Finding the Jordan form and basis for a matrix
I have this matrix $A=left[ {begin{array}{ccc}
-3&3&-2\
-7&6&-3\
1&-1&2\
end{array} } right]$
I computed the characteristic polynomial $C_A(t)=-(t-2)^2(t-1)$
When I go to try and find the eigenvectors and generalized eigenvectors I compute $A-2I=left[ {begin{array}{ccc}
-5&3&-2\
-7&4&-3\
1&-1&0\
end{array} } right]$
Using row reduction $left[ {begin{array}{ccc}
1&-1&0\
0&1&1\
0&0&0\
end{array} } right]x=0$ gives me eigenvectors having the form $e=rleft[ {begin{array}{c}
1\
1\
-1\
end{array} } right]$
$v=left[ {begin{array}{c}
1\
1\
-1\
end{array} } right]$
Is the only eigenvector up to a constant. So I know there is 1 generalized eigenvector and I simply chose this eigenvector to be my initial vector in the cycle.
At this point I know the Jordan form is $J=left[ {begin{array}{ccc}
2&1&0\
0&2&0\
0&0&1\
end{array} } right]$
Now to find the generalized eigenvector I can take an augmented matrix
$left(begin{array}{ccc|c}
5&3&-2&1\
-7&4&1&1\
1&-1&0&-1
end{array}right)$
I row reduced this to the form
$left(begin{array}{ccc|c}
1 & -1 & 0&-1\
0 & 1 & 1&2\
0&0&0&0
end{array}right)$
So I need a vector that satisfies $x_1-x_2=-1$ and $x_2+x_3=2$ I let $x_3=1$
Which gives me a vector $x=left(begin{array}{c}
0\
1\
1
end{array}right)$ which seems to work, so I have a basis for the generalized eigenspace for $lambda=2$
then I need to find a solution for $(A-I)x=0$ for $lambda=1$.
I row reduce $A-I=left[begin{array}{ccc}
-4&3&-2\
-7&5&1\
1&-1&1
end{array}right]$ to get $left[begin{array}{ccc}
1&0&-1\
0&1&-2\
0&0&0
end{array}right]$ and using $(A-I)x=0$ found that all eigenvectors for $lambda=1$ have the form $v=sleft(begin{array}{c}
1\
2\
1
end{array}right)$
So I can make a matrix $Q=left[begin{array}{ccc}
1&1&0\
2&1&1\
1&-1&1
end{array}right]$ s.t $A=QJQ^{-1}$ where the columns of $Q$ are a jordan basis for $J$
linear-algebra matrices
asked Dec 8 at 17:36
AColoredReptile
1788
|
show 5 more comments
I have this matrix $A=left[ {begin{array}{ccc}
-3&3&-2\
-7&6&-3\
1&-1&2\
end{array} } right]$
I computed the characteristic polynomial $C_A(t)=-(t-2)^2(t-1)$
When I go to try and find the eigenvectors and generalized eigenvectors I compute $A-2I=left[ {begin{array}{ccc}
-5&3&-2\
-7&4&-3\
1&-1&0\
end{array} } right]$
Using row reduction $left[ {begin{array}{ccc}
1&-1&0\
0&1&1\
0&0&0\
end{array} } right]x=0$ gives me eigenvectors having the form $e=rleft[ {begin{array}{c}
1\
1\
-1\
end{array} } right]$
$v=left[ {begin{array}{c}
1\
1\
-1\
end{array} } right]$
Is the only eigenvector up to a constant. So I know there is 1 generalized eigenvector and I simply chose this eigenvector to be my initial vector in the cycle.
At this point I know the Jordan form is $J=left[ {begin{array}{ccc}
2&1&0\
0&2&0\
0&0&1\
end{array} } right]$
Now to find the generalized eigenvector I can take an augmented matrix
$left(begin{array}{ccc|c}
5&3&-2&1\
-7&4&1&1\
1&-1&0&-1
end{array}right)$
I row reduced this to the form
$left(begin{array}{ccc|c}
1 & -1 & 0&-1\
0 & 1 & 1&2\
0&0&0&0
end{array}right)$
So I need a vector that satisfies $x_1-x_2=-1$ and $x_2+x_3=2$ I let $x_3=1$
Which gives me a vector $x=left(begin{array}{c}
0\
1\
1
end{array}right)$ which seems to work, so I have a basis for the generalized eigenspace for $lambda=2$
then I need to find a solution for $(A-I)x=0$ for $lambda=1$.
