How do I find the bases of the Jordan Canonical Form of $C$?Jordan canonical forms and deficiency indicesFinding Jordan Canonical form for 3x3 matrixFind the Jordan form of a 4 x 4 matrixCondition on matrix to ensure nontrivial Jordan canonical formFind the Jordan canonical form of AJordan Canonical Forms of an endomorphism in $mathbbR^3$Jordan normal form which depends on parameterJordan Canonical form with 3 eigenvalues =0?Jordan canonical form of strictly upper or lower triangular matrices of even order n and rank n-1.Jordan Normal Form: Two times the same basis vector?!
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How do I find the bases of the Jordan Canonical Form of $C$?
Jordan canonical forms and deficiency indicesFinding Jordan Canonical form for 3x3 matrixFind the Jordan form of a 4 x 4 matrixCondition on matrix to ensure nontrivial Jordan canonical formFind the Jordan canonical form of AJordan Canonical Forms of an endomorphism in $mathbbR^3$Jordan normal form which depends on parameterJordan Canonical form with 3 eigenvalues =0?Jordan canonical form of strictly upper or lower triangular matrices of even order n and rank n-1.Jordan Normal Form: Two times the same basis vector?!
$begingroup$
Let $$C = left[ beginarraycccc
0 & -1 & -2 & 3 \
0 & 0 & -2 & 3 \
0 & 1 & 1 & -1 \
0 & 0 & -1 & 2
endarray right].$$ What is the Jordan canonical form of C?
We know that the characteristic polynomial
$$chi_C(lambda) = (z - 1)^4$$
So the algebraic multiplicity of $1$ is $4$.
- How do I find the geometric multiplicity of $1$?
- If the geometric multiplicity of $1$ is less than $4$, how do I find the bases that give me the Jordan canonical form?
linear-algebra matrices jordan-normal-form
edited Mar 14 at 21:02
Rodrigo de Azevedo
13.2k41960
asked Mar 14 at 5:44
Tomislav OstojichTomislav Ostojich
759718
$endgroup$
add a comment |
$begingroup$
Let $$C = left[ beginarraycccc
0 & -1 & -2 & 3 \
0 & 0 & -2 & 3 \
0 & 1 & 1 & -1 \
0 & 0 & -1 & 2
endarray right].$$ What is the Jordan canonical form of C?
We know that the characteristic polynomial
$$chi_C(lambda) = (z - 1)^4$$
So the algebraic multiplicity of $1$ is $4$.
- How do I find the geometric multiplicity of $1$?
- If the geometric multiplicity of $1$ is less than $4$, how do I find the bases that give me the Jordan canonical form?
linear-algebra matrices jordan-normal-form
edited Mar 14 at 21:02
Rodrigo de Azevedo
13.2k41960
asked Mar 14 at 5:44
Tomislav OstojichTomislav Ostojich
759718
$endgroup$
1
$begingroup$
Geo-multi = $dim (mathrm Ker ,(C - 1I)) = n - mathrm rank (C - 1I)$, hence you could find $mathrm rank (C-1I)$ via RREF or some other thing.
$endgroup$
– xbh
Mar 14 at 5:51
$begingroup$
$C$ is not invertible, so shouldn’t it’s characteristic polynomial be divisible by $z$?
$endgroup$
– Joppy
Mar 14 at 6:41
add a comment |
$begingroup$
Let $$C = left[ beginarraycccc
0 & -1 & -2 & 3 \
0 & 0 & -2 & 3 \
0 & 1 & 1 & -1 \
0 & 0 & -1 & 2
endarray right].$$ What is the Jordan canonical form of C?
We know that the characteristic polynomial
$$chi_C(lambda) = (z - 1)^4$$
So the algebraic multiplicity of $1$ is $4$.
- How do I find the geometric multiplicity of $1$?
- If the geometric multiplicity of $1$ is less than $4$, how do I find the bases that give me the Jordan canonical form?
linear-algebra matrices jordan-normal-form
edited Mar 14 at 21:02
Rodrigo de Azevedo
13.2k41960
asked Mar 14 at 5:44
Tomislav OstojichTomislav Ostojich
759718
$endgroup$
Let $$C = left[ beginarraycccc
0 & -1 & -2 & 3 \
0 & 0 & -2 & 3 \
0 & 1 & 1 & -1 \
0 & 0 & -1 & 2
endarray right].$$ What is the Jordan canonical form of C?
We know that the characteristic polynomial
$$chi_C(lambda) = (z - 1)^4$$
So the algebraic multiplicity of $1$ is $4$.
