Mixing
Find a Jordan's matrix and basis
$begingroup$
I have a matrix: $A={begin{bmatrix}3&-1&0&1\0&3&4&4\0&0&-5&-8\0&0&4&7end{bmatrix}}$.
1) I calculate characteristic polynomial. It is: $p_{A}(lambda)=(lambda-3)^{3}(lambda+1)$
2) So exist Jordan's matrix: $J={begin{bmatrix}3&?&0&0\0&3&?&0\0&0&3&0\0&0&0&-1end{bmatrix}}$
3)I find own subspace:
$V_{3}=ker(A-3I)=dots= operatorname{span}({begin{bmatrix}1&0&0&0end{bmatrix}}^{T},{begin{bmatrix}0&1&-1&1end{bmatrix}}^{T}$
$V_{-1}=ker(A+I)=dots= operatorname{span}({begin{bmatrix}0&1&-2&1end{bmatrix}}^{T}$
4) Jordan's matrix is: $J={begin{bmatrix}3&0&0&0\0&3&1&0\0&0&3&0\0&0&0&-1end{bmatrix}}$
However I have a problem with a basis:
I read that in basis are vectors from span $V_{3}$ and $V_{-1}$ and the next I must find one more vector. So I can use my vectors from span and I have for example:${begin{bmatrix}0&-1&0&1&|&1\0&0&4&4&|&0\0&0&-8&-8&|&0\0&0&4&4&|&0end{bmatrix}}$ and I have for example $(0,d-1,-d,d)^{T}$ and I change for example $d=5$ so my last vectors in basic is $(0,4,-5,5)^{T}$.
Hovewer on my lectures we use vectors from image but I don't understand it and I am afraid that my sollution is not good.
Can you help me?
linear-algebra
edited Feb 3 at 12:49
Bernard
124k742117
asked Feb 3 at 12:10
MP3129MP3129
869211
$endgroup$
add a comment |
$begingroup$
I have a matrix: $A={begin{bmatrix}3&-1&0&1\0&3&4&4\0&0&-5&-8\0&0&4&7end{bmatrix}}$.
1) I calculate characteristic polynomial. It is: $p_{A}(lambda)=(lambda-3)^{3}(lambda+1)$
2) So exist Jordan's matrix: $J={begin{bmatrix}3&?&0&0\0&3&?&0\0&0&3&0\0&0&0&-1end{bmatrix}}$
3)I find own subspace:
$V_{3}=ker(A-3I)=dots= operatorname{span}({begin{bmatrix}1&0&0&0end{bmatrix}}^{T},{begin{bmatrix}0&1&-1&1end{bmatrix}}^{T}$
$V_{-1}=ker(A+I)=dots= operatorname{span}({begin{bmatrix}0&1&-2&1end{bmatrix}}^{T}$
4) Jordan's matrix is: $J={begin{bmatrix}3&0&0&0\0&3&1&0\0&0&3&0\0&0&0&-1end{bmatrix}}$
However I have a problem with a basis:
I read that in basis are vectors from span $V_{3}$ and $V_{-1}$ and the next I must find one more vector. So I can use my vectors from span and I have for example:${begin{bmatrix}0&-1&0&1&|&1\0&0&4&4&|&0\0&0&-8&-8&|&0\0&0&4&4&|&0end{bmatrix}}$ and I have for example $(0,d-1,-d,d)^{T}$ and I change for example $d=5$ so my last vectors in basic is $(0,4,-5,5)^{T}$.
Hovewer on my lectures we use vectors from image but I don't understand it and I am afraid that my sollution is not good.
Can you help me?
linear-algebra
edited Feb 3 at 12:49
Bernard
124k742117
asked Feb 3 at 12:10
MP3129MP3129
869211
$endgroup$
1
$begingroup$
I think that you may want to google generalized eigenvector, which is what you need to complete the two lin. ind. eigenvectors of $;3;$ to a basis of $;V_3;$ ....
$endgroup$
– DonAntonio
Feb 3 at 12:20
$begingroup$
I have the same problem too, can @DonAntonio show us how to deal with such problem?
$endgroup$
– VirtualUser
Feb 3 at 12:27
$begingroup$
@VirtualUser I've posted a rather complete answer now. Hope it helps.
$endgroup$
– DonAntonio
Feb 3 at 16:40
add a comment |
$begingroup$
I have a matrix: $A={begin{bmatrix}3&-1&0&1\0&3&4&4\0&0&-5&-8\0&0&4&7end{bmatrix}}$.