I row reduce $A-I=left[begin{array}{ccc}
-4&3&-2\
-7&5&1\
1&-1&1
end{array}right]$ to get $left[begin{array}{ccc}
1&0&-1\
0&1&-2\
0&0&0
end{array}right]$ and using $(A-I)x=0$ found that all eigenvectors for $lambda=1$ have the form $v=sleft(begin{array}{c}
1\
2\
1
end{array}right)$
So I can make a matrix $Q=left[begin{array}{ccc}
1&1&0\
2&1&1\
1&-1&1
end{array}right]$ s.t $A=QJQ^{-1}$ where the columns of $Q$ are a jordan basis for $J$
linear-algebra matrices
asked Dec 8 at 17:36
AColoredReptile
1788
1
I stopped reading at the first unclear point: given the characteristic polynomial, how can you be sure that the Jordan normal form is what you have provided? Can't it be the diagonal matrix with $2,2,1$ in the diagonal?
– A. Pongrácz
Dec 8 at 17:38
Yes I fixed this. I dont know that until I know the geometric multiplicity of that eigenvalue is 1.
– AColoredReptile
Dec 8 at 17:42
Which you can figure out by solving the system of linear equations $(A-2I)x=0$. Try that! (Or compute the rank of $A-2I$ at least, that also helps.)
– A. Pongrácz
Dec 8 at 17:48
I found an eigenvector. My problem is finding the generalized eigenvector. I tried to use this eigenvector I found , $v$ and then compute $(A-2I)(x)=v$.
– AColoredReptile
Dec 8 at 17:51
You are not focusing on the main point of my comment: compute the rank, and if it is $2$, you need to find $2$ independent eigenvectors, not just one. In that case, the Jordan normal form is diagonal. (You still seem to work under the assumption that the Jordan form has a $2times 2$ block, which it might not...)
– A. Pongrácz
Dec 8 at 17:54
|
show 5 more comments
I have this matrix $A=left[ {begin{array}{ccc}
-3&3&-2\
-7&6&-3\
1&-1&2\
end{array} } right]$
I computed the characteristic polynomial $C_A(t)=-(t-2)^2(t-1)$
When I go to try and find the eigenvectors and generalized eigenvectors I compute $A-2I=left[ {begin{array}{ccc}
-5&3&-2\
-7&4&-3\
1&-1&0\
end{array} } right]$
Using row reduction $left[ {begin{array}{ccc}
1&-1&0\
0&1&1\
0&0&0\
end{array} } right]x=0$ gives me eigenvectors having the form $e=rleft[ {begin{array}{c}
1\
1\
-1\
end{array} } right]$
$v=left[ {begin{array}{c}
1\
1\
-1\
end{array} } right]$
Is the only eigenvector up to a constant. So I know there is 1 generalized eigenvector and I simply chose this eigenvector to be my initial vector in the cycle.
At this point I know the Jordan form is $J=left[ {begin{array}{ccc}
2&1&0\
0&2&0\
0&0&1\
end{array} } right]$
Now to find the generalized eigenvector I can take an augmented matrix
$left(begin{array}{ccc|c}
5&3&-2&1\
-7&4&1&1\
1&-1&0&-1
end{array}right)$
I row reduced this to the form
$left(begin{array}{ccc|c}
1 & -1 & 0&-1\
0 & 1 & 1&2\
0&0&0&0
end{array}right)$
So I need a vector that satisfies $x_1-x_2=-1$ and $x_2+x_3=2$ I let $x_3=1$
Which gives me a vector $x=left(begin{array}{c}
0\
1\
1
end{array}right)$ which seems to work, so I have a basis for the generalized eigenspace for $lambda=2$
then I need to find a solution for $(A-I)x=0$ for $lambda=1$.
