Jordan Normal Form: Two times the same basis vector?!
$begingroup$
I have 3 dimensional matrix $$A =
left(begin{array}{c} 2 & 1 & 0 \ -1 & 0 & 1 \ 1 & 3 & 1end{array}right)$$
and want to find a Jordan Form for it and a basis for the Jordan Form. My procedure: I calculated the characteristic polynomial $chi_A(lambda) = -(2-lambda)^2(1+lambda)$ and found the roots $lambda_1 = 2$ with algebraic multiplicity $mu_1 = 2$ and $lambda_2 = -1$ with algebraic multiplicity $mu_2 = -1$, respectively. Then, for $lambda_1$, I found that a basis for the kernel of $A - 2 I$ is the vector $left(begin{array}{c} 1 \ 0 \ 1end{array}right).$ Clearly, this has subspace has dimension $gamma_{11} = 1$ which is less than $mu_1 = 2$, so I have to continue and calculate the kernel of $(A - 2I)^2$. A basis for this space is given by $left(begin{array}{c} 1 \ 0 \ 1end{array}right), left(begin{array}{c} 1 \ 1 \ 0end{array}right).$ Since now the geometric multiplicity equals the algebraic multiplicity, I am finished with calculating kernels. Now I have to pick some vector $w_{12}$ in the kernel of $(A-2I)^2$ which is not in $(A-2 I)$. An obvious choice is $w_{12} = left(begin{array}{c} 1 \ 1 \ 0end{array}right)$. Then: $$w_{11} = (A - 2I) = left(begin{array}{c}1 \ -3 \ 4end{array}right).$$
Now turning to $lambda_2$, a basis for the kernel is $left(begin{array}{c} 1 \ -3 \ 4end{array}right)$.
But then I get stuck because I have two times the exact same vector in my basis which of course is not enough to span a 3 dimensional space. I cannot see what I did wrong or where my mistake comes from. What do I do in such a situation?
linear-algebra jordan-normal-form
asked Dec 12 '18 at 23:05
Syd AmerikanerSyd Amerikaner
138213
$endgroup$
add a comment |
$begingroup$
I have 3 dimensional matrix $$A =
left(begin{array}{c} 2 & 1 & 0 \ -1 & 0 & 1 \ 1 & 3 & 1end{array}right)$$
and want to find a Jordan Form for it and a basis for the Jordan Form. My procedure: I calculated the characteristic polynomial $chi_A(lambda) = -(2-lambda)^2(1+lambda)$ and found the roots $lambda_1 = 2$ with algebraic multiplicity $mu_1 = 2$ and $lambda_2 = -1$ with algebraic multiplicity $mu_2 = -1$, respectively. Then, for $lambda_1$, I found that a basis for the kernel of $A - 2 I$ is the vector $left(begin{array}{c} 1 \ 0 \ 1end{array}right).$ Clearly, this has subspace has dimension $gamma_{11} = 1$ which is less than $mu_1 = 2$, so I have to continue and calculate the kernel of $(A - 2I)^2$. A basis for this space is given by $left(begin{array}{c} 1 \ 0 \ 1end{array}right), left(begin{array}{c} 1 \ 1 \ 0end{array}right).$ Since now the geometric multiplicity equals the algebraic multiplicity, I am finished with calculating kernels. Now I have to pick some vector $w_{12}$ in the kernel of $(A-2I)^2$ which is not in $(A-2 I)$. An obvious choice is $w_{12} = left(begin{array}{c} 1 \ 1 \ 0end{array}right)$. Then: $$w_{11} = (A - 2I) = left(begin{array}{c}1 \ -3 \ 4end{array}right).$$
Now turning to $lambda_2$, a basis for the kernel is $left(begin{array}{c} 1 \ -3 \ 4end{array}right)$.
