Finding the Jordan form and basis for a matrix
$begingroup$
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 '18 at 17:36
AColoredReptileAColoredReptile
23728
$endgroup$
|
show 5 more comments
$begingroup$
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 '18 at 17:36
AColoredReptileAColoredReptile
23728
$endgroup$
1
$begingroup$
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?
$endgroup$
– A. Pongrácz
Dec 8 '18 at 17:38
$begingroup$
Yes I fixed this. I dont know that until I know the geometric multiplicity of that eigenvalue is 1.
$endgroup$
– AColoredReptile
Dec 8 '18 at 17:42
$begingroup$
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.)
$endgroup$
– A. Pongrácz
Dec 8 '18 at 17:48
$begingroup$
I found an eigenvector. My problem is finding the generalized eigenvector. I tried to use this eigenvector I found , $v$ and then compute $(A-2I)(x)=v$.
$endgroup$
– AColoredReptile
Dec 8 '18 at 17:51
$begingroup$
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...)
$endgroup$
– A. Pongrácz
Dec 8 '18 at 17:54
|
show 5 more comments
$begingroup$
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 '18 at 17:36
AColoredReptileAColoredReptile
23728
$endgroup$
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 '18 at 17:36
AColoredReptileAColoredReptile
23728
asked Dec 8 '18 at 17:36
AColoredReptileAColoredReptile
23728
edited Dec 8 '18 at 18:30
AColoredReptile
asked Dec 8 '18 at 17:36
AColoredReptileAColoredReptile
23728
asked Dec 8 '18 at 17:36
AColoredReptileAColoredReptile
23728
asked Dec 8 '18 at 17:36
AColoredReptileAColoredReptile
23728
23728
1
$begingroup$
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?
$endgroup$
– A. Pongrácz
Dec 8 '18 at 17:38
$begingroup$
Yes I fixed this. I dont know that until I know the geometric multiplicity of that eigenvalue is 1.
$endgroup$
– AColoredReptile
Dec 8 '18 at 17:42
$begingroup$
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.)
$endgroup$
– A. Pongrácz
Dec 8 '18 at 17:48
$begingroup$
I found an eigenvector. My problem is finding the generalized eigenvector. I tried to use this eigenvector I found , $v$ and then compute $(A-2I)(x)=v$.
$endgroup$
– AColoredReptile
Dec 8 '18 at 17:51
$begingroup$
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...)
$endgroup$
– A. Pongrácz
Dec 8 '18 at 17:54
|
show 5 more comments
1
$begingroup$
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?
$endgroup$
– A. Pongrácz
Dec 8 '18 at 17:38
$begingroup$
Yes I fixed this. I dont know that until I know the geometric multiplicity of that eigenvalue is 1.
$endgroup$
– AColoredReptile
Dec 8 '18 at 17:42
$begingroup$
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.)
$endgroup$
– A. Pongrácz
Dec 8 '18 at 17:48
$begingroup$
I found an eigenvector. My problem is finding the generalized eigenvector. I tried to use this eigenvector I found , $v$ and then compute $(A-2I)(x)=v$.
$endgroup$
– AColoredReptile
Dec 8 '18 at 17:51
$begingroup$
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...)
$endgroup$
– A. Pongrácz
Dec 8 '18 at 17:54
1
1
$begingroup$
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?
$endgroup$
– A. Pongrácz
Dec 8 '18 at 17:38
$begingroup$
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?
$endgroup$
– A. Pongrácz
Dec 8 '18 at 17:38
$begingroup$
Yes I fixed this. I dont know that until I know the geometric multiplicity of that eigenvalue is 1.
$endgroup$
– AColoredReptile
Dec 8 '18 at 17:42
$begingroup$
Yes I fixed this. I dont know that until I know the geometric multiplicity of that eigenvalue is 1.
$endgroup$
– AColoredReptile
Dec 8 '18 at 17:42
$begingroup$
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.)
$endgroup$
– A. Pongrácz
Dec 8 '18 at 17:48
$begingroup$
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.)
$endgroup$
– A. Pongrácz
Dec 8 '18 at 17:48
$begingroup$
I found an eigenvector. My problem is finding the generalized eigenvector. I tried to use this eigenvector I found , $v$ and then compute $(A-2I)(x)=v$.
$endgroup$
– AColoredReptile
Dec 8 '18 at 17:51
$begingroup$
I found an eigenvector. My problem is finding the generalized eigenvector. I tried to use this eigenvector I found , $v$ and then compute $(A-2I)(x)=v$.
$endgroup$
– AColoredReptile
Dec 8 '18 at 17:51
$begingroup$
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...)
$endgroup$
– A. Pongrácz
Dec 8 '18 at 17:54
$begingroup$
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...)
$endgroup$
– A. Pongrácz
Dec 8 '18 at 17:54
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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 '18 at 18:25
Adam CartisanoAdam Cartisano
1764
$endgroup$
$begingroup$
I believe I finished.
$endgroup$
– AColoredReptile
Dec 8 '18 at 18:31
add a comment |
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$begingroup$
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 '18 at 18:25
Adam CartisanoAdam Cartisano
1764
$endgroup$
$begingroup$
I believe I finished.
$endgroup$
– AColoredReptile
Dec 8 '18 at 18:31
add a comment |
$begingroup$
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 '18 at 18:25
Adam CartisanoAdam Cartisano
1764
$endgroup$
$begingroup$
I believe I finished.
$endgroup$
– AColoredReptile
Dec 8 '18 at 18:31
add a comment |
$begingroup$
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 '18 at 18:25
Adam CartisanoAdam Cartisano
1764
$endgroup$
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 '18 at 18:25
Adam CartisanoAdam Cartisano
1764
answered Dec 8 '18 at 18:25
Adam CartisanoAdam Cartisano
1764
answered Dec 8 '18 at 18:25
Adam CartisanoAdam Cartisano
1764
answered Dec 8 '18 at 18:25
Adam CartisanoAdam Cartisano
1764
1764
$begingroup$
I believe I finished.
$endgroup$
– AColoredReptile
Dec 8 '18 at 18:31
add a comment |
$begingroup$
I believe I finished.
$endgroup$
– AColoredReptile
Dec 8 '18 at 18:31
$begingroup$
I believe I finished.
$endgroup$
– AColoredReptile
Dec 8 '18 at 18:31
$begingroup$
I believe I finished.
$endgroup$
– AColoredReptile
Dec 8 '18 at 18:31
add a comment |
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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?
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– A. Pongrácz
Dec 8 '18 at 17:38
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Yes I fixed this. I dont know that until I know the geometric multiplicity of that eigenvalue is 1.
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– AColoredReptile
Dec 8 '18 at 17:42
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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.)
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– A. Pongrácz
Dec 8 '18 at 17:48
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I found an eigenvector. My problem is finding the generalized eigenvector. I tried to use this eigenvector I found , $v$ and then compute $(A-2I)(x)=v$.
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– AColoredReptile
Dec 8 '18 at 17:51
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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...)
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– A. Pongrácz
Dec 8 '18 at 17:54