- How do I find the geometric multiplicity of $1$?
- If the geometric multiplicity of $1$ is less than $4$, how do I find the bases that give me the Jordan canonical form?
linear-algebra matrices jordan-normal-form
linear-algebra matrices jordan-normal-form
edited Mar 14 at 21:02
Rodrigo de Azevedo
13.2k41960
asked Mar 14 at 5:44
Tomislav OstojichTomislav Ostojich
759718
edited Mar 14 at 21:02
Rodrigo de Azevedo
13.2k41960
asked Mar 14 at 5:44
Tomislav OstojichTomislav Ostojich
759718
edited Mar 14 at 21:02
Rodrigo de Azevedo
13.2k41960
edited Mar 14 at 21:02
Rodrigo de Azevedo
13.2k41960
edited Mar 14 at 21:02
Rodrigo de Azevedo
13.2k41960
13.2k41960
asked Mar 14 at 5:44
Tomislav OstojichTomislav Ostojich
759718
asked Mar 14 at 5:44
Tomislav OstojichTomislav Ostojich
759718
asked Mar 14 at 5:44
Tomislav OstojichTomislav Ostojich
759718
759718
1
$begingroup$
Geo-multi = $dim (mathrm Ker ,(C - 1I)) = n - mathrm rank (C - 1I)$, hence you could find $mathrm rank (C-1I)$ via RREF or some other thing.
$endgroup$
– xbh
Mar 14 at 5:51
$begingroup$
$C$ is not invertible, so shouldn’t it’s characteristic polynomial be divisible by $z$?
$endgroup$
– Joppy
Mar 14 at 6:41
add a comment |
1
$begingroup$
Geo-multi = $dim (mathrm Ker ,(C - 1I)) = n - mathrm rank (C - 1I)$, hence you could find $mathrm rank (C-1I)$ via RREF or some other thing.
$endgroup$
– xbh
Mar 14 at 5:51
$begingroup$
$C$ is not invertible, so shouldn’t it’s characteristic polynomial be divisible by $z$?
$endgroup$
– Joppy
Mar 14 at 6:41
1
1
$begingroup$
Geo-multi = $dim (mathrm Ker ,(C - 1I)) = n - mathrm rank (C - 1I)$, hence you could find $mathrm rank (C-1I)$ via RREF or some other thing.
$endgroup$
– xbh
Mar 14 at 5:51
$begingroup$
Geo-multi = $dim (mathrm Ker ,(C - 1I)) = n - mathrm rank (C - 1I)$, hence you could find $mathrm rank (C-1I)$ via RREF or some other thing.
$endgroup$
– xbh
Mar 14 at 5:51
$begingroup$
$C$ is not invertible, so shouldn’t it’s characteristic polynomial be divisible by $z$?
$endgroup$
– Joppy
Mar 14 at 6:41
$begingroup$
$C$ is not invertible, so shouldn’t it’s characteristic polynomial be divisible by $z$?
$endgroup$
– Joppy
Mar 14 at 6:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's follow this algorithm described by Stefan Friedl.
A little work shows that the characteristic polynomial of $C$ is
$$
chi_C(t)
= t cdot (t - 1)^3
$$
which gives a table of eigenvalues
$$
beginarrayc
lambda & operatornameam_C(lambda) & operatornamegm_C(lambda) \ hline
0 & 1 & ? \
1 & 3 & ?
endarray
$$
Here, $operatornameam_C(lambda)$ is the algebraic multiplicity of $lambda$ as an eigenvalue of $C$ and $operatornamegm_C(lambda)$ is the geometric multiplicity.
Our factorization of the characteristic polynomial allowed us to fill in the algebraic multiplicities in our table. The geometric multiplicities can be computed from the definition $operatornamegm_C(lambda)=operatornamenullity(lambdacdot I-C)$. In our case, we have
$$
beginarrayc
lambda & operatornameam_C(lambda) & operatornamegm_C(lambda) \ hline
0 & 1 & 1 \
1 & 3 & 1
endarray
$$
Note that $operatornamegm_C(0)$ can also be quickly inferred from the inequality $1leqoperatornamegm_C(0)leqoperatornameam_C(0)=1$.