But then I get stuck because I have two times the exact same vector in my basis which of course is not enough to span a 3 dimensional space. I cannot see what I did wrong or where my mistake comes from. What do I do in such a situation?
linear-algebra jordan-normal-form
asked Dec 12 '18 at 23:05
Syd AmerikanerSyd Amerikaner
138213
$endgroup$
$begingroup$
You’ve clearly made an error somewhere, since you can’t have both $A(1,0,1)^T = (2,0,2)^T$ and $A(1,0,1)^T = -(1,0,1)^T$, which is what you’re claiming if it’s an eigenvector of both eigenvalues.
$endgroup$
– amd
Dec 13 '18 at 0:49
add a comment |
$begingroup$
I have 3 dimensional matrix $$A =
left(begin{array}{c} 2 & 1 & 0 \ -1 & 0 & 1 \ 1 & 3 & 1end{array}right)$$
and want to find a Jordan Form for it and a basis for the Jordan Form. My procedure: I calculated the characteristic polynomial $chi_A(lambda) = -(2-lambda)^2(1+lambda)$ and found the roots $lambda_1 = 2$ with algebraic multiplicity $mu_1 = 2$ and $lambda_2 = -1$ with algebraic multiplicity $mu_2 = -1$, respectively. Then, for $lambda_1$, I found that a basis for the kernel of $A - 2 I$ is the vector $left(begin{array}{c} 1 \ 0 \ 1end{array}right).$ Clearly, this has subspace has dimension $gamma_{11} = 1$ which is less than $mu_1 = 2$, so I have to continue and calculate the kernel of $(A - 2I)^2$. A basis for this space is given by $left(begin{array}{c} 1 \ 0 \ 1end{array}right), left(begin{array}{c} 1 \ 1 \ 0end{array}right).$ Since now the geometric multiplicity equals the algebraic multiplicity, I am finished with calculating kernels. Now I have to pick some vector $w_{12}$ in the kernel of $(A-2I)^2$ which is not in $(A-2 I)$. An obvious choice is $w_{12} = left(begin{array}{c} 1 \ 1 \ 0end{array}right)$. Then: $$w_{11} = (A - 2I) = left(begin{array}{c}1 \ -3 \ 4end{array}right).$$
Now turning to $lambda_2$, a basis for the kernel is $left(begin{array}{c} 1 \ -3 \ 4end{array}right)$.
But then I get stuck because I have two times the exact same vector in my basis which of course is not enough to span a 3 dimensional space. I cannot see what I did wrong or where my mistake comes from. What do I do in such a situation?
linear-algebra jordan-normal-form
asked Dec 12 '18 at 23:05
Syd AmerikanerSyd Amerikaner
138213
$endgroup$
I have 3 dimensional matrix $$A =
left(begin{array}{c} 2 & 1 & 0 \ -1 & 0 & 1 \ 1 & 3 & 1end{array}right)$$
and want to find a Jordan Form for it and a basis for the Jordan Form. My procedure: I calculated the characteristic polynomial $chi_A(lambda) = -(2-lambda)^2(1+lambda)$ and found the roots $lambda_1 = 2$ with algebraic multiplicity $mu_1 = 2$ and $lambda_2 = -1$ with algebraic multiplicity $mu_2 = -1$, respectively. Then, for $lambda_1$, I found that a basis for the kernel of $A - 2 I$ is the vector $left(begin{array}{c} 1 \ 0 \ 1end{array}right).$ Clearly, this has subspace has dimension $gamma_{11} = 1$ which is less than $mu_1 = 2$, so I have to continue and calculate the kernel of $(A - 2I)^2$. A basis for this space is given by $left(begin{array}{c} 1 \ 0 \ 1end{array}right), left(begin{array}{c} 1 \ 1 \ 0end{array}right).$ Since now the geometric multiplicity equals the algebraic multiplicity, I am finished with calculating kernels. Now I have to pick some vector $w_{12}$ in the kernel of $(A-2I)^2$ which is not in $(A-2 I)$. An obvious choice is $w_{12} = left(begin{array}{c} 1 \ 1 \ 0end{array}right)$. Then: $$w_{11} = (A - 2I) = left(begin{array}{c}1 \ -3 \ 4end{array}right).$$
Now turning to $lambda_2$, a basis for the kernel is $left(begin{array}{c} 1 \ -3 \ 4end{array}right)$.