I row reduce $A-I=left[begin{array}{ccc}
-4&3&-2\
-7&5&1\
1&-1&1
end{array}right]$ to get $left[begin{array}{ccc}
1&0&-1\
0&1&-2\
0&0&0
end{array}right]$ and using $(A-I)x=0$ found that all eigenvectors for $lambda=1$ have the form $v=sleft(begin{array}{c}
1\
2\
1
end{array}right)$
So I can make a matrix $Q=left[begin{array}{ccc}
1&1&0\
2&1&1\
1&-1&1
end{array}right]$ s.t $A=QJQ^{-1}$ where the columns of $Q$ are a jordan basis for $J$
linear-algebra matrices
asked Dec 8 at 17:36
AColoredReptile
1788
I have this matrix $A=left[ {begin{array}{ccc}
-3&3&-2\
-7&6&-3\
1&-1&2\
end{array} } right]$
I computed the characteristic polynomial $C_A(t)=-(t-2)^2(t-1)$
When I go to try and find the eigenvectors and generalized eigenvectors I compute $A-2I=left[ {begin{array}{ccc}
-5&3&-2\
-7&4&-3\
1&-1&0\
end{array} } right]$
Using row reduction $left[ {begin{array}{ccc}
1&-1&0\
0&1&1\
0&0&0\
end{array} } right]x=0$ gives me eigenvectors having the form $e=rleft[ {begin{array}{c}
1\
1\
-1\
end{array} } right]$
$v=left[ {begin{array}{c}
1\
1\
-1\
end{array} } right]$
Is the only eigenvector up to a constant. So I know there is 1 generalized eigenvector and I simply chose this eigenvector to be my initial vector in the cycle.
At this point I know the Jordan form is $J=left[ {begin{array}{ccc}
2&1&0\
0&2&0\
0&0&1\
end{array} } right]$
Now to find the generalized eigenvector I can take an augmented matrix
$left(begin{array}{ccc|c}
5&3&-2&1\
-7&4&1&1\
1&-1&0&-1
end{array}right)$
I row reduced this to the form
$left(begin{array}{ccc|c}
1 & -1 & 0&-1\
0 & 1 & 1&2\
0&0&0&0
end{array}right)$
So I need a vector that satisfies $x_1-x_2=-1$ and $x_2+x_3=2$ I let $x_3=1$
Which gives me a vector $x=left(begin{array}{c}
0\
1\
1
end{array}right)$ which seems to work, so I have a basis for the generalized eigenspace for $lambda=2$
then I need to find a solution for $(A-I)x=0$ for $lambda=1$.
I row reduce $A-I=left[begin{array}{ccc}
-4&3&-2\
-7&5&1\
1&-1&1
end{array}right]$ to get $left[begin{array}{ccc}
1&0&-1\
0&1&-2\
0&0&0
end{array}right]$ and using $(A-I)x=0$ found that all eigenvectors for $lambda=1$ have the form $v=sleft(begin{array}{c}
1\
2\
1
end{array}right)$
So I can make a matrix $Q=left[begin{array}{ccc}
1&1&0\
2&1&1\
1&-1&1
end{array}right]$ s.t $A=QJQ^{-1}$ where the columns of $Q$ are a jordan basis for $J$
linear-algebra matrices
linear-algebra matrices
asked Dec 8 at 17:36
AColoredReptile
1788
asked Dec 8 at 17:36
AColoredReptile
1788
edited Dec 8 at 18:30
asked Dec 8 at 17:36
AColoredReptile
1788
asked Dec 8 at 17:36
AColoredReptile
1788
asked Dec 8 at 17:36
AColoredReptile
1788
1788
1
I stopped reading at the first unclear point: given the characteristic polynomial, how can you be sure that the Jordan normal form is what you have provided? Can't it be the diagonal matrix with $2,2,1$ in the diagonal?
– A. Pongrácz
Dec 8 at 17:38
Yes I fixed this. I dont know that until I know the geometric multiplicity of that eigenvalue is 1.
– AColoredReptile
Dec 8 at 17:42
Which you can figure out by solving the system of linear equations $(A-2I)x=0$. Try that! (Or compute the rank of $A-2I$ at least, that also helps.)
– A. Pongrácz
Dec 8 at 17:48
I found an eigenvector. My problem is finding the generalized eigenvector. I tried to use this eigenvector I found , $v$ and then compute $(A-2I)(x)=v$.