At this stage, we can infer the Jordan form $J$ of $C$. Recall the interpretations of the multiplicities of the eigenvalues as
beginalign*
operatornameam_C(lambda) &= textnumber of $lambda$'s on the diagonal of $J$ \
operatornamegm_C(lambda) &= textsize of the largest Jordan block corresponding to $lambda$ inside $J$
endalign*
In general, knowing these multiplicities is not enough to infer $J$. However, in our case we can see that
$$
J=left[beginarrayr
0 & 0 & 0 & 0 \
hline
0 & 1 & 1 & 0 \
0 & 0 & 1 & 1 \
0 & 0 & 0 & 1
endarrayright]
$$
Now, we proceed to compute the change of basis matrix $P$. The easiest way to start is to note that
$$
E_0 = operatornameSpanlangle1, 0, 0, 0rangle
$$
This gives our first column of $P$, so
$$
P=
left[beginarrayrrrr
1 & ? & ? & ? \
0 & ? & ? & ? \
0 & ? & ? & ? \
0 & ? & ? & ?
endarrayright]
$$
Now, to build the other three columns, we compute the numbers
$$
d_k=operatornamenullity((lambdacdot I-C)^k)-operatornamenullity((lambdacdot I-C)^k-1)
$$
for $1leq kleqoperatornamegm_C(lambda)$ where $lambda=1$. For us, these numbers turn out to be
beginalign*
d_1 &= 1 & d_2 &= 1 & d_3 &= 1
endalign*
We now take these numbers and build a diagram of empty boxes
$$
beginarrayc
Box\ Box\ Box
endarray
$$
The algorithm we're following demands that we start at the bottom of this diagram and fill the boxes in row $k$ with linearly independent vectors that belong to $operatornameNull((lambdacdot I-C)^k)$ but not $operatornameNull((lambdacdot I-C)^k-1)$. Once a box in the diagram is filled with a vector $vecv$, the box immediately above it is filled with $(lambdacdot I-C)vecv$.
In our situation, we have
beginalign*
(I-C)^2 &= left[beginarrayrrrr
1 & 0 & 1 & -1 \
0 & -1 & -1 & 2 \
0 & -1 & -1 & 2 \
0 & -1 & -1 & 2
endarrayright] & (I-C)^3 &= left[beginarrayrrrr
1 & 0 & 1 & -1 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0
endarrayright]
endalign*
We easily see that $langle0,1,0,0rangleinoperatornameNull((I-C)^3)$ but $langle0,1,0,0ranglenotinoperatornameNull((I-C)^2)$. This allows us to fill out our diagram
$$
beginarrayc
fbox$leftlangle0,,-1,,-1,,-1rightrangle$\
fbox$leftlangle-1,,-1,,1,,0rightrangle$ \
fbox$leftlangle0,,1,,0,,0rightrangle$
endarray
$$
This defines our matrix $P$ as
$$
P = left[beginarrayrrrr
1 & 0 & -1 & 0 \
0 & -1 & -1 & 1 \
0 & -1 & 1 & 0 \
0 & -1 & 0 & 0
endarrayright]
$$
We can verify ourselves that this is indeed correct
$$
oversetCleft[beginarrayrrrr
0 & -1 & -2 & 3 \
0 & 0 & -2 & 3 \
0 & 1 & 1 & -1 \
0 & 0 & -1 & 2
endarrayright]
=
oversetPleft[beginarrayrrrr
1 & 0 & -1 & 0 \
0 & -1 & -1 & 1 \
0 & -1 & 1 & 0 \
0 & -1 & 0 & 0
endarrayright]
oversetJleft[beginarrayr
0 & 0 & 0 & 0 \
hline
0 & 1 & 1 & 0 \
0 & 0 & 1 & 1 \
0 & 0 & 0 & 1
endarrayright]
oversetP^-1left[beginarrayrrrr
1 & 0 & 1 & -1 \
0 & 0 & 0 & -1 \
0 & 0 & 1 & -1 \
0 & 1 & 1 & -2
endarrayright]
$$
answered Mar 14 at 23:53
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
$endgroup$
$begingroup$
"In general, knowing these multiplicities is not enough to infer 𝐽." What do we do in that case? I would greatly appreciate you if you could tell me.
$endgroup$
– Tomislav Ostojich
Mar 17 at 3:17
add a comment |
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's follow this algorithm described by Stefan Friedl.
A little work shows that the characteristic polynomial of $C$ is
$$
chi_C(t)
= t cdot (t - 1)^3
$$
which gives a table of eigenvalues
$$
beginarrayc
lambda & operatornameam_C(lambda) & operatornamegm_C(lambda) \ hline
0 & 1 & ? \
1 & 3 & ?
endarray
$$
Here, $operatornameam_C(lambda)$ is the algebraic multiplicity of $lambda$ as an eigenvalue of $C$ and $operatornamegm_C(lambda)$ is the geometric multiplicity.