But then I get stuck because I have two times the exact same vector in my basis which of course is not enough to span a 3 dimensional space. I cannot see what I did wrong or where my mistake comes from. What do I do in such a situation?
linear-algebra jordan-normal-form
linear-algebra jordan-normal-form
asked Dec 12 '18 at 23:05
Syd AmerikanerSyd Amerikaner
138213
asked Dec 12 '18 at 23:05
Syd AmerikanerSyd Amerikaner
138213
asked Dec 12 '18 at 23:05
Syd AmerikanerSyd Amerikaner
138213
asked Dec 12 '18 at 23:05
Syd AmerikanerSyd Amerikaner
138213
asked Dec 12 '18 at 23:05
Syd AmerikanerSyd Amerikaner
138213
138213
$begingroup$
You’ve clearly made an error somewhere, since you can’t have both $A(1,0,1)^T = (2,0,2)^T$ and $A(1,0,1)^T = -(1,0,1)^T$, which is what you’re claiming if it’s an eigenvector of both eigenvalues.
$endgroup$
– amd
Dec 13 '18 at 0:49
add a comment |
$begingroup$
You’ve clearly made an error somewhere, since you can’t have both $A(1,0,1)^T = (2,0,2)^T$ and $A(1,0,1)^T = -(1,0,1)^T$, which is what you’re claiming if it’s an eigenvector of both eigenvalues.
$endgroup$
– amd
Dec 13 '18 at 0:49
$begingroup$
You’ve clearly made an error somewhere, since you can’t have both $A(1,0,1)^T = (2,0,2)^T$ and $A(1,0,1)^T = -(1,0,1)^T$, which is what you’re claiming if it’s an eigenvector of both eigenvalues.
$endgroup$
– amd
Dec 13 '18 at 0:49
$begingroup$
You’ve clearly made an error somewhere, since you can’t have both $A(1,0,1)^T = (2,0,2)^T$ and $A(1,0,1)^T = -(1,0,1)^T$, which is what you’re claiming if it’s an eigenvector of both eigenvalues.
$endgroup$
– amd
Dec 13 '18 at 0:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$ (A-2I)^2 =
left(
begin{array}{ccc}
-1&-2&1 \
3&6&-3 \
-4&-8&4 \
end{array}
right)
$$
of rank one, with row echelon form
$$ (A-2I)^2 Longrightarrow
left(
begin{array}{ccc}
1&2&-1 \
0&0&0 \
0&0&0 \
end{array}
right)
$$
Your vector $w_{12}$ is not in the kernel of $ (A-2I)^2 ; ; ;$ your basis for that kernel is wrong.
answered Dec 12 '18 at 23:22
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
@Bill, it's alright. I did not particularly expect the OP to pay any attention; something like four hours passed. Meanwhile, I had not noticed your hint, as the display of comments stops at about five unless I make a point of requesting to see all the comments
$endgroup$
– Will Jagy
Dec 15 '18 at 18:49
add a comment |
$begingroup$
Where do you get $pmatrix{1\1\0}$?
What do you have for $(A-2I)^2$?
I have $(A-2I)^2 = pmatrix{-1&-2&1\3&6&-3\-4&-8&4}$
And in addition to $pmatrix{1\0\1}$, I see $pmatrix{0\1\2}$ as a candidate for the second eigenvector.
$Apmatrix{0\1\2} = pmatrix{1\2\5} = 2pmatrix{0\1\2}+pmatrix{1\0\1}$ which is exactly what we were hoping for.
