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Prove that upper triangular matrices are closed under inversion
Clash Royale CLAN TAG#URR8PPP
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Prove that group of upper triangle matrices $G$ is a subgroup of $GL_n(F)$
Contains identity: $I_nin G$ since $I_ij=0$ when $i>j$
Closed under multi: $C=AB, c_ij=sum_k=1^n=a_ikb_kj,i=1,2..,n,j=1,2..,n$ for all $c_ij,i>j$ either $ige k>j$ or $i>kge j$. One factor in each addend is $0$: either $a_ik=0$, or $b_kj=0$ thus $c_ij=0$ if $i>jrightarrow Cin G$
Closed under inversion: We know that a unique inverse exist since G is atleast a subset of $GL_n(F)$. Any tips on how do I proceed form here on? I thought about using my proof above to prove that $A^-1in G$ How do I know there is not a fringe case where $AA^-1=I, A^-1notin G$
abstract-algebra group-theory
asked 3 hours ago
Wen2400
257
add a comment |Â
up vote
2
down vote
favorite
Prove that group of upper triangle matrices $G$ is a subgroup of $GL_n(F)$
Contains identity: $I_nin G$ since $I_ij=0$ when $i>j$
Closed under multi: $C=AB, c_ij=sum_k=1^n=a_ikb_kj,i=1,2..,n,j=1,2..,n$ for all $c_ij,i>j$ either $ige k>j$ or $i>kge j$. One factor in each addend is $0$: either $a_ik=0$, or $b_kj=0$ thus $c_ij=0$ if $i>jrightarrow Cin G$
Closed under inversion: We know that a unique inverse exist since G is atleast a subset of $GL_n(F)$. Any tips on how do I proceed form here on? I thought about using my proof above to prove that $A^-1in G$ How do I know there is not a fringe case where $AA^-1=I, A^-1notin G$
abstract-algebra group-theory
asked 3 hours ago
Wen2400
257
1
Try with the expression of $A^-1$ in term of its comatrix.
– AlexL
2 hours ago
Is the zero matrix also an upper triangular matrix?
– dan_fulea
2 hours ago
Yes, I believe so.
– AnotherJohnDoe
2 hours ago
Intuitively, in performing Gaussian elimination to convert the augmented matrix $beginbmatrix A & | & I endbmatrix$ to $beginbmatrix I & | & A^-1 endbmatrix$, you can restrict to operations which keep both halves of the augmented matrix upper triangular.
– Daniel Schepler
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Prove that group of upper triangle matrices $G$ is a subgroup of $GL_n(F)$
Contains identity: $I_nin G$ since $I_ij=0$ when $i>j$
Closed under multi: $C=AB, c_ij=sum_k=1^n=a_ikb_kj,i=1,2..,n,j=1,2..,n$ for all $c_ij,i>j$ either $ige k>j$ or $i>kge j$. One factor in each addend is $0$: either $a_ik=0$, or $b_kj=0$ thus $c_ij=0$ if $i>jrightarrow Cin G$
Closed under inversion: We know that a unique inverse exist since G is atleast a subset of $GL_n(F)$. Any tips on how do I proceed form here on? I thought about using my proof above to prove that $A^-1in G$ How do I know there is not a fringe case where $AA^-1=I, A^-1notin G$
abstract-algebra group-theory
asked 3 hours ago
Wen2400
257
Prove that group of upper triangle matrices $G$ is a subgroup of $GL_n(F)$
Contains identity: $I_nin G$ since $I_ij=0$ when $i>j$
Closed under multi: $C=AB, c_ij=sum_k=1^n=a_ikb_kj,i=1,2..,n,j=1,2..,n$ for all $c_ij,i>j$ either $ige k>j$ or $i>kge j$. One factor in each addend is $0$: either $a_ik=0$, or $b_kj=0$ thus $c_ij=0$ if $i>jrightarrow Cin G$
Closed under inversion: We know that a unique inverse exist since G is atleast a subset of $GL_n(F)$. Any tips on how do I proceed form here on? I thought about using my proof above to prove that $A^-1in G$ How do I know there is not a fringe case where $AA^-1=I, A^-1notin G$
abstract-algebra group-theory
abstract-algebra group-theory
asked 3 hours ago
Wen2400
257
asked 3 hours ago
Wen2400
257
asked 3 hours ago
Wen2400
257
asked 3 hours ago
Wen2400
257
asked 3 hours ago
Wen2400
257
257
1
Try with the expression of $A^-1$ in term of its comatrix.