Our factorization of the characteristic polynomial allowed us to fill in the algebraic multiplicities in our table. The geometric multiplicities can be computed from the definition $operatornamegm_C(lambda)=operatornamenullity(lambdacdot I-C)$. In our case, we have
$$
beginarrayc
lambda & operatornameam_C(lambda) & operatornamegm_C(lambda) \ hline
0 & 1 & 1 \
1 & 3 & 1
endarray
$$
Note that $operatornamegm_C(0)$ can also be quickly inferred from the inequality $1leqoperatornamegm_C(0)leqoperatornameam_C(0)=1$.
At this stage, we can infer the Jordan form $J$ of $C$. Recall the interpretations of the multiplicities of the eigenvalues as
beginalign*
operatornameam_C(lambda) &= textnumber of $lambda$'s on the diagonal of $J$ \
operatornamegm_C(lambda) &= textsize of the largest Jordan block corresponding to $lambda$ inside $J$
endalign*
In general, knowing these multiplicities is not enough to infer $J$. However, in our case we can see that
$$
J=left[beginarrayr
0 & 0 & 0 & 0 \
hline
0 & 1 & 1 & 0 \
0 & 0 & 1 & 1 \
0 & 0 & 0 & 1
endarrayright]
$$
Now, we proceed to compute the change of basis matrix $P$. The easiest way to start is to note that
$$
E_0 = operatornameSpanlangle1, 0, 0, 0rangle
$$
This gives our first column of $P$, so
$$
P=
left[beginarrayrrrr
1 & ? & ? & ? \
0 & ? & ? & ? \
0 & ? & ? & ? \
0 & ? & ? & ?
endarrayright]
$$
Now, to build the other three columns, we compute the numbers
$$
d_k=operatornamenullity((lambdacdot I-C)^k)-operatornamenullity((lambdacdot I-C)^k-1)
$$
for $1leq kleqoperatornamegm_C(lambda)$ where $lambda=1$. For us, these numbers turn out to be
beginalign*
d_1 &= 1 & d_2 &= 1 & d_3 &= 1
endalign*
We now take these numbers and build a diagram of empty boxes
$$
beginarrayc
Box\ Box\ Box
endarray
$$
The algorithm we're following demands that we start at the bottom of this diagram and fill the boxes in row $k$ with linearly independent vectors that belong to $operatornameNull((lambdacdot I-C)^k)$ but not $operatornameNull((lambdacdot I-C)^k-1)$. Once a box in the diagram is filled with a vector $vecv$, the box immediately above it is filled with $(lambdacdot I-C)vecv$.
In our situation, we have
beginalign*
(I-C)^2 &= left[beginarrayrrrr
1 & 0 & 1 & -1 \
0 & -1 & -1 & 2 \
0 & -1 & -1 & 2 \
0 & -1 & -1 & 2
endarrayright] & (I-C)^3 &= left[beginarrayrrrr
1 & 0 & 1 & -1 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0
endarrayright]
endalign*
We easily see that $langle0,1,0,0rangleinoperatornameNull((I-C)^3)$ but $langle0,1,0,0ranglenotinoperatornameNull((I-C)^2)$. This allows us to fill out our diagram
$$
beginarrayc
fbox$leftlangle0,,-1,,-1,,-1rightrangle$\
fbox$leftlangle-1,,-1,,1,,0rightrangle$ \
fbox$leftlangle0,,1,,0,,0rightrangle$
endarray
$$
This defines our matrix $P$ as
$$
P = left[beginarrayrrrr
1 & 0 & -1 & 0 \
0 & -1 & -1 & 1 \
0 & -1 & 1 & 0 \
0 & -1 & 0 & 0
endarrayright]
$$
We can verify ourselves that this is indeed correct
$$
oversetCleft[beginarrayrrrr
0 & -1 & -2 & 3 \
0 & 0 & -2 & 3 \
0 & 1 & 1 & -1 \
0 & 0 & -1 & 2
endarrayright]
=
oversetPleft[beginarrayrrrr
1 & 0 & -1 & 0 \
0 & -1 & -1 & 1 \
0 & -1 & 1 & 0 \
0 & -1 & 0 & 0
endarrayright]
oversetJleft[beginarrayr
0 & 0 & 0 & 0 \
hline
0 & 1 & 1 & 0 \
0 & 0 & 1 & 1 \
0 & 0 & 0 & 1
endarrayright]
oversetP^-1left[beginarrayrrrr
1 & 0 & 1 & -1 \
0 & 0 & 0 & -1 \
0 & 0 & 1 & -1 \
0 & 1 & 1 & -2
endarrayright]
$$
answered Mar 14 at 23:53
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
$endgroup$
$begingroup$
"In general, knowing these multiplicities is not enough to infer 𝐽." What do we do in that case? I would greatly appreciate you if you could tell me.