$Apmatrix{1&1&0\-3&0&1\4&2&1} = pmatrix{1&1&0\-3&0&1\4&2&1}pmatrix{-1&0&0\0&2&1\0&0&2}$
answered Dec 13 '18 at 0:38
Doug MDoug M
45.4k31954
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$ (A-2I)^2 =
left(
begin{array}{ccc}
-1&-2&1 \
3&6&-3 \
-4&-8&4 \
end{array}
right)
$$
of rank one, with row echelon form
$$ (A-2I)^2 Longrightarrow
left(
begin{array}{ccc}
1&2&-1 \
0&0&0 \
0&0&0 \
end{array}
right)
$$
Your vector $w_{12}$ is not in the kernel of $ (A-2I)^2 ; ; ;$ your basis for that kernel is wrong.
answered Dec 12 '18 at 23:22
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
@Bill, it's alright. I did not particularly expect the OP to pay any attention; something like four hours passed. Meanwhile, I had not noticed your hint, as the display of comments stops at about five unless I make a point of requesting to see all the comments
$endgroup$
– Will Jagy
Dec 15 '18 at 18:49
add a comment |
$begingroup$
$$ (A-2I)^2 =
left(
begin{array}{ccc}
-1&-2&1 \
3&6&-3 \
-4&-8&4 \
end{array}
right)
$$
of rank one, with row echelon form
$$ (A-2I)^2 Longrightarrow
left(
begin{array}{ccc}
1&2&-1 \
0&0&0 \
0&0&0 \
end{array}
right)
$$
Your vector $w_{12}$ is not in the kernel of $ (A-2I)^2 ; ; ;$ your basis for that kernel is wrong.
answered Dec 12 '18 at 23:22
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
@Bill, it's alright. I did not particularly expect the OP to pay any attention; something like four hours passed. Meanwhile, I had not noticed your hint, as the display of comments stops at about five unless I make a point of requesting to see all the comments
$endgroup$
– Will Jagy
Dec 15 '18 at 18:49
add a comment |
$begingroup$
$$ (A-2I)^2 =
left(
begin{array}{ccc}
-1&-2&1 \
3&6&-3 \
-4&-8&4 \
end{array}
right)
$$
of rank one, with row echelon form
$$ (A-2I)^2 Longrightarrow
left(
begin{array}{ccc}
1&2&-1 \
0&0&0 \
0&0&0 \
end{array}
right)
$$
Your vector $w_{12}$ is not in the kernel of $ (A-2I)^2 ; ; ;$ your basis for that kernel is wrong.
answered Dec 12 '18 at 23:22
Will JagyWill Jagy
104k5102201
$endgroup$
$$ (A-2I)^2 =
left(
begin{array}{ccc}
-1&-2&1 \
3&6&-3 \
-4&-8&4 \
end{array}
right)
$$
of rank one, with row echelon form
$$ (A-2I)^2 Longrightarrow
left(
begin{array}{ccc}
1&2&-1 \
0&0&0 \
0&0&0 \
end{array}
right)
$$
Your vector $w_{12}$ is not in the kernel of $ (A-2I)^2 ; ; ;$ your basis for that kernel is wrong.
answered Dec 12 '18 at 23:22
Will JagyWill Jagy
104k5102201
answered Dec 12 '18 at 23:22
Will JagyWill Jagy
104k5102201
answered Dec 12 '18 at 23:22
Will JagyWill Jagy
104k5102201
answered Dec 12 '18 at 23:22
Will JagyWill Jagy
104k5102201
104k5102201
$begingroup$
@Bill, it's alright. I did not particularly expect the OP to pay any attention; something like four hours passed. Meanwhile, I had not noticed your hint, as the display of comments stops at about five unless I make a point of requesting to see all the comments
$endgroup$
– Will Jagy
Dec 15 '18 at 18:49
add a comment |
$begingroup$
@Bill, it's alright. I did not particularly expect the OP to pay any attention; something like four hours passed. Meanwhile, I had not noticed your hint, as the display of comments stops at about five unless I make a point of requesting to see all the comments
$endgroup$
– Will Jagy
Dec 15 '18 at 18:49
$begingroup$
@Bill, it's alright. I did not particularly expect the OP to pay any attention; something like four hours passed. Meanwhile, I had not noticed your hint, as the display of comments stops at about five unless I make a point of requesting to see all the comments
$endgroup$
– Will Jagy
Dec 15 '18 at 18:49
$begingroup$
@Bill, it's alright. I did not particularly expect the OP to pay any attention; something like four hours passed. Meanwhile, I had not noticed your hint, as the display of comments stops at about five unless I make a point of requesting to see all the comments
$endgroup$
– Will Jagy
Dec 15 '18 at 18:49
add a comment |
$begingroup$
Where do you get $pmatrix{1\1\0}$?