– AlexL
2 hours ago
Is the zero matrix also an upper triangular matrix?
– dan_fulea
2 hours ago
Yes, I believe so.
– AnotherJohnDoe
2 hours ago
Intuitively, in performing Gaussian elimination to convert the augmented matrix $beginbmatrix A & | & I endbmatrix$ to $beginbmatrix I & | & A^-1 endbmatrix$, you can restrict to operations which keep both halves of the augmented matrix upper triangular.
– Daniel Schepler
1 hour ago
add a comment |Â
1
Try with the expression of $A^-1$ in term of its comatrix.
– AlexL
2 hours ago
Is the zero matrix also an upper triangular matrix?
– dan_fulea
2 hours ago
Yes, I believe so.
– AnotherJohnDoe
2 hours ago
Intuitively, in performing Gaussian elimination to convert the augmented matrix $beginbmatrix A & | & I endbmatrix$ to $beginbmatrix I & | & A^-1 endbmatrix$, you can restrict to operations which keep both halves of the augmented matrix upper triangular.
– Daniel Schepler
1 hour ago
1
1
Try with the expression of $A^-1$ in term of its comatrix.
– AlexL
2 hours ago
Try with the expression of $A^-1$ in term of its comatrix.
– AlexL
2 hours ago
Is the zero matrix also an upper triangular matrix?
– dan_fulea
2 hours ago
Is the zero matrix also an upper triangular matrix?
– dan_fulea
2 hours ago
Yes, I believe so.
– AnotherJohnDoe
2 hours ago
Yes, I believe so.
– AnotherJohnDoe
2 hours ago
Intuitively, in performing Gaussian elimination to convert the augmented matrix $beginbmatrix A & | & I endbmatrix$ to $beginbmatrix I & | & A^-1 endbmatrix$, you can restrict to operations which keep both halves of the augmented matrix upper triangular.
– Daniel Schepler
1 hour ago
Intuitively, in performing Gaussian elimination to convert the augmented matrix $beginbmatrix A & | & I endbmatrix$ to $beginbmatrix I & | & A^-1 endbmatrix$, you can restrict to operations which keep both halves of the augmented matrix upper triangular.
– Daniel Schepler
1 hour ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
Suppose $A^-1$ is not upper triangular. Consider any one of the rows, say $i$, of $A^-1$ such that it has non - zero entries before it's diagonal entry $a_ii$. Let the first of these be in the $j^th$ column.
Finally, a hint - consider the element of the product of $A^-1A$ in the position $(i,j)$
answered 2 hours ago
AnotherJohnDoe
824212
It seems like I can construct a combination of numbers where non zero terms cancel eachother out. Therefore i was reluctant using similair proof as I did above
– Wen2400
2 hours ago
But, in particular, what happens to the element in position $(i,j)?$
– AnotherJohnDoe
2 hours ago
add a comment |Â
up vote
2
down vote
If $A$ is an invertible matrix, then its inverse $A^-1$ can be expressed as a polynomial in $A$.
Therefore, if $A$ is upper triangular, so is $A^-1$, since the product of upper triangular matrices is upper triangular, as you have already proved.
answered 49 mins ago
lhf
158k9161375
When I did some testing on $3times 3$ matrix, I found that $I_32=0=a_31^-1a_12+a_32^-1a_22+a_33^-1a_32$ reduces to $0=a_31^-1a_12+a_32^-1a_22$ Which got me worried that somewhere out there there might exist a fringe case where $a_31^-1,a_32^-1neq 0$
– Wen2400
32 mins ago
+1 Even better!
– rschwieb
7 mins ago
add a comment |Â
up vote
1
down vote
Conceptually, the algebra of upper triangular matrices $UT_n(F)$ is the set of transformations which fixes a complete flag of subspaces of $F^n$.