$endgroup$
– Tomislav Ostojich
Mar 17 at 3:17
add a comment |
$begingroup$
Let's follow this algorithm described by Stefan Friedl.
A little work shows that the characteristic polynomial of $C$ is
$$
chi_C(t)
= t cdot (t - 1)^3
$$
which gives a table of eigenvalues
$$
beginarrayc
lambda & operatornameam_C(lambda) & operatornamegm_C(lambda) \ hline
0 & 1 & ? \
1 & 3 & ?
endarray
$$
Here, $operatornameam_C(lambda)$ is the algebraic multiplicity of $lambda$ as an eigenvalue of $C$ and $operatornamegm_C(lambda)$ is the geometric multiplicity.
Our factorization of the characteristic polynomial allowed us to fill in the algebraic multiplicities in our table. The geometric multiplicities can be computed from the definition $operatornamegm_C(lambda)=operatornamenullity(lambdacdot I-C)$. In our case, we have
$$
beginarrayc
lambda & operatornameam_C(lambda) & operatornamegm_C(lambda) \ hline
0 & 1 & 1 \
1 & 3 & 1
endarray
$$
Note that $operatornamegm_C(0)$ can also be quickly inferred from the inequality $1leqoperatornamegm_C(0)leqoperatornameam_C(0)=1$.
At this stage, we can infer the Jordan form $J$ of $C$. Recall the interpretations of the multiplicities of the eigenvalues as
beginalign*
operatornameam_C(lambda) &= textnumber of $lambda$'s on the diagonal of $J$ \
operatornamegm_C(lambda) &= textsize of the largest Jordan block corresponding to $lambda$ inside $J$
endalign*
In general, knowing these multiplicities is not enough to infer $J$. However, in our case we can see that
$$
J=left[beginarrayr
0 & 0 & 0 & 0 \
hline
0 & 1 & 1 & 0 \
0 & 0 & 1 & 1 \
0 & 0 & 0 & 1
endarrayright]
$$
Now, we proceed to compute the change of basis matrix $P$. The easiest way to start is to note that
$$
E_0 = operatornameSpanlangle1, 0, 0, 0rangle
$$
This gives our first column of $P$, so
$$
P=
left[beginarrayrrrr
1 & ? & ? & ? \
0 & ? & ? & ? \
0 & ? & ? & ? \
0 & ? & ? & ?
endarrayright]
$$
Now, to build the other three columns, we compute the numbers
$$
d_k=operatornamenullity((lambdacdot I-C)^k)-operatornamenullity((lambdacdot I-C)^k-1)
$$
for $1leq kleqoperatornamegm_C(lambda)$ where $lambda=1$. For us, these numbers turn out to be
beginalign*
d_1 &= 1 & d_2 &= 1 & d_3 &= 1
endalign*
We now take these numbers and build a diagram of empty boxes
$$
beginarrayc
Box\ Box\ Box
endarray
$$
The algorithm we're following demands that we start at the bottom of this diagram and fill the boxes in row $k$ with linearly independent vectors that belong to $operatornameNull((lambdacdot I-C)^k)$ but not $operatornameNull((lambdacdot I-C)^k-1)$. Once a box in the diagram is filled with a vector $vecv$, the box immediately above it is filled with $(lambdacdot I-C)vecv$.
