Mixing
Excercise dealing with subgroups of $text{GL}_n(K)$
$begingroup$
$U_n=$$left( begin{array}{rrrr}
1 & * & cdots & * \
0 & ddots & * & vdots \
vdots & 0 & ddots & * \
0 & cdots & 0 & 1 \
end{array}right) $$T_n=$$left( begin{array}{rrrr}
* & * & cdots & * \
0 & ddots & * & vdots \
vdots & 0 & ddots & * \
0 & cdots & 0 & * \
end{array}right) $$D_n=$
$left( begin{array}{rrrr}
* & 0 & cdots & 0 \
0 & ddots & 0 & vdots \
vdots & 0 & ddots & 0 \
0 & cdots & 0 & * \
end{array}right) $
$U_n,T_n,D_n$ are all sets which describe Matrices of the forms above. Furthermore they are all subsets of $text{GL}_n(K)$. I have to Show that a Matrix $A$ of the set $T_n$ can be describes as a product of a Matrix $U_A$ from the set $U_n$ with another Matrix $D_A$ from the set $D_n$.
Is there a way to prove it by induction?
I have looked at the Problem for $n=2$
$begin{pmatrix}
*_{11} & *_{12}\
0 & *_{22} \
end{pmatrix} $
=$begin{pmatrix}
1 & *_{12}*_{22}^{-1}\
0 & 1 \
end{pmatrix} $$begin{pmatrix}
*_{11} & 0\
0 & *_{22} \
end{pmatrix} $
I know that I have to set the trace-elements of $D_A$ equal to the trace Elements of $A$.
And from my Observation my guess for the Elements which are not within the trace of $U_A$ (i.e. they are not $1$) is that they can be calculated inductively by the element in $A$ which is in the same Position as the element we want to calculate and the inverse of the element which is below the element we want to calculate.
How can I formalize this thought, and how can I then prove this property inductively. Would appreciate an Approach which does not make use of bloxmatrices.
linear-algebra matrices
asked Jan 28 at 14:54
RM777RM777
38312
$endgroup$
add a comment |
$begingroup$
$U_n=$$left( begin{array}{rrrr}
1 & * & cdots & * \
0 & ddots & * & vdots \
vdots & 0 & ddots & * \
0 & cdots & 0 & 1 \
end{array}right) $$T_n=$$left( begin{array}{rrrr}
* & * & cdots & * \
0 & ddots & * & vdots \
vdots & 0 & ddots & * \
0 & cdots & 0 & * \
end{array}right) $$D_n=$
$left( begin{array}{rrrr}
* & 0 & cdots & 0 \
0 & ddots & 0 & vdots \
vdots & 0 & ddots & 0 \
0 & cdots & 0 & * \
end{array}right) $
$U_n,T_n,D_n$ are all sets which describe Matrices of the forms above. Furthermore they are all subsets of $text{GL}_n(K)$. I have to Show that a Matrix $A$ of the set $T_n$ can be describes as a product of a Matrix $U_A$ from the set $U_n$ with another Matrix $D_A$ from the set $D_n$.
Is there a way to prove it by induction?
I have looked at the Problem for $n=2$
$begin{pmatrix}
*_{11} & *_{12}\
0 & *_{22} \
end{pmatrix} $
=$begin{pmatrix}
1 & *_{12}*_{22}^{-1}\
0 & 1 \
end{pmatrix} $$begin{pmatrix}
*_{11} & 0\
0 & *_{22} \
end{pmatrix} $
I know that I have to set the trace-elements of $D_A$ equal to the trace Elements of $A$.
And from my Observation my guess for the Elements which are not within the trace of $U_A$ (i.e. they are not $1$) is that they can be calculated inductively by the element in $A$ which is in the same Position as the element we want to calculate and the inverse of the element which is below the element we want to calculate.