1) I calculate characteristic polynomial. It is: $p_{A}(lambda)=(lambda-3)^{3}(lambda+1)$
2) So exist Jordan's matrix: $J={begin{bmatrix}3&?&0&0\0&3&?&0\0&0&3&0\0&0&0&-1end{bmatrix}}$
3)I find own subspace:
$V_{3}=ker(A-3I)=dots= operatorname{span}({begin{bmatrix}1&0&0&0end{bmatrix}}^{T},{begin{bmatrix}0&1&-1&1end{bmatrix}}^{T}$
$V_{-1}=ker(A+I)=dots= operatorname{span}({begin{bmatrix}0&1&-2&1end{bmatrix}}^{T}$
4) Jordan's matrix is: $J={begin{bmatrix}3&0&0&0\0&3&1&0\0&0&3&0\0&0&0&-1end{bmatrix}}$
However I have a problem with a basis:
I read that in basis are vectors from span $V_{3}$ and $V_{-1}$ and the next I must find one more vector. So I can use my vectors from span and I have for example:${begin{bmatrix}0&-1&0&1&|&1\0&0&4&4&|&0\0&0&-8&-8&|&0\0&0&4&4&|&0end{bmatrix}}$ and I have for example $(0,d-1,-d,d)^{T}$ and I change for example $d=5$ so my last vectors in basic is $(0,4,-5,5)^{T}$.
Hovewer on my lectures we use vectors from image but I don't understand it and I am afraid that my sollution is not good.
Can you help me?
linear-algebra
edited Feb 3 at 12:49
Bernard
124k742117
asked Feb 3 at 12:10
MP3129MP3129
869211
$endgroup$
I have a matrix: $A={begin{bmatrix}3&-1&0&1\0&3&4&4\0&0&-5&-8\0&0&4&7end{bmatrix}}$.
1) I calculate characteristic polynomial. It is: $p_{A}(lambda)=(lambda-3)^{3}(lambda+1)$
2) So exist Jordan's matrix: $J={begin{bmatrix}3&?&0&0\0&3&?&0\0&0&3&0\0&0&0&-1end{bmatrix}}$
3)I find own subspace:
$V_{3}=ker(A-3I)=dots= operatorname{span}({begin{bmatrix}1&0&0&0end{bmatrix}}^{T},{begin{bmatrix}0&1&-1&1end{bmatrix}}^{T}$
$V_{-1}=ker(A+I)=dots= operatorname{span}({begin{bmatrix}0&1&-2&1end{bmatrix}}^{T}$
4) Jordan's matrix is: $J={begin{bmatrix}3&0&0&0\0&3&1&0\0&0&3&0\0&0&0&-1end{bmatrix}}$
However I have a problem with a basis:
I read that in basis are vectors from span $V_{3}$ and $V_{-1}$ and the next I must find one more vector. So I can use my vectors from span and I have for example:${begin{bmatrix}0&-1&0&1&|&1\0&0&4&4&|&0\0&0&-8&-8&|&0\0&0&4&4&|&0end{bmatrix}}$ and I have for example $(0,d-1,-d,d)^{T}$ and I change for example $d=5$ so my last vectors in basic is $(0,4,-5,5)^{T}$.
Hovewer on my lectures we use vectors from image but I don't understand it and I am afraid that my sollution is not good.
Can you help me?
linear-algebra
linear-algebra
edited Feb 3 at 12:49
Bernard
124k742117
asked Feb 3 at 12:10
MP3129MP3129
869211
edited Feb 3 at 12:49
Bernard
124k742117
asked Feb 3 at 12:10
MP3129MP3129
869211
edited Feb 3 at 12:49
Bernard
124k742117
edited Feb 3 at 12:49
Bernard
124k742117
edited Feb 3 at 12:49
Bernard
124k742117
124k742117
asked Feb 3 at 12:10
MP3129MP3129
869211
asked Feb 3 at 12:10
MP3129MP3129
869211
asked Feb 3 at 12:10
MP3129MP3129
869211
869211
1
$begingroup$
I think that you may want to google generalized eigenvector, which is what you need to complete the two lin. ind. eigenvectors of $;3;$ to a basis of $;V_3;$ ....
$endgroup$
– DonAntonio
Feb 3 at 12:20
$begingroup$
I have the same problem too, can @DonAntonio show us how to deal with such problem?
