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Show that the determinant of jordan normal form wtih spatial weight matrix by its eigenvalues
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I try to figure out the proof of determinant of matrix by eigen decomposition,
$$det(I_n-lambda W)=det(QQ^-1(I_n-lambda W))$$
$$=det(Q(I_n-lambda W)Q^-1)$$
$$=(1-lambda upsilon_1)(1-lambda upsilon_2)...(1-lambda upsilon_n)$$
where $lambda$ is spatial autoregressive coefficient, $W$ is normalized weight matrix, $upsilon$ is eigenvalues of $W$, $Q$ is the matrix that trinagularizes $W$.
1.i can't figure out how changes taking place of the $Q^-1$ matrix in front of $(I_n-lambda W)$ in the first equation to behind of $(I_n-lambda W)$ in the second equation? is there any proporties for that?
2.why do need $W$ is triangularized? can it be proof without triangularized?
I would really appreciate some help.
linear-algebra eigenvalues-eigenvectors determinant jordan-normal-form
asked Jul 25 at 14:07
v.yildirim
32
add a comment |Â
up vote
0
down vote
favorite
I try to figure out the proof of determinant of matrix by eigen decomposition,
$$det(I_n-lambda W)=det(QQ^-1(I_n-lambda W))$$
$$=det(Q(I_n-lambda W)Q^-1)$$
$$=(1-lambda upsilon_1)(1-lambda upsilon_2)...(1-lambda upsilon_n)$$
where $lambda$ is spatial autoregressive coefficient, $W$ is normalized weight matrix, $upsilon$ is eigenvalues of $W$, $Q$ is the matrix that trinagularizes $W$.
1.i can't figure out how changes taking place of the $Q^-1$ matrix in front of $(I_n-lambda W)$ in the first equation to behind of $(I_n-lambda W)$ in the second equation? is there any proporties for that?
2.why do need $W$ is triangularized? can it be proof without triangularized?
I would really appreciate some help.
linear-algebra eigenvalues-eigenvectors determinant jordan-normal-form
asked Jul 25 at 14:07
v.yildirim
32
Note that the determinant is multiplicative i.e. $det(Q(I_n-lambda W)Q^-1)=det(Q)det(I_n-lambda W)det(Q^-1)$ and the values of the determinant commute
– Peter Melech
Jul 25 at 14:14
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I try to figure out the proof of determinant of matrix by eigen decomposition,
$$det(I_n-lambda W)=det(QQ^-1(I_n-lambda W))$$
$$=det(Q(I_n-lambda W)Q^-1)$$
$$=(1-lambda upsilon_1)(1-lambda upsilon_2)...(1-lambda upsilon_n)$$
where $lambda$ is spatial autoregressive coefficient, $W$ is normalized weight matrix, $upsilon$ is eigenvalues of $W$, $Q$ is the matrix that trinagularizes $W$.
1.i can't figure out how changes taking place of the $Q^-1$ matrix in front of $(I_n-lambda W)$ in the first equation to behind of $(I_n-lambda W)$ in the second equation? is there any proporties for that?
2.why do need $W$ is triangularized? can it be proof without triangularized?
I would really appreciate some help.
linear-algebra eigenvalues-eigenvectors determinant jordan-normal-form
asked Jul 25 at 14:07
v.yildirim
32
I try to figure out the proof of determinant of matrix by eigen decomposition,
$$det(I_n-lambda W)=det(QQ^-1(I_n-lambda W))$$
$$=det(Q(I_n-lambda W)Q^-1)$$
$$=(1-lambda upsilon_1)(1-lambda upsilon_2)...(1-lambda upsilon_n)$$
where $lambda$ is spatial autoregressive coefficient, $W$ is normalized weight matrix, $upsilon$ is eigenvalues of $W$, $Q$ is the matrix that trinagularizes $W$.
1.i can't figure out how changes taking place of the $Q^-1$ matrix in front of $(I_n-lambda W)$ in the first equation to behind of $(I_n-lambda W)$ in the second equation? is there any proporties for that?
2.why do need $W$ is triangularized? can it be proof without triangularized?