How can I formalize this thought, and how can I then prove this property inductively. Would appreciate an Approach which does not make use of bloxmatrices.
linear-algebra matrices
asked Jan 28 at 14:54
RM777RM777
38312
$endgroup$
$begingroup$
do you know quotients of groups by subsets? I would use that, since as soon as you can proof that $D_n$ is a normal subgroup, you could just devide it out and be done, (respectively, you could do that either way, just a little less nice)
$endgroup$
– Enkidu
Jan 28 at 15:04
$begingroup$
I have heard something About normal Groups we have defined them as subgroups of a Group where the left coset is Always equal to the Right coset. They create a Quotient set
$endgroup$
– RM777
Jan 28 at 15:07
$begingroup$
yep, then just use that $D_n subset T_n$, kill it, and proof that you can identify this with $U_n$ (and as you actually do not need a groupstructure on the quotient, you can ignore the reauirements on the cosets).
$endgroup$
– Enkidu
Jan 28 at 15:08
add a comment |
$begingroup$
$U_n=$$left( begin{array}{rrrr}
1 & * & cdots & * \
0 & ddots & * & vdots \
vdots & 0 & ddots & * \
0 & cdots & 0 & 1 \
end{array}right) $$T_n=$$left( begin{array}{rrrr}
* & * & cdots & * \
0 & ddots & * & vdots \
vdots & 0 & ddots & * \
0 & cdots & 0 & * \
end{array}right) $$D_n=$
$left( begin{array}{rrrr}
* & 0 & cdots & 0 \
0 & ddots & 0 & vdots \
vdots & 0 & ddots & 0 \
0 & cdots & 0 & * \
end{array}right) $
$U_n,T_n,D_n$ are all sets which describe Matrices of the forms above. Furthermore they are all subsets of $text{GL}_n(K)$. I have to Show that a Matrix $A$ of the set $T_n$ can be describes as a product of a Matrix $U_A$ from the set $U_n$ with another Matrix $D_A$ from the set $D_n$.
Is there a way to prove it by induction?
I have looked at the Problem for $n=2$
$begin{pmatrix}
*_{11} & *_{12}\
0 & *_{22} \
end{pmatrix} $
=$begin{pmatrix}
1 & *_{12}*_{22}^{-1}\
0 & 1 \
end{pmatrix} $$begin{pmatrix}
*_{11} & 0\
0 & *_{22} \
end{pmatrix} $
I know that I have to set the trace-elements of $D_A$ equal to the trace Elements of $A$.
And from my Observation my guess for the Elements which are not within the trace of $U_A$ (i.e. they are not $1$) is that they can be calculated inductively by the element in $A$ which is in the same Position as the element we want to calculate and the inverse of the element which is below the element we want to calculate.
How can I formalize this thought, and how can I then prove this property inductively. Would appreciate an Approach which does not make use of bloxmatrices.
linear-algebra matrices
asked Jan 28 at 14:54
RM777RM777
38312
$endgroup$
$U_n=$$left( begin{array}{rrrr}
1 & * & cdots & * \
0 & ddots & * & vdots \
vdots & 0 & ddots & * \
0 & cdots & 0 & 1 \
end{array}right) $$T_n=$$left( begin{array}{rrrr}
* & * & cdots & * \
0 & ddots & * & vdots \
vdots & 0 & ddots & * \
0 & cdots & 0 & * \
end{array}right) $$D_n=$
$left( begin{array}{rrrr}
* & 0 & cdots & 0 \
0 & ddots & 0 & vdots \
vdots & 0 & ddots & 0 \
0 & cdots & 0 & * \
end{array}right) $
$U_n,T_n,D_n$ are all sets which describe Matrices of the forms above. Furthermore they are all subsets of $text{GL}_n(K)$. I have to Show that a Matrix $A$ of the set $T_n$ can be describes as a product of a Matrix $U_A$ from the set $U_n$ with another Matrix $D_A$ from the set $D_n$.
Is there a way to prove it by induction?