$endgroup$
– VirtualUser
Feb 3 at 12:27
$begingroup$
@VirtualUser I've posted a rather complete answer now. Hope it helps.
$endgroup$
– DonAntonio
Feb 3 at 16:40
add a comment |
1
$begingroup$
I think that you may want to google generalized eigenvector, which is what you need to complete the two lin. ind. eigenvectors of $;3;$ to a basis of $;V_3;$ ....
$endgroup$
– DonAntonio
Feb 3 at 12:20
$begingroup$
I have the same problem too, can @DonAntonio show us how to deal with such problem?
$endgroup$
– VirtualUser
Feb 3 at 12:27
$begingroup$
@VirtualUser I've posted a rather complete answer now. Hope it helps.
$endgroup$
– DonAntonio
Feb 3 at 16:40
1
1
$begingroup$
I think that you may want to google generalized eigenvector, which is what you need to complete the two lin. ind. eigenvectors of $;3;$ to a basis of $;V_3;$ ....
$endgroup$
– DonAntonio
Feb 3 at 12:20
$begingroup$
I think that you may want to google generalized eigenvector, which is what you need to complete the two lin. ind. eigenvectors of $;3;$ to a basis of $;V_3;$ ....
$endgroup$
– DonAntonio
Feb 3 at 12:20
$begingroup$
I have the same problem too, can @DonAntonio show us how to deal with such problem?
$endgroup$
– VirtualUser
Feb 3 at 12:27
$begingroup$
I have the same problem too, can @DonAntonio show us how to deal with such problem?
$endgroup$
– VirtualUser
Feb 3 at 12:27
$begingroup$
@VirtualUser I've posted a rather complete answer now. Hope it helps.
$endgroup$
– DonAntonio
Feb 3 at 16:40
$begingroup$
@VirtualUser I've posted a rather complete answer now. Hope it helps.
$endgroup$
– DonAntonio
Feb 3 at 16:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I'll write down an algorithm to find a basis with eigenvectors / generalized eigenvectors, when the wanted vector will be denoted by $;w;$ .
First, we must find out what the eigenvalues are. You already did that, they are $;-1,3;$, with the second one of geometric multiplicity $2$ . Now
$$(A-(-1)I)w=0iff begin{pmatrix}4&-1&0&1\0&4&4&4\0&0&-4&-8\0&0&4&8end{pmatrix}begin{pmatrix}w_1\w_2\w_3\w_4end{pmatrix}=begin{pmatrix}0\0\0\0end{pmatrix}implies w_1=begin{pmatrix}0\1\-2\1end{pmatrix}$$$${}$$
$$(A-3I)w=begin{pmatrix}0\0\0\0end{pmatrix}iffbegin{pmatrix}0&-1&0&1\0&0&4&4\0&0&-8&-8\0&0&4&4end{pmatrix}begin{pmatrix}w_1\w_2\w_3\w_4end{pmatrix}=begin{pmatrix}0\0\0\0end{pmatrix}implies w_2=begin{pmatrix}1\0\0\0end{pmatrix},,,,w_3=begin{pmatrix}0\1\!!-1\1end{pmatrix}$$$${}$$
with $;w_1,w_2,w_3;$ eigenvectors of $;lambda=-1;$ (the first one), and of $;lambda=3;$ (the last two) .
We're missing one vector for a basis for $;Bbb R^4;$, and since there are only two linearly ind. eigenvectors of $;3;$ , we're going to calculate a generalized eigenvector of $;3;$ as follows:
$$(A-3I)w=w_2;text{or};w_3$$ (we only need one of these to work), so:
$$(A-3I)w=begin{pmatrix}1\0\0\0end{pmatrix}=w_2iffbegin{pmatrix}0&-1&0&1\0&0&4&4\0&0&-8&-8\0&0&4&4end{pmatrix}begin{pmatrix}w_1\w_2\w_3\w_4end{pmatrix}=begin{pmatrix}1\0\0\0end{pmatrix}implies$$$${}$$
$$begin{cases}-w_2+w_4=1\
w_3=-w_4end{cases};;implies w_4=begin{pmatrix}0\-1\0\0end{pmatrix}$$
and we thus have a basis for $;Bbb R^4;:;;{w_1,w_2,w_3,w_4}$
In this case is pretty easy to find out the JCF of $;A;$ , since $;dim V_3=2implies;$ there are two blocks for $;lambda=3;$ , and thus the JCF (upt to similarity) of $;A;$ is
$$J_A=begin{pmatrix}3&1&0&0\0&3&0&0\0&0&3&0\0&0&0&!!-1end{pmatrix}$$
But ...If you want to get the above by similarity, take the basis $;{w_1,w_2,color{red}{w_4},w_3};$ and form with it the columns of matrix $;P;$ . Then you can check that $;P^{-1}AP=J_A;$ . The reason is that we "arrange" the basis with "close" vectors to each other. Thus, first the eigenvector $;w_2;$ and then the generalized eigenvector $;w_4;$ which was obtained with the help of $;w_2;$ , and at the end $;w_3;$ .