– AColoredReptile
Dec 8 at 17:51
You are not focusing on the main point of my comment: compute the rank, and if it is $2$, you need to find $2$ independent eigenvectors, not just one. In that case, the Jordan normal form is diagonal. (You still seem to work under the assumption that the Jordan form has a $2times 2$ block, which it might not...)
– A. Pongrácz
Dec 8 at 17:54
|
show 5 more comments
1
I stopped reading at the first unclear point: given the characteristic polynomial, how can you be sure that the Jordan normal form is what you have provided? Can't it be the diagonal matrix with $2,2,1$ in the diagonal?
– A. Pongrácz
Dec 8 at 17:38
Yes I fixed this. I dont know that until I know the geometric multiplicity of that eigenvalue is 1.
– AColoredReptile
Dec 8 at 17:42
Which you can figure out by solving the system of linear equations $(A-2I)x=0$. Try that! (Or compute the rank of $A-2I$ at least, that also helps.)
– A. Pongrácz
Dec 8 at 17:48
I found an eigenvector. My problem is finding the generalized eigenvector. I tried to use this eigenvector I found , $v$ and then compute $(A-2I)(x)=v$.
– AColoredReptile
Dec 8 at 17:51
You are not focusing on the main point of my comment: compute the rank, and if it is $2$, you need to find $2$ independent eigenvectors, not just one. In that case, the Jordan normal form is diagonal. (You still seem to work under the assumption that the Jordan form has a $2times 2$ block, which it might not...)
– A. Pongrácz
Dec 8 at 17:54
1
1
I stopped reading at the first unclear point: given the characteristic polynomial, how can you be sure that the Jordan normal form is what you have provided? Can't it be the diagonal matrix with $2,2,1$ in the diagonal?
– A. Pongrácz
Dec 8 at 17:38
I stopped reading at the first unclear point: given the characteristic polynomial, how can you be sure that the Jordan normal form is what you have provided? Can't it be the diagonal matrix with $2,2,1$ in the diagonal?
– A. Pongrácz
Dec 8 at 17:38
Yes I fixed this. I dont know that until I know the geometric multiplicity of that eigenvalue is 1.
– AColoredReptile
Dec 8 at 17:42
Yes I fixed this. I dont know that until I know the geometric multiplicity of that eigenvalue is 1.
– AColoredReptile
Dec 8 at 17:42
Which you can figure out by solving the system of linear equations $(A-2I)x=0$. Try that! (Or compute the rank of $A-2I$ at least, that also helps.)
– A. Pongrácz
Dec 8 at 17:48
Which you can figure out by solving the system of linear equations $(A-2I)x=0$. Try that! (Or compute the rank of $A-2I$ at least, that also helps.)
– A. Pongrácz
Dec 8 at 17:48
I found an eigenvector. My problem is finding the generalized eigenvector. I tried to use this eigenvector I found , $v$ and then compute $(A-2I)(x)=v$.
– AColoredReptile
Dec 8 at 17:51
I found an eigenvector. My problem is finding the generalized eigenvector. I tried to use this eigenvector I found , $v$ and then compute $(A-2I)(x)=v$.
– AColoredReptile
Dec 8 at 17:51
You are not focusing on the main point of my comment: compute the rank, and if it is $2$, you need to find $2$ independent eigenvectors, not just one. In that case, the Jordan normal form is diagonal. (You still seem to work under the assumption that the Jordan form has a $2times 2$ block, which it might not...)
– A. Pongrácz
Dec 8 at 17:54
You are not focusing on the main point of my comment: compute the rank, and if it is $2$, you need to find $2$ independent eigenvectors, not just one. In that case, the Jordan normal form is diagonal. (You still seem to work under the assumption that the Jordan form has a $2times 2$ block, which it might not...)
– A. Pongrácz
Dec 8 at 17:54
|
show 5 more comments
1 Answer
1
active
oldest
votes
Because you are looking for a Jordan canonical basis, $Ax = 2x+v$, where $x$ is the generalized eigenvector you're looking for. Equivalently, $(A-2I)x = v$. Applying $A-2I$ one more time to each side will give you the following statement: $(A-2I)^{2}x = 0$.