I would really appreciate some help.
linear-algebra eigenvalues-eigenvectors determinant jordan-normal-form
asked Jul 25 at 14:07
v.yildirim
32
asked Jul 25 at 14:07
v.yildirim
32
asked Jul 25 at 14:07
v.yildirim
32
asked Jul 25 at 14:07
v.yildirim
32
32
Note that the determinant is multiplicative i.e. $det(Q(I_n-lambda W)Q^-1)=det(Q)det(I_n-lambda W)det(Q^-1)$ and the values of the determinant commute
– Peter Melech
Jul 25 at 14:14
add a comment |Â
Note that the determinant is multiplicative i.e. $det(Q(I_n-lambda W)Q^-1)=det(Q)det(I_n-lambda W)det(Q^-1)$ and the values of the determinant commute
– Peter Melech
Jul 25 at 14:14
Note that the determinant is multiplicative i.e. $det(Q(I_n-lambda W)Q^-1)=det(Q)det(I_n-lambda W)det(Q^-1)$ and the values of the determinant commute
– Peter Melech
Jul 25 at 14:14
Note that the determinant is multiplicative i.e. $det(Q(I_n-lambda W)Q^-1)=det(Q)det(I_n-lambda W)det(Q^-1)$ and the values of the determinant commute
– Peter Melech
Jul 25 at 14:14
add a comment |Â
2 Answers
2
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So we know (given) that there is a $Q$ such that $QWQ^-1$ is triangular. The first step is to introduce $Q$
$$det(I - lambda W) = det( Q^-1Q(I-lambda W))$$
Second let's use the property $det(AB) = det(BA)$ where $A = Q^-1$ and $B = Q(I-lambda W)$ hence
$$det(I - lambda W) = det(Q(I-lambda W) Q^-1)$$
Third, expand the above
$$det(I - lambda W) = det(QQ^-1-lambda Q W Q^-1 )$$
Notice that $QQ^-1 = I$ again so
$$det(I - lambda W) = det(I-lambda Q W Q^-1 )$$
Fifth denote our triangular matrix $QWQ^-1 = T$ hence
$$det(I - lambda W) = det(I-lambda T )$$
Notice that $I - lambda T$ is also triangular with diagonal elements $1 - lambda v_1$, where $v_k$ appear on the diagonal entries of $T$. Using the fact that the determinant of a triangular matrix is nothing other than the product of it's diagonal entries, then
$$det(I - lambda W) =(1 - lambda v_1)ldots(1-lambda v_n)$$
answered Jul 25 at 14:20
Ahmad Bazzi
2,6271417
Thank you detailed explanation. It is cleary seen.
– v.yildirim
Jul 26 at 7:20
I'm glad it helped. If you found it useful, you could mark the answer as correct and upvote to help other viewers know that it was helpful @v.yildirim
– Ahmad Bazzi
Jul 26 at 13:31
add a comment |Â
up vote
0
down vote
- Just think of a theorem saying that $$det(AB)=det(A)det(B)$$so we have $$det(AB)=det(BA)$$2. sure it can be proved in such way. Probably in your problem the matrix is particularly triangular but it doesn't require to remain so. To show that the key idea is that $$textthe determinant of a matrix is multiplication of all its eigenvalues$$can you finish this one?
answered Jul 25 at 15:36
Mostafa Ayaz
8,5373630
So i can see how it can change and i think it is easy to show dererminant with eigenvalues by triangularized matrix. Thank you
– v.yildirim
Jul 26 at 7:17
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
So we know (given) that there is a $Q$ such that $QWQ^-1$ is triangular. The first step is to introduce $Q$
$$det(I - lambda W) = det( Q^-1Q(I-lambda W))$$
Second let's use the property $det(AB) = det(BA)$ where $A = Q^-1$ and $B = Q(I-lambda W)$ hence
$$det(I - lambda W) = det(Q(I-lambda W) Q^-1)$$
Third, expand the above
$$det(I - lambda W) = det(QQ^-1-lambda Q W Q^-1 )$$
Notice that $QQ^-1 = I$ again so
$$det(I - lambda W) = det(I-lambda Q W Q^-1 )$$
Fifth denote our triangular matrix $QWQ^-1 = T$ hence
$$det(I - lambda W) = det(I-lambda T )$$
Notice that $I - lambda T$ is also triangular with diagonal elements $1 - lambda v_1$, where $v_k$ appear on the diagonal entries of $T$. Using the fact that the determinant of a triangular matrix is nothing other than the product of it's diagonal entries, then
$$det(I - lambda W) =(1 - lambda v_1)ldots(1-lambda v_n)$$
answered Jul 25 at 14:20
Ahmad Bazzi
2,6271417
Thank you detailed explanation. It is cleary seen.