answered Feb 3 at 16:39
DonAntonioDonAntonio
180k1495233
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add a comment |
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Let $v_1=(1,0,0,0)$ and let $v_2=(0,1,-1,1)$. Then both $v_1$ and $v_2$ are eigenvectors with eigenvalue $3$. Now, consider the equations $A.v=3v_1+v$ and $A.v=3v_2+v$. You can check that $v_3=(0,0,-1,1)$ is a solution of the first equation, whereas the second one has no solutions. So, if $v_4=(0,1-,2,1)$, then the basis that you're after is ${v_2,v_1,v_3,v_4}$.
edited Feb 3 at 12:50
Bernard
124k742117
answered Feb 3 at 12:43
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
$begingroup$
Why $A.v=3v_1+v$ and $A.v=3v_2+v$.?
$endgroup$
– MP3129
Feb 3 at 13:06
$begingroup$
Look at the matrix $J$ that you're after. It's third column is $[0 1 3 0]^T$, right? What that means is that $A$ times the third vector of your basis will be the second vector of your basis plus $3$ times the third vector of your basis. So, in order to solve the problem, I have to solve one of the equations that I mentioned.
$endgroup$
– José Carlos Santos
Feb 3 at 13:36
add a comment |
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2 Answers
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active
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2 Answers
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$begingroup$
I'll write down an algorithm to find a basis with eigenvectors / generalized eigenvectors, when the wanted vector will be denoted by $;w;$ .
First, we must find out what the eigenvalues are. You already did that, they are $;-1,3;$, with the second one of geometric multiplicity $2$ . Now
$$(A-(-1)I)w=0iff begin{pmatrix}4&-1&0&1\0&4&4&4\0&0&-4&-8\0&0&4&8end{pmatrix}begin{pmatrix}w_1\w_2\w_3\w_4end{pmatrix}=begin{pmatrix}0\0\0\0end{pmatrix}implies w_1=begin{pmatrix}0\1\-2\1end{pmatrix}$$$${}$$
$$(A-3I)w=begin{pmatrix}0\0\0\0end{pmatrix}iffbegin{pmatrix}0&-1&0&1\0&0&4&4\0&0&-8&-8\0&0&4&4end{pmatrix}begin{pmatrix}w_1\w_2\w_3\w_4end{pmatrix}=begin{pmatrix}0\0\0\0end{pmatrix}implies w_2=begin{pmatrix}1\0\0\0end{pmatrix},,,,w_3=begin{pmatrix}0\1\!!-1\1end{pmatrix}$$$${}$$
with $;w_1,w_2,w_3;$ eigenvectors of $;lambda=-1;$ (the first one), and of $;lambda=3;$ (the last two) .
We're missing one vector for a basis for $;Bbb R^4;$, and since there are only two linearly ind. eigenvectors of $;3;$ , we're going to calculate a generalized eigenvector of $;3;$ as follows:
$$(A-3I)w=w_2;text{or};w_3$$ (we only need one of these to work), so:
$$(A-3I)w=begin{pmatrix}1\0\0\0end{pmatrix}=w_2iffbegin{pmatrix}0&-1&0&1\0&0&4&4\0&0&-8&-8\0&0&4&4end{pmatrix}begin{pmatrix}w_1\w_2\w_3\w_4end{pmatrix}=begin{pmatrix}1\0\0\0end{pmatrix}implies$$$${}$$
$$begin{cases}-w_2+w_4=1\
w_3=-w_4end{cases};;implies w_4=begin{pmatrix}0\-1\0\0end{pmatrix}$$
and we thus have a basis for $;Bbb R^4;:;;{w_1,w_2,w_3,w_4}$
In this case is pretty easy to find out the JCF of $;A;$ , since $;dim V_3=2implies;$ there are two blocks for $;lambda=3;$ , and thus the JCF (upt to similarity) of $;A;$ is
$$J_A=begin{pmatrix}3&1&0&0\0&3&0&0\0&0&3&0\0&0&0&!!-1end{pmatrix}$$
But ...If you want to get the above by similarity, take the basis $;{w_1,w_2,color{red}{w_4},w_3};$ and form with it the columns of matrix $;P;$ . Then you can check that $;P^{-1}AP=J_A;$ . The reason is that we "arrange" the basis with "close" vectors to each other. Thus, first the eigenvector $;w_2;$ and then the generalized eigenvector $;w_4;$ which was obtained with the help of $;w_2;$ , and at the end $;w_3;$ .
