Jordan Form of a 3x3 Matrix with an eigenvalue of multiplicity 3…
$begingroup$
Let
$$A=
begin{bmatrix}
2&2&3\
1&3&3\
-1&-2&-2
end{bmatrix}
.
$$
Find the Jordan Form, $J$, of this matrix, and an invertible matrix $Q$ such that $A = QJQ^{-1}$. I have already found the Jordan Form of this matrix, that is,
$$J = begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}.$$
The part that I am confused about is finding the matrix $Q$. I know that the columns of $Q$ will consist of the eigenvectors, and generalized eigenvectors of $A - lambda I$. The characteristic polynomial of $A$ is
$$p_A(lambda) = lambda^3 - 3lambda^2 + 3lambda + 1 = (lambda - 1)^3.$$
I have found the eigenvector associated with $lambda = 1$ to be
$$v = (-5, 1, 1).$$
However, $(A - I)^2 = 0$, so I am confused on how to find the generalized eigenvectors. Thanks in advance!
linear-algebra jordan-normal-form
asked Jan 7 at 20:18
Taylor McMillanTaylor McMillan
695
$endgroup$
add a comment |
$begingroup$
Let
$$A=
begin{bmatrix}
2&2&3\
1&3&3\
-1&-2&-2
end{bmatrix}
.
$$
Find the Jordan Form, $J$, of this matrix, and an invertible matrix $Q$ such that $A = QJQ^{-1}$. I have already found the Jordan Form of this matrix, that is,
$$J = begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}.$$
The part that I am confused about is finding the matrix $Q$. I know that the columns of $Q$ will consist of the eigenvectors, and generalized eigenvectors of $A - lambda I$. The characteristic polynomial of $A$ is
$$p_A(lambda) = lambda^3 - 3lambda^2 + 3lambda + 1 = (lambda - 1)^3.$$
I have found the eigenvector associated with $lambda = 1$ to be
$$v = (-5, 1, 1).$$
However, $(A - I)^2 = 0$, so I am confused on how to find the generalized eigenvectors. Thanks in advance!
linear-algebra jordan-normal-form
asked Jan 7 at 20:18
Taylor McMillanTaylor McMillan
695
$endgroup$
2
$begingroup$
This has been asked many times at this site, for example here, or here, or here, etc. I think, you could follow the steps and explanations given at this site.
$endgroup$
– Dietrich Burde
Jan 7 at 20:32
add a comment |
$begingroup$
Let
$$A=
begin{bmatrix}
2&2&3\
1&3&3\
-1&-2&-2
end{bmatrix}
.
$$
Find the Jordan Form, $J$, of this matrix, and an invertible matrix $Q$ such that $A = QJQ^{-1}$. I have already found the Jordan Form of this matrix, that is,
$$J = begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}.$$
The part that I am confused about is finding the matrix $Q$. I know that the columns of $Q$ will consist of the eigenvectors, and generalized eigenvectors of $A - lambda I$. The characteristic polynomial of $A$ is
$$p_A(lambda) = lambda^3 - 3lambda^2 + 3lambda + 1 = (lambda - 1)^3.$$
I have found the eigenvector associated with $lambda = 1$ to be
$$v = (-5, 1, 1).$$
However, $(A - I)^2 = 0$, so I am confused on how to find the generalized eigenvectors. Thanks in advance!
linear-algebra jordan-normal-form
asked Jan 7 at 20:18
Taylor McMillanTaylor McMillan
695
$endgroup$
Let
$$A=
begin{bmatrix}
2&2&3\
1&3&3\
-1&-2&-2
end{bmatrix}
.
$$
Find the Jordan Form, $J$, of this matrix, and an invertible matrix $Q$ such that $A = QJQ^{-1}$. I have already found the Jordan Form of this matrix, that is,
$$J = begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}.$$
The part that I am confused about is finding the matrix $Q$. I know that the columns of $Q$ will consist of the eigenvectors, and generalized eigenvectors of $A - lambda I$. The characteristic polynomial of $A$ is
$$p_A(lambda) = lambda^3 - 3lambda^2 + 3lambda + 1 = (lambda - 1)^3.$$
I have found the eigenvector associated with $lambda = 1$ to be
$$v = (-5, 1, 1).$$
However, $(A - I)^2 = 0$, so I am confused on how to find the generalized eigenvectors. Thanks in advance!
linear-algebra jordan-normal-form
linear-algebra jordan-normal-form
asked Jan 7 at 20:18
Taylor McMillanTaylor McMillan
695
asked Jan 7 at 20:18
Taylor McMillanTaylor McMillan
695
asked Jan 7 at 20:18
Taylor McMillanTaylor McMillan
695
asked Jan 7 at 20:18
Taylor McMillanTaylor McMillan
695
asked Jan 7 at 20:18
Taylor McMillanTaylor McMillan
695
695
2
$begingroup$
This has been asked many times at this site, for example here, or here, or here, etc. I think, you could follow the steps and explanations given at this site.