That is, for all upper triangular matrices, there is a chain of subspaces of $F^n$ written as $0subseteq V_1subseteq V_2subseteqldotssubseteq V_n$, where $dim_F(V_i)=i$, such that for every upper triangular matrix $A$, and each index $i$, $A(V_i)subseteq V_i$. Conversely given any such flag, you can choose a basis such that the flag-preserving transformations are upper triangular. You should have no problem describing such a flag of subspaces of $F^n$ such that the transformations are the upper triangular matrices.
If $A$ is invertible, then it preserves dimension, so $A(V_i)=V_i$ for all $i$. The inverse map, if it exists, would map $V_i$ right back into $V_i$, so the inverse is also an upper triangular matrix.
answered 55 mins ago
rschwieb
101k1198234
add a comment |Â
up vote
0
down vote
Thanks for all the help, I will now attempt to solve this using AnotherJohnDoe's hint. Any pointers or feedback to make this proof more elegant are appreciated.
Let $BA=I,Ain G$ then I choose to look at entries $I_i1,i=2,3,..$ then $I_i1=0=sum_k=1^nb_ika_k1$ since $a_k1=0,k>1$ the expression reduces to $I_i1=0=b_i1a_11$. Determinant of an upper triangle matrix is the product of the diagonal entries, in our case none of the diagonal entries can be zero. Thus $b_i1=0$.
Moving on to $I_i2,i=3,4...$ then $I_i2=0=sum_k=1^nb_ika_k2$ since $a_k2=0,k>2$ the expression reduces to $I_i2=0=b_i1a_12+b_i2a_22$ since $b_i1=0,a_22neq0$ Only solution is $b_i2=0$
Now using induction to complete the proof. (Which I might give it a go at a later date)
answered 1 hour ago
Wen2400
257
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Suppose $A^-1$ is not upper triangular. Consider any one of the rows, say $i$, of $A^-1$ such that it has non - zero entries before it's diagonal entry $a_ii$. Let the first of these be in the $j^th$ column.
Finally, a hint - consider the element of the product of $A^-1A$ in the position $(i,j)$
answered 2 hours ago
AnotherJohnDoe
824212
It seems like I can construct a combination of numbers where non zero terms cancel eachother out. Therefore i was reluctant using similair proof as I did above
– Wen2400
2 hours ago
But, in particular, what happens to the element in position $(i,j)?$
– AnotherJohnDoe
2 hours ago
add a comment |Â
up vote
3
down vote
accepted
Suppose $A^-1$ is not upper triangular. Consider any one of the rows, say $i$, of $A^-1$ such that it has non - zero entries before it's diagonal entry $a_ii$. Let the first of these be in the $j^th$ column.
Finally, a hint - consider the element of the product of $A^-1A$ in the position $(i,j)$
answered 2 hours ago
AnotherJohnDoe
824212
It seems like I can construct a combination of numbers where non zero terms cancel eachother out. Therefore i was reluctant using similair proof as I did above
– Wen2400
2 hours ago
But, in particular, what happens to the element in position $(i,j)?$
– AnotherJohnDoe
2 hours ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Suppose $A^-1$ is not upper triangular. Consider any one of the rows, say $i$, of $A^-1$ such that it has non - zero entries before it's diagonal entry $a_ii$. Let the first of these be in the $j^th$ column.
Finally, a hint - consider the element of the product of $A^-1A$ in the position $(i,j)$
answered 2 hours ago
AnotherJohnDoe
824212
Suppose $A^-1$ is not upper triangular. Consider any one of the rows, say $i$, of $A^-1$ such that it has non - zero entries before it's diagonal entry $a_ii$. Let the first of these be in the $j^th$ column.