answered Feb 3 at 16:39
DonAntonioDonAntonio
180k1495233
$endgroup$
add a comment |
$begingroup$
I'll write down an algorithm to find a basis with eigenvectors / generalized eigenvectors, when the wanted vector will be denoted by $;w;$ .
First, we must find out what the eigenvalues are. You already did that, they are $;-1,3;$, with the second one of geometric multiplicity $2$ . Now
$$(A-(-1)I)w=0iff begin{pmatrix}4&-1&0&1\0&4&4&4\0&0&-4&-8\0&0&4&8end{pmatrix}begin{pmatrix}w_1\w_2\w_3\w_4end{pmatrix}=begin{pmatrix}0\0\0\0end{pmatrix}implies w_1=begin{pmatrix}0\1\-2\1end{pmatrix}$$$${}$$
$$(A-3I)w=begin{pmatrix}0\0\0\0end{pmatrix}iffbegin{pmatrix}0&-1&0&1\0&0&4&4\0&0&-8&-8\0&0&4&4end{pmatrix}begin{pmatrix}w_1\w_2\w_3\w_4end{pmatrix}=begin{pmatrix}0\0\0\0end{pmatrix}implies w_2=begin{pmatrix}1\0\0\0end{pmatrix},,,,w_3=begin{pmatrix}0\1\!!-1\1end{pmatrix}$$$${}$$
with $;w_1,w_2,w_3;$ eigenvectors of $;lambda=-1;$ (the first one), and of $;lambda=3;$ (the last two) .
We're missing one vector for a basis for $;Bbb R^4;$, and since there are only two linearly ind. eigenvectors of $;3;$ , we're going to calculate a generalized eigenvector of $;3;$ as follows:
$$(A-3I)w=w_2;text{or};w_3$$ (we only need one of these to work), so:
$$(A-3I)w=begin{pmatrix}1\0\0\0end{pmatrix}=w_2iffbegin{pmatrix}0&-1&0&1\0&0&4&4\0&0&-8&-8\0&0&4&4end{pmatrix}begin{pmatrix}w_1\w_2\w_3\w_4end{pmatrix}=begin{pmatrix}1\0\0\0end{pmatrix}implies$$$${}$$
$$begin{cases}-w_2+w_4=1\
w_3=-w_4end{cases};;implies w_4=begin{pmatrix}0\-1\0\0end{pmatrix}$$
and we thus have a basis for $;Bbb R^4;:;;{w_1,w_2,w_3,w_4}$
In this case is pretty easy to find out the JCF of $;A;$ , since $;dim V_3=2implies;$ there are two blocks for $;lambda=3;$ , and thus the JCF (upt to similarity) of $;A;$ is
$$J_A=begin{pmatrix}3&1&0&0\0&3&0&0\0&0&3&0\0&0&0&!!-1end{pmatrix}$$
But ...If you want to get the above by similarity, take the basis $;{w_1,w_2,color{red}{w_4},w_3};$ and form with it the columns of matrix $;P;$ . Then you can check that $;P^{-1}AP=J_A;$ . The reason is that we "arrange" the basis with "close" vectors to each other. Thus, first the eigenvector $;w_2;$ and then the generalized eigenvector $;w_4;$ which was obtained with the help of $;w_2;$ , and at the end $;w_3;$ .
answered Feb 3 at 16:39
DonAntonioDonAntonio
180k1495233
$endgroup$
add a comment |
$begingroup$
I'll write down an algorithm to find a basis with eigenvectors / generalized eigenvectors, when the wanted vector will be denoted by $;w;$ .