$endgroup$
– Dietrich Burde
Jan 7 at 20:32
add a comment |
2
$begingroup$
This has been asked many times at this site, for example here, or here, or here, etc. I think, you could follow the steps and explanations given at this site.
$endgroup$
– Dietrich Burde
Jan 7 at 20:32
2
2
$begingroup$
This has been asked many times at this site, for example here, or here, or here, etc. I think, you could follow the steps and explanations given at this site.
$endgroup$
– Dietrich Burde
Jan 7 at 20:32
$begingroup$
This has been asked many times at this site, for example here, or here, or here, etc. I think, you could follow the steps and explanations given at this site.
$endgroup$
– Dietrich Burde
Jan 7 at 20:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
You should proceed backwards:
- take any vector $u_3$ in $ker(A-I)^2smallsetminusker(A-I)$, i.e. any vector in $mathbf R^3$ which does not satisfy the equation $;x+2y+3z=0$, e.g. $u_3=(1,0,0)$.
- set $u_2=(A-I)u_3$. This vector is an eigenvector.
- complete $u_2$ with a linearly independent vector $u_1$, so as to obtain a basis of the eigenspace.
answered Jan 7 at 21:13
BernardBernard
121k740116
$endgroup$
add a comment |
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$begingroup$
Hint:
You should proceed backwards:
- take any vector $u_3$ in $ker(A-I)^2smallsetminusker(A-I)$, i.e. any vector in $mathbf R^3$ which does not satisfy the equation $;x+2y+3z=0$, e.g. $u_3=(1,0,0)$.
- set $u_2=(A-I)u_3$. This vector is an eigenvector.
- complete $u_2$ with a linearly independent vector $u_1$, so as to obtain a basis of the eigenspace.
answered Jan 7 at 21:13
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
Hint:
You should proceed backwards:
- take any vector $u_3$ in $ker(A-I)^2smallsetminusker(A-I)$, i.e. any vector in $mathbf R^3$ which does not satisfy the equation $;x+2y+3z=0$, e.g. $u_3=(1,0,0)$.
- set $u_2=(A-I)u_3$. This vector is an eigenvector.
- complete $u_2$ with a linearly independent vector $u_1$, so as to obtain a basis of the eigenspace.
answered Jan 7 at 21:13
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
Hint:
You should proceed backwards:
- take any vector $u_3$ in $ker(A-I)^2smallsetminusker(A-I)$, i.e. any vector in $mathbf R^3$ which does not satisfy the equation $;x+2y+3z=0$, e.g. $u_3=(1,0,0)$.
- set $u_2=(A-I)u_3$. This vector is an eigenvector.
- complete $u_2$ with a linearly independent vector $u_1$, so as to obtain a basis of the eigenspace.
answered Jan 7 at 21:13
BernardBernard
121k740116
$endgroup$
Hint:
You should proceed backwards:
- take any vector $u_3$ in $ker(A-I)^2smallsetminusker(A-I)$, i.e. any vector in $mathbf R^3$ which does not satisfy the equation $;x+2y+3z=0$, e.g. $u_3=(1,0,0)$.
- set $u_2=(A-I)u_3$. This vector is an eigenvector.
- complete $u_2$ with a linearly independent vector $u_1$, so as to obtain a basis of the eigenspace.
answered Jan 7 at 21:13
BernardBernard
121k740116
answered Jan 7 at 21:13
BernardBernard
121k740116
answered Jan 7 at 21:13
BernardBernard
121k740116
answered Jan 7 at 21:13
BernardBernard
121k740116
121k740116
add a comment |
add a comment |
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$begingroup$
This has been asked many times at this site, for example here, or here, or here, etc. I think, you could follow the steps and explanations given at this site.
$endgroup$
– Dietrich Burde
Jan 7 at 20:32