In our situation, we have
beginalign*
(I-C)^2 &= left[beginarrayrrrr
1 & 0 & 1 & -1 \
0 & -1 & -1 & 2 \
0 & -1 & -1 & 2 \
0 & -1 & -1 & 2
endarrayright] & (I-C)^3 &= left[beginarrayrrrr
1 & 0 & 1 & -1 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0
endarrayright]
endalign*
We easily see that $langle0,1,0,0rangleinoperatornameNull((I-C)^3)$ but $langle0,1,0,0ranglenotinoperatornameNull((I-C)^2)$. This allows us to fill out our diagram
$$
beginarrayc
fbox$leftlangle0,,-1,,-1,,-1rightrangle$\
fbox$leftlangle-1,,-1,,1,,0rightrangle$ \
fbox$leftlangle0,,1,,0,,0rightrangle$
endarray
$$
This defines our matrix $P$ as
$$
P = left[beginarrayrrrr
1 & 0 & -1 & 0 \
0 & -1 & -1 & 1 \
0 & -1 & 1 & 0 \
0 & -1 & 0 & 0
endarrayright]
$$
We can verify ourselves that this is indeed correct
$$
oversetCleft[beginarrayrrrr
0 & -1 & -2 & 3 \
0 & 0 & -2 & 3 \
0 & 1 & 1 & -1 \
0 & 0 & -1 & 2
endarrayright]
=
oversetPleft[beginarrayrrrr
1 & 0 & -1 & 0 \
0 & -1 & -1 & 1 \
0 & -1 & 1 & 0 \
0 & -1 & 0 & 0
endarrayright]
oversetJleft[beginarrayr
0 & 0 & 0 & 0 \
hline
0 & 1 & 1 & 0 \
0 & 0 & 1 & 1 \
0 & 0 & 0 & 1
endarrayright]
oversetP^-1left[beginarrayrrrr
1 & 0 & 1 & -1 \
0 & 0 & 0 & -1 \
0 & 0 & 1 & -1 \
0 & 1 & 1 & -2
endarrayright]
$$
answered Mar 14 at 23:53
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
$endgroup$
$begingroup$
"In general, knowing these multiplicities is not enough to infer 𝐽." What do we do in that case? I would greatly appreciate you if you could tell me.
$endgroup$
– Tomislav Ostojich
Mar 17 at 3:17
add a comment |
$begingroup$
Let's follow this algorithm described by Stefan Friedl.
A little work shows that the characteristic polynomial of $C$ is
$$
chi_C(t)
= t cdot (t - 1)^3
$$
which gives a table of eigenvalues
$$
beginarrayc
lambda & operatornameam_C(lambda) & operatornamegm_C(lambda) \ hline
0 & 1 & ? \
1 & 3 & ?
endarray
$$
Here, $operatornameam_C(lambda)$ is the algebraic multiplicity of $lambda$ as an eigenvalue of $C$ and $operatornamegm_C(lambda)$ is the geometric multiplicity.
Our factorization of the characteristic polynomial allowed us to fill in the algebraic multiplicities in our table. The geometric multiplicities can be computed from the definition $operatornamegm_C(lambda)=operatornamenullity(lambdacdot I-C)$. In our case, we have
$$
beginarrayc
lambda & operatornameam_C(lambda) & operatornamegm_C(lambda) \ hline
0 & 1 & 1 \
1 & 3 & 1
endarray
$$
Note that $operatornamegm_C(0)$ can also be quickly inferred from the inequality $1leqoperatornamegm_C(0)leqoperatornameam_C(0)=1$.
At this stage, we can infer the Jordan form $J$ of $C$. Recall the interpretations of the multiplicities of the eigenvalues as
beginalign*
operatornameam_C(lambda) &= textnumber of $lambda$'s on the diagonal of $J$ \
operatornamegm_C(lambda) &= textsize of the largest Jordan block corresponding to $lambda$ inside $J$
endalign*
In general, knowing these multiplicities is not enough to infer $J$. However, in our case we can see that
$$
J=left[beginarrayr
0 & 0 & 0 & 0 \
hline
0 & 1 & 1 & 0 \
0 & 0 & 1 & 1 \
0 & 0 & 0 & 1
endarrayright]
$$
Now, we proceed to compute the change of basis matrix $P$. The easiest way to start is to note that
$$
E_0 = operatornameSpanlangle1, 0, 0, 0rangle
$$
This gives our first column of $P$, so
$$
P=
left[beginarrayrrrr
1 & ? & ? & ? \
0 & ? & ? & ? \
0 & ? & ? & ? \
0 & ? & ? & ?
endarrayright]
$$
Now, to build the other three columns, we compute the numbers
$$
d_k=operatornamenullity((lambdacdot I-C)^k)-operatornamenullity((lambdacdot I-C)^k-1)
$$
for $1leq kleqoperatornamegm_C(lambda)$ where $lambda=1$. For us, these numbers turn out to be
beginalign*
d_1 &= 1 & d_2 &= 1 & d_3 &= 1
endalign*
We now take these numbers and build a diagram of empty boxes
$$
beginarrayc
Box\ Box\ Box
endarray
$$
The algorithm we're following demands that we start at the bottom of this diagram and fill the boxes in row $k$ with linearly independent vectors that belong to $operatornameNull((lambdacdot I-C)^k)$ but not $operatornameNull((lambdacdot I-C)^k-1)$. Once a box in the diagram is filled with a vector $vecv$, the box immediately above it is filled with $(lambdacdot I-C)vecv$.