I have looked at the Problem for $n=2$
$begin{pmatrix}
*_{11} & *_{12}\
0 & *_{22} \
end{pmatrix} $
=$begin{pmatrix}
1 & *_{12}*_{22}^{-1}\
0 & 1 \
end{pmatrix} $$begin{pmatrix}
*_{11} & 0\
0 & *_{22} \
end{pmatrix} $
I know that I have to set the trace-elements of $D_A$ equal to the trace Elements of $A$.
And from my Observation my guess for the Elements which are not within the trace of $U_A$ (i.e. they are not $1$) is that they can be calculated inductively by the element in $A$ which is in the same Position as the element we want to calculate and the inverse of the element which is below the element we want to calculate.
How can I formalize this thought, and how can I then prove this property inductively. Would appreciate an Approach which does not make use of bloxmatrices.
linear-algebra matrices
linear-algebra matrices
asked Jan 28 at 14:54
RM777RM777
38312
asked Jan 28 at 14:54
RM777RM777
38312
asked Jan 28 at 14:54
RM777RM777
38312
asked Jan 28 at 14:54
RM777RM777
38312
asked Jan 28 at 14:54
RM777RM777
38312
38312
$begingroup$
do you know quotients of groups by subsets? I would use that, since as soon as you can proof that $D_n$ is a normal subgroup, you could just devide it out and be done, (respectively, you could do that either way, just a little less nice)
$endgroup$
– Enkidu
Jan 28 at 15:04
$begingroup$
I have heard something About normal Groups we have defined them as subgroups of a Group where the left coset is Always equal to the Right coset. They create a Quotient set
$endgroup$
– RM777
Jan 28 at 15:07
$begingroup$
yep, then just use that $D_n subset T_n$, kill it, and proof that you can identify this with $U_n$ (and as you actually do not need a groupstructure on the quotient, you can ignore the reauirements on the cosets).
$endgroup$
– Enkidu
Jan 28 at 15:08
add a comment |
$begingroup$
do you know quotients of groups by subsets? I would use that, since as soon as you can proof that $D_n$ is a normal subgroup, you could just devide it out and be done, (respectively, you could do that either way, just a little less nice)
$endgroup$
– Enkidu
Jan 28 at 15:04
$begingroup$
I have heard something About normal Groups we have defined them as subgroups of a Group where the left coset is Always equal to the Right coset. They create a Quotient set
$endgroup$
– RM777
Jan 28 at 15:07
$begingroup$
yep, then just use that $D_n subset T_n$, kill it, and proof that you can identify this with $U_n$ (and as you actually do not need a groupstructure on the quotient, you can ignore the reauirements on the cosets).
$endgroup$
– Enkidu
Jan 28 at 15:08
$begingroup$
do you know quotients of groups by subsets? I would use that, since as soon as you can proof that $D_n$ is a normal subgroup, you could just devide it out and be done, (respectively, you could do that either way, just a little less nice)
$endgroup$
– Enkidu
Jan 28 at 15:04
$begingroup$
do you know quotients of groups by subsets? I would use that, since as soon as you can proof that $D_n$ is a normal subgroup, you could just devide it out and be done, (respectively, you could do that either way, just a little less nice)
$endgroup$
– Enkidu
Jan 28 at 15:04
$begingroup$
I have heard something About normal Groups we have defined them as subgroups of a Group where the left coset is Always equal to the Right coset. They create a Quotient set
$endgroup$
– RM777
Jan 28 at 15:07
$begingroup$
I have heard something About normal Groups we have defined them as subgroups of a Group where the left coset is Always equal to the Right coset. They create a Quotient set
$endgroup$
– RM777
Jan 28 at 15:07
$begingroup$
yep, then just use that $D_n subset T_n$, kill it, and proof that you can identify this with $U_n$ (and as you actually do not need a groupstructure on the quotient, you can ignore the reauirements on the cosets).