Finally, a hint - consider the element of the product of $A^-1A$ in the position $(i,j)$
answered 2 hours ago
AnotherJohnDoe
824212
answered 2 hours ago
AnotherJohnDoe
824212
answered 2 hours ago
AnotherJohnDoe
824212
answered 2 hours ago
AnotherJohnDoe
824212
824212
It seems like I can construct a combination of numbers where non zero terms cancel eachother out. Therefore i was reluctant using similair proof as I did above
– Wen2400
2 hours ago
But, in particular, what happens to the element in position $(i,j)?$
– AnotherJohnDoe
2 hours ago
add a comment |Â
It seems like I can construct a combination of numbers where non zero terms cancel eachother out. Therefore i was reluctant using similair proof as I did above
– Wen2400
2 hours ago
But, in particular, what happens to the element in position $(i,j)?$
– AnotherJohnDoe
2 hours ago
It seems like I can construct a combination of numbers where non zero terms cancel eachother out. Therefore i was reluctant using similair proof as I did above
– Wen2400
2 hours ago
It seems like I can construct a combination of numbers where non zero terms cancel eachother out. Therefore i was reluctant using similair proof as I did above
– Wen2400
2 hours ago
But, in particular, what happens to the element in position $(i,j)?$
– AnotherJohnDoe
2 hours ago
But, in particular, what happens to the element in position $(i,j)?$
– AnotherJohnDoe
2 hours ago
add a comment |Â
up vote
2
down vote
If $A$ is an invertible matrix, then its inverse $A^-1$ can be expressed as a polynomial in $A$.
Therefore, if $A$ is upper triangular, so is $A^-1$, since the product of upper triangular matrices is upper triangular, as you have already proved.
answered 49 mins ago
lhf
158k9161375
When I did some testing on $3times 3$ matrix, I found that $I_32=0=a_31^-1a_12+a_32^-1a_22+a_33^-1a_32$ reduces to $0=a_31^-1a_12+a_32^-1a_22$ Which got me worried that somewhere out there there might exist a fringe case where $a_31^-1,a_32^-1neq 0$
– Wen2400
32 mins ago
+1 Even better!
– rschwieb
7 mins ago
add a comment |Â
up vote
2
down vote
If $A$ is an invertible matrix, then its inverse $A^-1$ can be expressed as a polynomial in $A$.
Therefore, if $A$ is upper triangular, so is $A^-1$, since the product of upper triangular matrices is upper triangular, as you have already proved.
answered 49 mins ago
lhf
158k9161375
When I did some testing on $3times 3$ matrix, I found that $I_32=0=a_31^-1a_12+a_32^-1a_22+a_33^-1a_32$ reduces to $0=a_31^-1a_12+a_32^-1a_22$ Which got me worried that somewhere out there there might exist a fringe case where $a_31^-1,a_32^-1neq 0$
– Wen2400
32 mins ago
+1 Even better!
– rschwieb
7 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If $A$ is an invertible matrix, then its inverse $A^-1$ can be expressed as a polynomial in $A$.
Therefore, if $A$ is upper triangular, so is $A^-1$, since the product of upper triangular matrices is upper triangular, as you have already proved.
answered 49 mins ago
lhf
158k9161375
If $A$ is an invertible matrix, then its inverse $A^-1$ can be expressed as a polynomial in $A$.
Therefore, if $A$ is upper triangular, so is $A^-1$, since the product of upper triangular matrices is upper triangular, as you have already proved.
answered 49 mins ago
lhf
158k9161375
answered 49 mins ago
lhf
158k9161375
answered 49 mins ago
lhf
158k9161375
answered 49 mins ago
lhf
158k9161375
158k9161375
When I did some testing on $3times 3$ matrix, I found that $I_32=0=a_31^-1a_12+a_32^-1a_22+a_33^-1a_32$ reduces to $0=a_31^-1a_12+a_32^-1a_22$ Which got me worried that somewhere out there there might exist a fringe case where $a_31^-1,a_32^-1neq 0$
– Wen2400
32 mins ago
+1 Even better!
– rschwieb
7 mins ago
add a comment |Â
When I did some testing on $3times 3$ matrix, I found that $I_32=0=a_31^-1a_12+a_32^-1a_22+a_33^-1a_32$ reduces to $0=a_31^-1a_12+a_32^-1a_22$ Which got me worried that somewhere out there there might exist a fringe case where $a_31^-1,a_32^-1neq 0$
– Wen2400
32 mins ago
+1 Even better!