First, we must find out what the eigenvalues are. You already did that, they are $;-1,3;$, with the second one of geometric multiplicity $2$ . Now
$$(A-(-1)I)w=0iff begin{pmatrix}4&-1&0&1\0&4&4&4\0&0&-4&-8\0&0&4&8end{pmatrix}begin{pmatrix}w_1\w_2\w_3\w_4end{pmatrix}=begin{pmatrix}0\0\0\0end{pmatrix}implies w_1=begin{pmatrix}0\1\-2\1end{pmatrix}$$$${}$$
$$(A-3I)w=begin{pmatrix}0\0\0\0end{pmatrix}iffbegin{pmatrix}0&-1&0&1\0&0&4&4\0&0&-8&-8\0&0&4&4end{pmatrix}begin{pmatrix}w_1\w_2\w_3\w_4end{pmatrix}=begin{pmatrix}0\0\0\0end{pmatrix}implies w_2=begin{pmatrix}1\0\0\0end{pmatrix},,,,w_3=begin{pmatrix}0\1\!!-1\1end{pmatrix}$$$${}$$
with $;w_1,w_2,w_3;$ eigenvectors of $;lambda=-1;$ (the first one), and of $;lambda=3;$ (the last two) .
We're missing one vector for a basis for $;Bbb R^4;$, and since there are only two linearly ind. eigenvectors of $;3;$ , we're going to calculate a generalized eigenvector of $;3;$ as follows:
$$(A-3I)w=w_2;text{or};w_3$$ (we only need one of these to work), so:
$$(A-3I)w=begin{pmatrix}1\0\0\0end{pmatrix}=w_2iffbegin{pmatrix}0&-1&0&1\0&0&4&4\0&0&-8&-8\0&0&4&4end{pmatrix}begin{pmatrix}w_1\w_2\w_3\w_4end{pmatrix}=begin{pmatrix}1\0\0\0end{pmatrix}implies$$$${}$$
$$begin{cases}-w_2+w_4=1\
w_3=-w_4end{cases};;implies w_4=begin{pmatrix}0\-1\0\0end{pmatrix}$$
and we thus have a basis for $;Bbb R^4;:;;{w_1,w_2,w_3,w_4}$
In this case is pretty easy to find out the JCF of $;A;$ , since $;dim V_3=2implies;$ there are two blocks for $;lambda=3;$ , and thus the JCF (upt to similarity) of $;A;$ is
$$J_A=begin{pmatrix}3&1&0&0\0&3&0&0\0&0&3&0\0&0&0&!!-1end{pmatrix}$$
But ...If you want to get the above by similarity, take the basis $;{w_1,w_2,color{red}{w_4},w_3};$ and form with it the columns of matrix $;P;$ . Then you can check that $;P^{-1}AP=J_A;$ . The reason is that we "arrange" the basis with "close" vectors to each other. Thus, first the eigenvector $;w_2;$ and then the generalized eigenvector $;w_4;$ which was obtained with the help of $;w_2;$ , and at the end $;w_3;$ .
answered Feb 3 at 16:39
DonAntonioDonAntonio
180k1495233
$endgroup$
I'll write down an algorithm to find a basis with eigenvectors / generalized eigenvectors, when the wanted vector will be denoted by $;w;$ .
First, we must find out what the eigenvalues are. You already did that, they are $;-1,3;$, with the second one of geometric multiplicity $2$ . Now
$$(A-(-1)I)w=0iff begin{pmatrix}4&-1&0&1\0&4&4&4\0&0&-4&-8\0&0&4&8end{pmatrix}begin{pmatrix}w_1\w_2\w_3\w_4end{pmatrix}=begin{pmatrix}0\0\0\0end{pmatrix}implies w_1=begin{pmatrix}0\1\-2\1end{pmatrix}$$$${}$$
$$(A-3I)w=begin{pmatrix}0\0\0\0end{pmatrix}iffbegin{pmatrix}0&-1&0&1\0&0&4&4\0&0&-8&-8\0&0&4&4end{pmatrix}begin{pmatrix}w_1\w_2\w_3\w_4end{pmatrix}=begin{pmatrix}0\0\0\0end{pmatrix}implies w_2=begin{pmatrix}1\0\0\0end{pmatrix},,,,w_3=begin{pmatrix}0\1\!!-1\1end{pmatrix}$$$${}$$
with $;w_1,w_2,w_3;$ eigenvectors of $;lambda=-1;$ (the first one), and of $;lambda=3;$ (the last two) .
