Mixing
If (A,B) is controllable, is $(A^2,B)$ controllable as well?
$begingroup$
I'm assuming a 3x3 matrix with controllability matrix as $[B AB A^2B]$. I feel that if A is a nilpotent matrix with n=4, then controllability of $(A^2,B)$ would be $[B A^2B A^4B]=[B A^2B 0]$ which would make the rank<3 and therefore uncontrollable. Am I right? Or am I missing something?
linear-algebra control-theory
asked Jan 31 at 0:52
Mahathi AnandMahathi Anand
31
$endgroup$
add a comment |
$begingroup$
I'm assuming a 3x3 matrix with controllability matrix as $[B AB A^2B]$. I feel that if A is a nilpotent matrix with n=4, then controllability of $(A^2,B)$ would be $[B A^2B A^4B]=[B A^2B 0]$ which would make the rank<3 and therefore uncontrollable. Am I right? Or am I missing something?
linear-algebra control-theory
asked Jan 31 at 0:52
Mahathi AnandMahathi Anand
31
$endgroup$
add a comment |
$begingroup$
I'm assuming a 3x3 matrix with controllability matrix as $[B AB A^2B]$. I feel that if A is a nilpotent matrix with n=4, then controllability of $(A^2,B)$ would be $[B A^2B A^4B]=[B A^2B 0]$ which would make the rank<3 and therefore uncontrollable. Am I right? Or am I missing something?
linear-algebra control-theory
asked Jan 31 at 0:52
Mahathi AnandMahathi Anand
31
$endgroup$
I'm assuming a 3x3 matrix with controllability matrix as $[B AB A^2B]$. I feel that if A is a nilpotent matrix with n=4, then controllability of $(A^2,B)$ would be $[B A^2B A^4B]=[B A^2B 0]$ which would make the rank<3 and therefore uncontrollable. Am I right? Or am I missing something?
linear-algebra control-theory
linear-algebra control-theory
asked Jan 31 at 0:52
Mahathi AnandMahathi Anand
31
asked Jan 31 at 0:52
Mahathi AnandMahathi Anand
31
asked Jan 31 at 0:52
Mahathi AnandMahathi Anand
31
asked Jan 31 at 0:52
Mahathi AnandMahathi Anand
31
asked Jan 31 at 0:52
Mahathi AnandMahathi Anand
31
31
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
An obvious example along the lines indicated in your question is the pair $(A,B)$ with $A=begin{bmatrix}0&1\0&0end{bmatrix}$ and $B=begin{bmatrix}0\1end{bmatrix}$. $(A,B)$ is controllable while $(A^2,B)$ is not.
answered Jan 31 at 7:33
DmitryDmitry
716618
$endgroup$
add a comment |
$begingroup$
It also depends on the rank of $B$. For example in the extreme case that $B=I$ then even $A=0$ would make $(A,B)$ controllable.
answered Jan 31 at 12:01
Kwin van der VeenKwin van der Veen
5,6502828
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3094346%2fif-a-b-is-controllable-is-a2-b-controllable-as-well%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An obvious example along the lines indicated in your question is the pair $(A,B)$ with $A=begin{bmatrix}0&1\0&0end{bmatrix}$ and $B=begin{bmatrix}0\1end{bmatrix}$. $(A,B)$ is controllable while $(A^2,B)$ is not.
answered Jan 31 at 7:33
DmitryDmitry
716618
$endgroup$
add a comment |
$begingroup$
An obvious example along the lines indicated in your question is the pair $(A,B)$ with $A=begin{bmatrix}0&1\0&0end{bmatrix}$ and $B=begin{bmatrix}0\1end{bmatrix}$. $(A,B)$ is controllable while $(A^2,B)$ is not.
answered Jan 31 at 7:33
DmitryDmitry
716618
$endgroup$
add a comment |
$begingroup$
An obvious example along the lines indicated in your question is the pair $(A,B)$ with $A=begin{bmatrix}0&1\0&0end{bmatrix}$ and $B=begin{bmatrix}0\1end{bmatrix}$. $(A,B)$ is controllable while $(A^2,B)$ is not.
answered Jan 31 at 7:33
DmitryDmitry
716618
$endgroup$
An obvious example along the lines indicated in your question is the pair $(A,B)$ with $A=begin{bmatrix}0&1\0&0end{bmatrix}$ and $B=begin{bmatrix}0\1end{bmatrix}$. $(A,B)$ is controllable while $(A^2,B)$ is not.
answered Jan 31 at 7:33
DmitryDmitry
716618
answered Jan 31 at 7:33
DmitryDmitry
716618
answered Jan 31 at 7:33
DmitryDmitry
716618
answered Jan 31 at 7:33
DmitryDmitry
716618
716618
add a comment |
add a comment |
$begingroup$
It also depends on the rank of $B$. For example in the extreme case that $B=I$ then even $A=0$ would make $(A,B)$ controllable.
answered Jan 31 at 12:01
Kwin van der VeenKwin van der Veen
5,6502828
$endgroup$
add a comment |
$begingroup$
It also depends on the rank of $B$. For example in the extreme case that $B=I$ then even $A=0$ would make $(A,B)$ controllable.
answered Jan 31 at 12:01
Kwin van der VeenKwin van der Veen
5,6502828
$endgroup$
add a comment |
$begingroup$
It also depends on the rank of $B$. For example in the extreme case that $B=I$ then even $A=0$ would make $(A,B)$ controllable.
answered Jan 31 at 12:01
Kwin van der VeenKwin van der Veen
5,6502828
$endgroup$
It also depends on the rank of $B$. For example in the extreme case that $B=I$ then even $A=0$ would make $(A,B)$ controllable.
answered Jan 31 at 12:01
Kwin van der VeenKwin van der Veen
5,6502828
answered Jan 31 at 12:01
Kwin van der VeenKwin van der Veen
5,6502828
answered Jan 31 at 12:01
Kwin van der VeenKwin van der Veen
5,6502828
answered Jan 31 at 12:01
Kwin van der VeenKwin van der Veen
5,6502828
5,6502828
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3094346%2fif-a-b-is-controllable-is-a2-b-controllable-as-well%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