– rschwieb
7 mins ago
When I did some testing on $3times 3$ matrix, I found that $I_32=0=a_31^-1a_12+a_32^-1a_22+a_33^-1a_32$ reduces to $0=a_31^-1a_12+a_32^-1a_22$ Which got me worried that somewhere out there there might exist a fringe case where $a_31^-1,a_32^-1neq 0$
– Wen2400
32 mins ago
When I did some testing on $3times 3$ matrix, I found that $I_32=0=a_31^-1a_12+a_32^-1a_22+a_33^-1a_32$ reduces to $0=a_31^-1a_12+a_32^-1a_22$ Which got me worried that somewhere out there there might exist a fringe case where $a_31^-1,a_32^-1neq 0$
– Wen2400
32 mins ago
+1 Even better!
– rschwieb
7 mins ago
+1 Even better!
– rschwieb
7 mins ago
add a comment |Â
up vote
1
down vote
Conceptually, the algebra of upper triangular matrices $UT_n(F)$ is the set of transformations which fixes a complete flag of subspaces of $F^n$.
That is, for all upper triangular matrices, there is a chain of subspaces of $F^n$ written as $0subseteq V_1subseteq V_2subseteqldotssubseteq V_n$, where $dim_F(V_i)=i$, such that for every upper triangular matrix $A$, and each index $i$, $A(V_i)subseteq V_i$. Conversely given any such flag, you can choose a basis such that the flag-preserving transformations are upper triangular. You should have no problem describing such a flag of subspaces of $F^n$ such that the transformations are the upper triangular matrices.
If $A$ is invertible, then it preserves dimension, so $A(V_i)=V_i$ for all $i$. The inverse map, if it exists, would map $V_i$ right back into $V_i$, so the inverse is also an upper triangular matrix.
answered 55 mins ago
rschwieb
101k1198234
add a comment |Â
up vote
1
down vote
Conceptually, the algebra of upper triangular matrices $UT_n(F)$ is the set of transformations which fixes a complete flag of subspaces of $F^n$.
That is, for all upper triangular matrices, there is a chain of subspaces of $F^n$ written as $0subseteq V_1subseteq V_2subseteqldotssubseteq V_n$, where $dim_F(V_i)=i$, such that for every upper triangular matrix $A$, and each index $i$, $A(V_i)subseteq V_i$. Conversely given any such flag, you can choose a basis such that the flag-preserving transformations are upper triangular. You should have no problem describing such a flag of subspaces of $F^n$ such that the transformations are the upper triangular matrices.
If $A$ is invertible, then it preserves dimension, so $A(V_i)=V_i$ for all $i$. The inverse map, if it exists, would map $V_i$ right back into $V_i$, so the inverse is also an upper triangular matrix.
answered 55 mins ago
rschwieb
101k1198234
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Conceptually, the algebra of upper triangular matrices $UT_n(F)$ is the set of transformations which fixes a complete flag of subspaces of $F^n$.
That is, for all upper triangular matrices, there is a chain of subspaces of $F^n$ written as $0subseteq V_1subseteq V_2subseteqldotssubseteq V_n$, where $dim_F(V_i)=i$, such that for every upper triangular matrix $A$, and each index $i$, $A(V_i)subseteq V_i$. Conversely given any such flag, you can choose a basis such that the flag-preserving transformations are upper triangular. You should have no problem describing such a flag of subspaces of $F^n$ such that the transformations are the upper triangular matrices.
If $A$ is invertible, then it preserves dimension, so $A(V_i)=V_i$ for all $i$. The inverse map, if it exists, would map $V_i$ right back into $V_i$, so the inverse is also an upper triangular matrix.
answered 55 mins ago
rschwieb
101k1198234
Conceptually, the algebra of upper triangular matrices $UT_n(F)$ is the set of transformations which fixes a complete flag of subspaces of $F^n$.
That is, for all upper triangular matrices, there is a chain of subspaces of $F^n$ written as $0subseteq V_1subseteq V_2subseteqldotssubseteq V_n$, where $dim_F(V_i)=i$, such that for every upper triangular matrix $A$, and each index $i$, $A(V_i)subseteq V_i$. Conversely given any such flag, you can choose a basis such that the flag-preserving transformations are upper triangular. You should have no problem describing such a flag of subspaces of $F^n$ such that the transformations are the upper triangular matrices.