We're missing one vector for a basis for $;Bbb R^4;$, and since there are only two linearly ind. eigenvectors of $;3;$ , we're going to calculate a generalized eigenvector of $;3;$ as follows:
$$(A-3I)w=w_2;text{or};w_3$$ (we only need one of these to work), so:
$$(A-3I)w=begin{pmatrix}1\0\0\0end{pmatrix}=w_2iffbegin{pmatrix}0&-1&0&1\0&0&4&4\0&0&-8&-8\0&0&4&4end{pmatrix}begin{pmatrix}w_1\w_2\w_3\w_4end{pmatrix}=begin{pmatrix}1\0\0\0end{pmatrix}implies$$$${}$$
$$begin{cases}-w_2+w_4=1\
w_3=-w_4end{cases};;implies w_4=begin{pmatrix}0\-1\0\0end{pmatrix}$$
and we thus have a basis for $;Bbb R^4;:;;{w_1,w_2,w_3,w_4}$
In this case is pretty easy to find out the JCF of $;A;$ , since $;dim V_3=2implies;$ there are two blocks for $;lambda=3;$ , and thus the JCF (upt to similarity) of $;A;$ is
$$J_A=begin{pmatrix}3&1&0&0\0&3&0&0\0&0&3&0\0&0&0&!!-1end{pmatrix}$$
But ...If you want to get the above by similarity, take the basis $;{w_1,w_2,color{red}{w_4},w_3};$ and form with it the columns of matrix $;P;$ . Then you can check that $;P^{-1}AP=J_A;$ . The reason is that we "arrange" the basis with "close" vectors to each other. Thus, first the eigenvector $;w_2;$ and then the generalized eigenvector $;w_4;$ which was obtained with the help of $;w_2;$ , and at the end $;w_3;$ .
answered Feb 3 at 16:39
DonAntonioDonAntonio
180k1495233
edited Feb 3 at 22:44
answered Feb 3 at 16:39
DonAntonioDonAntonio
180k1495233
answered Feb 3 at 16:39
DonAntonioDonAntonio
180k1495233
answered Feb 3 at 16:39
DonAntonioDonAntonio
180k1495233
180k1495233
add a comment |
add a comment |
$begingroup$
Let $v_1=(1,0,0,0)$ and let $v_2=(0,1,-1,1)$. Then both $v_1$ and $v_2$ are eigenvectors with eigenvalue $3$. Now, consider the equations $A.v=3v_1+v$ and $A.v=3v_2+v$. You can check that $v_3=(0,0,-1,1)$ is a solution of the first equation, whereas the second one has no solutions. So, if $v_4=(0,1-,2,1)$, then the basis that you're after is ${v_2,v_1,v_3,v_4}$.
edited Feb 3 at 12:50
Bernard
124k742117
answered Feb 3 at 12:43
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
$begingroup$
Why $A.v=3v_1+v$ and $A.v=3v_2+v$.?
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– MP3129
Feb 3 at 13:06
$begingroup$
Look at the matrix $J$ that you're after. It's third column is $[0 1 3 0]^T$, right? What that means is that $A$ times the third vector of your basis will be the second vector of your basis plus $3$ times the third vector of your basis. So, in order to solve the problem, I have to solve one of the equations that I mentioned.
$endgroup$
– José Carlos Santos
Feb 3 at 13:36
add a comment |
$begingroup$
Let $v_1=(1,0,0,0)$ and let $v_2=(0,1,-1,1)$. Then both $v_1$ and $v_2$ are eigenvectors with eigenvalue $3$. Now, consider the equations $A.v=3v_1+v$ and $A.v=3v_2+v$. You can check that $v_3=(0,0,-1,1)$ is a solution of the first equation, whereas the second one has no solutions. So, if $v_4=(0,1-,2,1)$, then the basis that you're after is ${v_2,v_1,v_3,v_4}$.
edited Feb 3 at 12:50
Bernard
124k742117
answered Feb 3 at 12:43
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
$begingroup$
Why $A.v=3v_1+v$ and $A.v=3v_2+v$.?
$endgroup$
– MP3129
Feb 3 at 13:06
$begingroup$
Look at the matrix $J$ that you're after. It's third column is $[0 1 3 0]^T$, right? What that means is that $A$ times the third vector of your basis will be the second vector of your basis plus $3$ times the third vector of your basis. So, in order to solve the problem, I have to solve one of the equations that I mentioned.