So, your generalized eigenvector is in the nullspace of $(A-2I)^2$, and it is not in the nullspace of $(A-2I)$. Can you finish from there?
answered Dec 8 at 18:25
Adam Cartisano
1614
I believe I finished.
– AColoredReptile
Dec 8 at 18:31
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%2f3031391%2ffinding-the-jordan-form-and-basis-for-a-matrix%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Because you are looking for a Jordan canonical basis, $Ax = 2x+v$, where $x$ is the generalized eigenvector you're looking for. Equivalently, $(A-2I)x = v$. Applying $A-2I$ one more time to each side will give you the following statement: $(A-2I)^{2}x = 0$.
So, your generalized eigenvector is in the nullspace of $(A-2I)^2$, and it is not in the nullspace of $(A-2I)$. Can you finish from there?
answered Dec 8 at 18:25
Adam Cartisano
1614
I believe I finished.
– AColoredReptile
Dec 8 at 18:31
add a comment |
Because you are looking for a Jordan canonical basis, $Ax = 2x+v$, where $x$ is the generalized eigenvector you're looking for. Equivalently, $(A-2I)x = v$. Applying $A-2I$ one more time to each side will give you the following statement: $(A-2I)^{2}x = 0$.
So, your generalized eigenvector is in the nullspace of $(A-2I)^2$, and it is not in the nullspace of $(A-2I)$. Can you finish from there?
answered Dec 8 at 18:25
Adam Cartisano
1614
I believe I finished.
– AColoredReptile
Dec 8 at 18:31
add a comment |
Because you are looking for a Jordan canonical basis, $Ax = 2x+v$, where $x$ is the generalized eigenvector you're looking for. Equivalently, $(A-2I)x = v$. Applying $A-2I$ one more time to each side will give you the following statement: $(A-2I)^{2}x = 0$.
So, your generalized eigenvector is in the nullspace of $(A-2I)^2$, and it is not in the nullspace of $(A-2I)$. Can you finish from there?
answered Dec 8 at 18:25
Adam Cartisano
1614
Because you are looking for a Jordan canonical basis, $Ax = 2x+v$, where $x$ is the generalized eigenvector you're looking for. Equivalently, $(A-2I)x = v$. Applying $A-2I$ one more time to each side will give you the following statement: $(A-2I)^{2}x = 0$.
So, your generalized eigenvector is in the nullspace of $(A-2I)^2$, and it is not in the nullspace of $(A-2I)$. Can you finish from there?
answered Dec 8 at 18:25
Adam Cartisano
1614
answered Dec 8 at 18:25
Adam Cartisano
1614
answered Dec 8 at 18:25
Adam Cartisano
1614
answered Dec 8 at 18:25
Adam Cartisano
1614
1614
I believe I finished.
– AColoredReptile
Dec 8 at 18:31
add a comment |
I believe I finished.
– AColoredReptile
Dec 8 at 18:31
I believe I finished.
– AColoredReptile
Dec 8 at 18:31
I believe I finished.
– AColoredReptile
Dec 8 at 18:31
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3031391%2ffinding-the-jordan-form-and-basis-for-a-matrix%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
1
I stopped reading at the first unclear point: given the characteristic polynomial, how can you be sure that the Jordan normal form is what you have provided? Can't it be the diagonal matrix with $2,2,1$ in the diagonal?
– A. Pongrácz
Dec 8 at 17:38
Yes I fixed this. I dont know that until I know the geometric multiplicity of that eigenvalue is 1.
– AColoredReptile
Dec 8 at 17:42
Which you can figure out by solving the system of linear equations $(A-2I)x=0$. Try that! (Or compute the rank of $A-2I$ at least, that also helps.)
– A. Pongrácz
Dec 8 at 17:48
I found an eigenvector. My problem is finding the generalized eigenvector. I tried to use this eigenvector I found , $v$ and then compute $(A-2I)(x)=v$.
– AColoredReptile
Dec 8 at 17:51
You are not focusing on the main point of my comment: compute the rank, and if it is $2$, you need to find $2$ independent eigenvectors, not just one. In that case, the Jordan normal form is diagonal. (You still seem to work under the assumption that the Jordan form has a $2times 2$ block, which it might not...)
– A. Pongrácz
Dec 8 at 17:54