$endgroup$
– Enkidu
Jan 28 at 15:08
$begingroup$
yep, then just use that $D_n subset T_n$, kill it, and proof that you can identify this with $U_n$ (and as you actually do not need a groupstructure on the quotient, you can ignore the reauirements on the cosets).
$endgroup$
– Enkidu
Jan 28 at 15:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Just set $D_n := text{diag}(T_n)$; that is, $D_n$ is the diagonal matrix consisting of the diagonal elements of $T_n$. Then, since multiplying on the right scales the rows of a matrix, set the elements of $U_n$ to be the elements of $T_n$ with each row divided by the corresponding diagonal. In other words, $(U_n)_{ij} = frac{(T_n)_{ij}}{(D_n)_{ii}}$.
answered Jan 28 at 15:06
OldGodzillaOldGodzilla
58227
$endgroup$
add a comment |
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$begingroup$
Just set $D_n := text{diag}(T_n)$; that is, $D_n$ is the diagonal matrix consisting of the diagonal elements of $T_n$. Then, since multiplying on the right scales the rows of a matrix, set the elements of $U_n$ to be the elements of $T_n$ with each row divided by the corresponding diagonal. In other words, $(U_n)_{ij} = frac{(T_n)_{ij}}{(D_n)_{ii}}$.
answered Jan 28 at 15:06
OldGodzillaOldGodzilla
58227
$endgroup$
add a comment |
$begingroup$
Just set $D_n := text{diag}(T_n)$; that is, $D_n$ is the diagonal matrix consisting of the diagonal elements of $T_n$. Then, since multiplying on the right scales the rows of a matrix, set the elements of $U_n$ to be the elements of $T_n$ with each row divided by the corresponding diagonal. In other words, $(U_n)_{ij} = frac{(T_n)_{ij}}{(D_n)_{ii}}$.
answered Jan 28 at 15:06
OldGodzillaOldGodzilla
58227
$endgroup$
add a comment |
$begingroup$
Just set $D_n := text{diag}(T_n)$; that is, $D_n$ is the diagonal matrix consisting of the diagonal elements of $T_n$. Then, since multiplying on the right scales the rows of a matrix, set the elements of $U_n$ to be the elements of $T_n$ with each row divided by the corresponding diagonal. In other words, $(U_n)_{ij} = frac{(T_n)_{ij}}{(D_n)_{ii}}$.
answered Jan 28 at 15:06
OldGodzillaOldGodzilla
58227
$endgroup$
Just set $D_n := text{diag}(T_n)$; that is, $D_n$ is the diagonal matrix consisting of the diagonal elements of $T_n$. Then, since multiplying on the right scales the rows of a matrix, set the elements of $U_n$ to be the elements of $T_n$ with each row divided by the corresponding diagonal. In other words, $(U_n)_{ij} = frac{(T_n)_{ij}}{(D_n)_{ii}}$.
answered Jan 28 at 15:06
OldGodzillaOldGodzilla
58227
answered Jan 28 at 15:06
OldGodzillaOldGodzilla
58227
answered Jan 28 at 15:06
OldGodzillaOldGodzilla
58227
answered Jan 28 at 15:06
OldGodzillaOldGodzilla
58227
58227
add a comment |
add a comment |
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$begingroup$
do you know quotients of groups by subsets? I would use that, since as soon as you can proof that $D_n$ is a normal subgroup, you could just devide it out and be done, (respectively, you could do that either way, just a little less nice)
$endgroup$
– Enkidu
Jan 28 at 15:04
$begingroup$
I have heard something About normal Groups we have defined them as subgroups of a Group where the left coset is Always equal to the Right coset. They create a Quotient set
$endgroup$
– RM777
Jan 28 at 15:07
$begingroup$
yep, then just use that $D_n subset T_n$, kill it, and proof that you can identify this with $U_n$ (and as you actually do not need a groupstructure on the quotient, you can ignore the reauirements on the cosets).
$endgroup$
– Enkidu
Jan 28 at 15:08