In our situation, we have
beginalign*
(I-C)^2 &= left[beginarrayrrrr
1 & 0 & 1 & -1 \
0 & -1 & -1 & 2 \
0 & -1 & -1 & 2 \
0 & -1 & -1 & 2
endarrayright] & (I-C)^3 &= left[beginarrayrrrr
1 & 0 & 1 & -1 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0
endarrayright]
endalign*
We easily see that $langle0,1,0,0rangleinoperatornameNull((I-C)^3)$ but $langle0,1,0,0ranglenotinoperatornameNull((I-C)^2)$. This allows us to fill out our diagram
$$
beginarrayc
fbox$leftlangle0,,-1,,-1,,-1rightrangle$\
fbox$leftlangle-1,,-1,,1,,0rightrangle$ \
fbox$leftlangle0,,1,,0,,0rightrangle$
endarray
$$
This defines our matrix $P$ as
$$
P = left[beginarrayrrrr
1 & 0 & -1 & 0 \
0 & -1 & -1 & 1 \
0 & -1 & 1 & 0 \
0 & -1 & 0 & 0
endarrayright]
$$
We can verify ourselves that this is indeed correct
$$
oversetCleft[beginarrayrrrr
0 & -1 & -2 & 3 \
0 & 0 & -2 & 3 \
0 & 1 & 1 & -1 \
0 & 0 & -1 & 2
endarrayright]
=
oversetPleft[beginarrayrrrr
1 & 0 & -1 & 0 \
0 & -1 & -1 & 1 \
0 & -1 & 1 & 0 \
0 & -1 & 0 & 0
endarrayright]
oversetJleft[beginarrayr
0 & 0 & 0 & 0 \
hline
0 & 1 & 1 & 0 \
0 & 0 & 1 & 1 \
0 & 0 & 0 & 1
endarrayright]
oversetP^-1left[beginarrayrrrr
1 & 0 & 1 & -1 \
0 & 0 & 0 & -1 \
0 & 0 & 1 & -1 \
0 & 1 & 1 & -2
endarrayright]
$$
answered Mar 14 at 23:53
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
$endgroup$
Let's follow this algorithm described by Stefan Friedl.
A little work shows that the characteristic polynomial of $C$ is
$$
chi_C(t)
= t cdot (t - 1)^3
$$
which gives a table of eigenvalues
$$
beginarrayc
lambda & operatornameam_C(lambda) & operatornamegm_C(lambda) \ hline
0 & 1 & ? \
1 & 3 & ?
endarray
$$
Here, $operatornameam_C(lambda)$ is the algebraic multiplicity of $lambda$ as an eigenvalue of $C$ and $operatornamegm_C(lambda)$ is the geometric multiplicity.
Our factorization of the characteristic polynomial allowed us to fill in the algebraic multiplicities in our table. The geometric multiplicities can be computed from the definition $operatornamegm_C(lambda)=operatornamenullity(lambdacdot I-C)$. In our case, we have
$$
beginarrayc
lambda & operatornameam_C(lambda) & operatornamegm_C(lambda) \ hline
0 & 1 & 1 \
1 & 3 & 1
endarray
$$
Note that $operatornamegm_C(0)$ can also be quickly inferred from the inequality $1leqoperatornamegm_C(0)leqoperatornameam_C(0)=1$.
At this stage, we can infer the Jordan form $J$ of $C$. Recall the interpretations of the multiplicities of the eigenvalues as
beginalign*
operatornameam_C(lambda) &= textnumber of $lambda$'s on the diagonal of $J$ \
operatornamegm_C(lambda) &= textsize of the largest Jordan block corresponding to $lambda$ inside $J$
endalign*
In general, knowing these multiplicities is not enough to infer $J$. However, in our case we can see that
$$
J=left[beginarrayr
0 & 0 & 0 & 0 \
hline
0 & 1 & 1 & 0 \
0 & 0 & 1 & 1 \
0 & 0 & 0 & 1
endarrayright]
$$
Now, we proceed to compute the change of basis matrix $P$. The easiest way to start is to note that
$$
E_0 = operatornameSpanlangle1, 0, 0, 0rangle
$$
This gives our first column of $P$, so
$$
P=
left[beginarrayrrrr
1 & ? & ? & ? \
0 & ? & ? & ? \
0 & ? & ? & ? \
0 & ? & ? & ?