– v.yildirim
Jul 26 at 7:20
I'm glad it helped. If you found it useful, you could mark the answer as correct and upvote to help other viewers know that it was helpful @v.yildirim
– Ahmad Bazzi
Jul 26 at 13:31
add a comment |Â
up vote
0
down vote
accepted
So we know (given) that there is a $Q$ such that $QWQ^-1$ is triangular. The first step is to introduce $Q$
$$det(I - lambda W) = det( Q^-1Q(I-lambda W))$$
Second let's use the property $det(AB) = det(BA)$ where $A = Q^-1$ and $B = Q(I-lambda W)$ hence
$$det(I - lambda W) = det(Q(I-lambda W) Q^-1)$$
Third, expand the above
$$det(I - lambda W) = det(QQ^-1-lambda Q W Q^-1 )$$
Notice that $QQ^-1 = I$ again so
$$det(I - lambda W) = det(I-lambda Q W Q^-1 )$$
Fifth denote our triangular matrix $QWQ^-1 = T$ hence
$$det(I - lambda W) = det(I-lambda T )$$
Notice that $I - lambda T$ is also triangular with diagonal elements $1 - lambda v_1$, where $v_k$ appear on the diagonal entries of $T$. Using the fact that the determinant of a triangular matrix is nothing other than the product of it's diagonal entries, then
$$det(I - lambda W) =(1 - lambda v_1)ldots(1-lambda v_n)$$
answered Jul 25 at 14:20
Ahmad Bazzi
2,6271417
Thank you detailed explanation. It is cleary seen.
– v.yildirim
Jul 26 at 7:20
I'm glad it helped. If you found it useful, you could mark the answer as correct and upvote to help other viewers know that it was helpful @v.yildirim
– Ahmad Bazzi
Jul 26 at 13:31
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
So we know (given) that there is a $Q$ such that $QWQ^-1$ is triangular. The first step is to introduce $Q$
$$det(I - lambda W) = det( Q^-1Q(I-lambda W))$$
Second let's use the property $det(AB) = det(BA)$ where $A = Q^-1$ and $B = Q(I-lambda W)$ hence
$$det(I - lambda W) = det(Q(I-lambda W) Q^-1)$$
Third, expand the above
$$det(I - lambda W) = det(QQ^-1-lambda Q W Q^-1 )$$
Notice that $QQ^-1 = I$ again so
$$det(I - lambda W) = det(I-lambda Q W Q^-1 )$$
Fifth denote our triangular matrix $QWQ^-1 = T$ hence
$$det(I - lambda W) = det(I-lambda T )$$
Notice that $I - lambda T$ is also triangular with diagonal elements $1 - lambda v_1$, where $v_k$ appear on the diagonal entries of $T$. Using the fact that the determinant of a triangular matrix is nothing other than the product of it's diagonal entries, then
$$det(I - lambda W) =(1 - lambda v_1)ldots(1-lambda v_n)$$
answered Jul 25 at 14:20
Ahmad Bazzi
2,6271417
So we know (given) that there is a $Q$ such that $QWQ^-1$ is triangular. The first step is to introduce $Q$
$$det(I - lambda W) = det( Q^-1Q(I-lambda W))$$
Second let's use the property $det(AB) = det(BA)$ where $A = Q^-1$ and $B = Q(I-lambda W)$ hence
$$det(I - lambda W) = det(Q(I-lambda W) Q^-1)$$
Third, expand the above
$$det(I - lambda W) = det(QQ^-1-lambda Q W Q^-1 )$$
Notice that $QQ^-1 = I$ again so
$$det(I - lambda W) = det(I-lambda Q W Q^-1 )$$
Fifth denote our triangular matrix $QWQ^-1 = T$ hence
$$det(I - lambda W) = det(I-lambda T )$$
Notice that $I - lambda T$ is also triangular with diagonal elements $1 - lambda v_1$, where $v_k$ appear on the diagonal entries of $T$. Using the fact that the determinant of a triangular matrix is nothing other than the product of it's diagonal entries, then
$$det(I - lambda W) =(1 - lambda v_1)ldots(1-lambda v_n)$$
answered Jul 25 at 14:20
Ahmad Bazzi
2,6271417
edited Jul 25 at 14:26
answered Jul 25 at 14:20
Ahmad Bazzi
2,6271417
answered Jul 25 at 14:20
Ahmad Bazzi
2,6271417
answered Jul 25 at 14:20
Ahmad Bazzi
2,6271417
2,6271417
Thank you detailed explanation. It is cleary seen.
