t)$.
– Rahul
Nov 16 at 13:26
What they mean by "$S(t)$ is the subspace where the homogeneous solutions proceed" is that for the homogeneous equation, $x(t)$ has to lie in $S(t)$. This is because otherwise it would not be possible to satisfy the homogeneous equation for any value of $x'(t)$.
– Rahul
Nov 16 at 13:26
@Rahul Sorry, i don't know,That "$S(t)$ is the subspace of the homogeneous equation solutions" is that $forall z in S(t)$ is the solution of $A(t)x'(t)+B(t)x(t)=0$?? And $A(t)z'+B(t)z=0, B(t)z=A(t)y$
– Hewie Ding
Nov 16 at 14:07
@Rahul Sorry, i don't know,That "$S(t)$ is the subspace of the homogeneous equation solutions" is that $forall z in S(t)$ is the solution of $A(t)x'(t)+B(t)x(t)=0$?? And $A(t)z'+B(t)z=0, B(t)z=A(t)y$
– Hewie Ding
Nov 16 at 14:07
add a comment |
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What they mean by "$S(t)$ is the subspace where the homogeneous solutions proceed" is that for the homogeneous equation, $x(t)$ has to lie in $S(t)$. This is because otherwise it would not be possible to satisfy the homogeneous equation for any value of $x'(t)$.
– Rahul
Nov 16 at 13:26
@Rahul Sorry, i don't know,That "$S(t)$ is the subspace of the homogeneous equation solutions" is that $forall z in S(t)$ is the solution of $A(t)x'(t)+B(t)x(t)=0$?? And $A(t)z'+B(t)z=0, B(t)z=A(t)y$
– Hewie Ding
Nov 16 at 14:07