$endgroup$
– José Carlos Santos
Feb 3 at 13:36
add a comment |
$begingroup$
Let $v_1=(1,0,0,0)$ and let $v_2=(0,1,-1,1)$. Then both $v_1$ and $v_2$ are eigenvectors with eigenvalue $3$. Now, consider the equations $A.v=3v_1+v$ and $A.v=3v_2+v$. You can check that $v_3=(0,0,-1,1)$ is a solution of the first equation, whereas the second one has no solutions. So, if $v_4=(0,1-,2,1)$, then the basis that you're after is ${v_2,v_1,v_3,v_4}$.
edited Feb 3 at 12:50
Bernard
124k742117
answered Feb 3 at 12:43
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
Let $v_1=(1,0,0,0)$ and let $v_2=(0,1,-1,1)$. Then both $v_1$ and $v_2$ are eigenvectors with eigenvalue $3$. Now, consider the equations $A.v=3v_1+v$ and $A.v=3v_2+v$. You can check that $v_3=(0,0,-1,1)$ is a solution of the first equation, whereas the second one has no solutions. So, if $v_4=(0,1-,2,1)$, then the basis that you're after is ${v_2,v_1,v_3,v_4}$.
edited Feb 3 at 12:50
Bernard
124k742117
answered Feb 3 at 12:43
José Carlos SantosJosé Carlos Santos
175k24134243
edited Feb 3 at 12:50
Bernard
124k742117
edited Feb 3 at 12:50
Bernard
124k742117
edited Feb 3 at 12:50
Bernard
124k742117
124k742117
answered Feb 3 at 12:43
José Carlos SantosJosé Carlos Santos
175k24134243
answered Feb 3 at 12:43
José Carlos SantosJosé Carlos Santos
175k24134243
answered Feb 3 at 12:43
José Carlos SantosJosé Carlos Santos
175k24134243
175k24134243
$begingroup$
Why $A.v=3v_1+v$ and $A.v=3v_2+v$.?
$endgroup$
– MP3129
Feb 3 at 13:06
$begingroup$
Look at the matrix $J$ that you're after. It's third column is $[0 1 3 0]^T$, right? What that means is that $A$ times the third vector of your basis will be the second vector of your basis plus $3$ times the third vector of your basis. So, in order to solve the problem, I have to solve one of the equations that I mentioned.
$endgroup$
– José Carlos Santos
Feb 3 at 13:36
add a comment |
$begingroup$
Why $A.v=3v_1+v$ and $A.v=3v_2+v$.?
$endgroup$
– MP3129
Feb 3 at 13:06
$begingroup$
Look at the matrix $J$ that you're after. It's third column is $[0 1 3 0]^T$, right? What that means is that $A$ times the third vector of your basis will be the second vector of your basis plus $3$ times the third vector of your basis. So, in order to solve the problem, I have to solve one of the equations that I mentioned.
$endgroup$
– José Carlos Santos
Feb 3 at 13:36
$begingroup$
Why $A.v=3v_1+v$ and $A.v=3v_2+v$.?
$endgroup$
– MP3129
Feb 3 at 13:06
$begingroup$
Why $A.v=3v_1+v$ and $A.v=3v_2+v$.?
$endgroup$
– MP3129
Feb 3 at 13:06
$begingroup$
Look at the matrix $J$ that you're after. It's third column is $[0 1 3 0]^T$, right? What that means is that $A$ times the third vector of your basis will be the second vector of your basis plus $3$ times the third vector of your basis. So, in order to solve the problem, I have to solve one of the equations that I mentioned.
$endgroup$
– José Carlos Santos
Feb 3 at 13:36
$begingroup$
Look at the matrix $J$ that you're after. It's third column is $[0 1 3 0]^T$, right? What that means is that $A$ times the third vector of your basis will be the second vector of your basis plus $3$ times the third vector of your basis. So, in order to solve the problem, I have to solve one of the equations that I mentioned.
$endgroup$
– José Carlos Santos
Feb 3 at 13:36
add a comment |
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$begingroup$
I think that you may want to google generalized eigenvector, which is what you need to complete the two lin. ind. eigenvectors of $;3;$ to a basis of $;V_3;$ ....
$endgroup$
– DonAntonio
Feb 3 at 12:20
$begingroup$
I have the same problem too, can @DonAntonio show us how to deal with such problem?
$endgroup$
– VirtualUser
Feb 3 at 12:27
$begingroup$
@VirtualUser I've posted a rather complete answer now. Hope it helps.
$endgroup$
– DonAntonio
Feb 3 at 16:40