What do you have for $(A-2I)^2$?
I have $(A-2I)^2 = pmatrix{-1&-2&1\3&6&-3\-4&-8&4}$
And in addition to $pmatrix{1\0\1}$, I see $pmatrix{0\1\2}$ as a candidate for the second eigenvector.
$Apmatrix{0\1\2} = pmatrix{1\2\5} = 2pmatrix{0\1\2}+pmatrix{1\0\1}$ which is exactly what we were hoping for.
$Apmatrix{1&1&0\-3&0&1\4&2&1} = pmatrix{1&1&0\-3&0&1\4&2&1}pmatrix{-1&0&0\0&2&1\0&0&2}$
answered Dec 13 '18 at 0:38
Doug MDoug M
45.4k31954
$endgroup$
add a comment |
$begingroup$
Where do you get $pmatrix{1\1\0}$?
What do you have for $(A-2I)^2$?
I have $(A-2I)^2 = pmatrix{-1&-2&1\3&6&-3\-4&-8&4}$
And in addition to $pmatrix{1\0\1}$, I see $pmatrix{0\1\2}$ as a candidate for the second eigenvector.
$Apmatrix{0\1\2} = pmatrix{1\2\5} = 2pmatrix{0\1\2}+pmatrix{1\0\1}$ which is exactly what we were hoping for.
$Apmatrix{1&1&0\-3&0&1\4&2&1} = pmatrix{1&1&0\-3&0&1\4&2&1}pmatrix{-1&0&0\0&2&1\0&0&2}$
answered Dec 13 '18 at 0:38
Doug MDoug M
45.4k31954
$endgroup$
add a comment |
$begingroup$
Where do you get $pmatrix{1\1\0}$?
What do you have for $(A-2I)^2$?
I have $(A-2I)^2 = pmatrix{-1&-2&1\3&6&-3\-4&-8&4}$
And in addition to $pmatrix{1\0\1}$, I see $pmatrix{0\1\2}$ as a candidate for the second eigenvector.
$Apmatrix{0\1\2} = pmatrix{1\2\5} = 2pmatrix{0\1\2}+pmatrix{1\0\1}$ which is exactly what we were hoping for.
$Apmatrix{1&1&0\-3&0&1\4&2&1} = pmatrix{1&1&0\-3&0&1\4&2&1}pmatrix{-1&0&0\0&2&1\0&0&2}$
answered Dec 13 '18 at 0:38
Doug MDoug M
45.4k31954
$endgroup$
Where do you get $pmatrix{1\1\0}$?
What do you have for $(A-2I)^2$?
I have $(A-2I)^2 = pmatrix{-1&-2&1\3&6&-3\-4&-8&4}$
And in addition to $pmatrix{1\0\1}$, I see $pmatrix{0\1\2}$ as a candidate for the second eigenvector.
$Apmatrix{0\1\2} = pmatrix{1\2\5} = 2pmatrix{0\1\2}+pmatrix{1\0\1}$ which is exactly what we were hoping for.
$Apmatrix{1&1&0\-3&0&1\4&2&1} = pmatrix{1&1&0\-3&0&1\4&2&1}pmatrix{-1&0&0\0&2&1\0&0&2}$
answered Dec 13 '18 at 0:38
Doug MDoug M
45.4k31954
answered Dec 13 '18 at 0:38
Doug MDoug M
45.4k31954
answered Dec 13 '18 at 0:38
Doug MDoug M
45.4k31954
answered Dec 13 '18 at 0:38
Doug MDoug M
45.4k31954
45.4k31954
add a comment |
add a comment |
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$begingroup$
You’ve clearly made an error somewhere, since you can’t have both $A(1,0,1)^T = (2,0,2)^T$ and $A(1,0,1)^T = -(1,0,1)^T$, which is what you’re claiming if it’s an eigenvector of both eigenvalues.
$endgroup$
– amd
Dec 13 '18 at 0:49