– v.yildirim
Jul 26 at 7:20
I'm glad it helped. If you found it useful, you could mark the answer as correct and upvote to help other viewers know that it was helpful @v.yildirim
– Ahmad Bazzi
Jul 26 at 13:31
add a comment |Â
Thank you detailed explanation. It is cleary seen.
– v.yildirim
Jul 26 at 7:20
I'm glad it helped. If you found it useful, you could mark the answer as correct and upvote to help other viewers know that it was helpful @v.yildirim
– Ahmad Bazzi
Jul 26 at 13:31
Thank you detailed explanation. It is cleary seen.
– v.yildirim
Jul 26 at 7:20
Thank you detailed explanation. It is cleary seen.
– v.yildirim
Jul 26 at 7:20
I'm glad it helped. If you found it useful, you could mark the answer as correct and upvote to help other viewers know that it was helpful @v.yildirim
– Ahmad Bazzi
Jul 26 at 13:31
I'm glad it helped. If you found it useful, you could mark the answer as correct and upvote to help other viewers know that it was helpful @v.yildirim
– Ahmad Bazzi
Jul 26 at 13:31
add a comment |Â
up vote
0
down vote
- Just think of a theorem saying that $$det(AB)=det(A)det(B)$$so we have $$det(AB)=det(BA)$$2. sure it can be proved in such way. Probably in your problem the matrix is particularly triangular but it doesn't require to remain so. To show that the key idea is that $$textthe determinant of a matrix is multiplication of all its eigenvalues$$can you finish this one?
answered Jul 25 at 15:36
Mostafa Ayaz
8,5373630
So i can see how it can change and i think it is easy to show dererminant with eigenvalues by triangularized matrix. Thank you
– v.yildirim
Jul 26 at 7:17
add a comment |Â
up vote
0
down vote
- Just think of a theorem saying that $$det(AB)=det(A)det(B)$$so we have $$det(AB)=det(BA)$$2. sure it can be proved in such way. Probably in your problem the matrix is particularly triangular but it doesn't require to remain so. To show that the key idea is that $$textthe determinant of a matrix is multiplication of all its eigenvalues$$can you finish this one?
answered Jul 25 at 15:36
Mostafa Ayaz
8,5373630
So i can see how it can change and i think it is easy to show dererminant with eigenvalues by triangularized matrix. Thank you
– v.yildirim
Jul 26 at 7:17
add a comment |Â
up vote
0
down vote
up vote
0
down vote
- Just think of a theorem saying that $$det(AB)=det(A)det(B)$$so we have $$det(AB)=det(BA)$$2. sure it can be proved in such way. Probably in your problem the matrix is particularly triangular but it doesn't require to remain so. To show that the key idea is that $$textthe determinant of a matrix is multiplication of all its eigenvalues$$can you finish this one?
answered Jul 25 at 15:36
Mostafa Ayaz
8,5373630
- Just think of a theorem saying that $$det(AB)=det(A)det(B)$$so we have $$det(AB)=det(BA)$$2. sure it can be proved in such way. Probably in your problem the matrix is particularly triangular but it doesn't require to remain so. To show that the key idea is that $$textthe determinant of a matrix is multiplication of all its eigenvalues$$can you finish this one?
answered Jul 25 at 15:36
Mostafa Ayaz
8,5373630
answered Jul 25 at 15:36
Mostafa Ayaz
8,5373630
answered Jul 25 at 15:36
Mostafa Ayaz
8,5373630
answered Jul 25 at 15:36
Mostafa Ayaz
8,5373630
8,5373630
So i can see how it can change and i think it is easy to show dererminant with eigenvalues by triangularized matrix. Thank you
– v.yildirim
Jul 26 at 7:17
add a comment |Â
So i can see how it can change and i think it is easy to show dererminant with eigenvalues by triangularized matrix. Thank you
– v.yildirim
Jul 26 at 7:17
So i can see how it can change and i think it is easy to show dererminant with eigenvalues by triangularized matrix. Thank you
– v.yildirim
Jul 26 at 7:17
So i can see how it can change and i think it is easy to show dererminant with eigenvalues by triangularized matrix. Thank you
– v.yildirim
Jul 26 at 7:17
add a comment |Â
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Note that the determinant is multiplicative i.e. $det(Q(I_n-lambda W)Q^-1)=det(Q)det(I_n-lambda W)det(Q^-1)$ and the values of the determinant commute
– Peter Melech
Jul 25 at 14:14