endarrayright]
$$
Now, to build the other three columns, we compute the numbers
$$
d_k=operatornamenullity((lambdacdot I-C)^k)-operatornamenullity((lambdacdot I-C)^k-1)
$$
for $1leq kleqoperatornamegm_C(lambda)$ where $lambda=1$. For us, these numbers turn out to be
beginalign*
d_1 &= 1 & d_2 &= 1 & d_3 &= 1
endalign*
We now take these numbers and build a diagram of empty boxes
$$
beginarrayc
Box\ Box\ Box
endarray
$$
The algorithm we're following demands that we start at the bottom of this diagram and fill the boxes in row $k$ with linearly independent vectors that belong to $operatornameNull((lambdacdot I-C)^k)$ but not $operatornameNull((lambdacdot I-C)^k-1)$. Once a box in the diagram is filled with a vector $vecv$, the box immediately above it is filled with $(lambdacdot I-C)vecv$.
In our situation, we have
beginalign*
(I-C)^2 &= left[beginarrayrrrr
1 & 0 & 1 & -1 \
0 & -1 & -1 & 2 \
0 & -1 & -1 & 2 \
0 & -1 & -1 & 2
endarrayright] & (I-C)^3 &= left[beginarrayrrrr
1 & 0 & 1 & -1 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0
endarrayright]
endalign*
We easily see that $langle0,1,0,0rangleinoperatornameNull((I-C)^3)$ but $langle0,1,0,0ranglenotinoperatornameNull((I-C)^2)$. This allows us to fill out our diagram
$$
beginarrayc
fbox$leftlangle0,,-1,,-1,,-1rightrangle$\
fbox$leftlangle-1,,-1,,1,,0rightrangle$ \
fbox$leftlangle0,,1,,0,,0rightrangle$
endarray
$$
This defines our matrix $P$ as
$$
P = left[beginarrayrrrr
1 & 0 & -1 & 0 \
0 & -1 & -1 & 1 \
0 & -1 & 1 & 0 \
0 & -1 & 0 & 0
endarrayright]
$$
We can verify ourselves that this is indeed correct
$$
oversetCleft[beginarrayrrrr
0 & -1 & -2 & 3 \
0 & 0 & -2 & 3 \
0 & 1 & 1 & -1 \
0 & 0 & -1 & 2
endarrayright]
=
oversetPleft[beginarrayrrrr
1 & 0 & -1 & 0 \
0 & -1 & -1 & 1 \
0 & -1 & 1 & 0 \
0 & -1 & 0 & 0
endarrayright]
oversetJleft[beginarrayr
0 & 0 & 0 & 0 \
hline
0 & 1 & 1 & 0 \
0 & 0 & 1 & 1 \
0 & 0 & 0 & 1
endarrayright]
oversetP^-1left[beginarrayrrrr
1 & 0 & 1 & -1 \
0 & 0 & 0 & -1 \
0 & 0 & 1 & -1 \
0 & 1 & 1 & -2
endarrayright]
$$
answered Mar 14 at 23:53
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
answered Mar 14 at 23:53
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
answered Mar 14 at 23:53
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
answered Mar 14 at 23:53
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
21.8k42959
$begingroup$
"In general, knowing these multiplicities is not enough to infer 𝐽." What do we do in that case? I would greatly appreciate you if you could tell me.
$endgroup$
– Tomislav Ostojich
Mar 17 at 3:17
add a comment |
$begingroup$
"In general, knowing these multiplicities is not enough to infer 𝐽." What do we do in that case? I would greatly appreciate you if you could tell me.
$endgroup$
– Tomislav Ostojich
Mar 17 at 3:17
$begingroup$
"In general, knowing these multiplicities is not enough to infer 𝐽." What do we do in that case? I would greatly appreciate you if you could tell me.
$endgroup$
– Tomislav Ostojich
Mar 17 at 3:17
$begingroup$
"In general, knowing these multiplicities is not enough to infer 𝐽." What do we do in that case? I would greatly appreciate you if you could tell me.
$endgroup$
– Tomislav Ostojich
Mar 17 at 3:17
add a comment |
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1
$begingroup$
Geo-multi = $dim (mathrm Ker ,(C - 1I)) = n - mathrm rank (C - 1I)$, hence you could find $mathrm rank (C-1I)$ via RREF or some other thing.
$endgroup$
– xbh
Mar 14 at 5:51
$begingroup$
$C$ is not invertible, so shouldn’t it’s characteristic polynomial be divisible by $z$?
$endgroup$
– Joppy
Mar 14 at 6:41