If $A$ is invertible, then it preserves dimension, so $A(V_i)=V_i$ for all $i$. The inverse map, if it exists, would map $V_i$ right back into $V_i$, so the inverse is also an upper triangular matrix.
answered 55 mins ago
rschwieb
101k1198234
answered 55 mins ago
rschwieb
101k1198234
answered 55 mins ago
rschwieb
101k1198234
answered 55 mins ago
rschwieb
101k1198234
101k1198234
add a comment |Â
add a comment |Â
up vote
0
down vote
Thanks for all the help, I will now attempt to solve this using AnotherJohnDoe's hint. Any pointers or feedback to make this proof more elegant are appreciated.
Let $BA=I,Ain G$ then I choose to look at entries $I_i1,i=2,3,..$ then $I_i1=0=sum_k=1^nb_ika_k1$ since $a_k1=0,k>1$ the expression reduces to $I_i1=0=b_i1a_11$. Determinant of an upper triangle matrix is the product of the diagonal entries, in our case none of the diagonal entries can be zero. Thus $b_i1=0$.
Moving on to $I_i2,i=3,4...$ then $I_i2=0=sum_k=1^nb_ika_k2$ since $a_k2=0,k>2$ the expression reduces to $I_i2=0=b_i1a_12+b_i2a_22$ since $b_i1=0,a_22neq0$ Only solution is $b_i2=0$
Now using induction to complete the proof. (Which I might give it a go at a later date)
answered 1 hour ago
Wen2400
257
add a comment |Â
up vote
0
down vote
Thanks for all the help, I will now attempt to solve this using AnotherJohnDoe's hint. Any pointers or feedback to make this proof more elegant are appreciated.
Let $BA=I,Ain G$ then I choose to look at entries $I_i1,i=2,3,..$ then $I_i1=0=sum_k=1^nb_ika_k1$ since $a_k1=0,k>1$ the expression reduces to $I_i1=0=b_i1a_11$. Determinant of an upper triangle matrix is the product of the diagonal entries, in our case none of the diagonal entries can be zero. Thus $b_i1=0$.
Moving on to $I_i2,i=3,4...$ then $I_i2=0=sum_k=1^nb_ika_k2$ since $a_k2=0,k>2$ the expression reduces to $I_i2=0=b_i1a_12+b_i2a_22$ since $b_i1=0,a_22neq0$ Only solution is $b_i2=0$
Now using induction to complete the proof. (Which I might give it a go at a later date)
answered 1 hour ago
Wen2400
257
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Thanks for all the help, I will now attempt to solve this using AnotherJohnDoe's hint. Any pointers or feedback to make this proof more elegant are appreciated.
Let $BA=I,Ain G$ then I choose to look at entries $I_i1,i=2,3,..$ then $I_i1=0=sum_k=1^nb_ika_k1$ since $a_k1=0,k>1$ the expression reduces to $I_i1=0=b_i1a_11$. Determinant of an upper triangle matrix is the product of the diagonal entries, in our case none of the diagonal entries can be zero. Thus $b_i1=0$.
Moving on to $I_i2,i=3,4...$ then $I_i2=0=sum_k=1^nb_ika_k2$ since $a_k2=0,k>2$ the expression reduces to $I_i2=0=b_i1a_12+b_i2a_22$ since $b_i1=0,a_22neq0$ Only solution is $b_i2=0$
Now using induction to complete the proof. (Which I might give it a go at a later date)
answered 1 hour ago
Wen2400
257
Thanks for all the help, I will now attempt to solve this using AnotherJohnDoe's hint. Any pointers or feedback to make this proof more elegant are appreciated.
Let $BA=I,Ain G$ then I choose to look at entries $I_i1,i=2,3,..$ then $I_i1=0=sum_k=1^nb_ika_k1$ since $a_k1=0,k>1$ the expression reduces to $I_i1=0=b_i1a_11$. Determinant of an upper triangle matrix is the product of the diagonal entries, in our case none of the diagonal entries can be zero. Thus $b_i1=0$.
Moving on to $I_i2,i=3,4...$ then $I_i2=0=sum_k=1^nb_ika_k2$ since $a_k2=0,k>2$ the expression reduces to $I_i2=0=b_i1a_12+b_i2a_22$ since $b_i1=0,a_22neq0$ Only solution is $b_i2=0$
Now using induction to complete the proof. (Which I might give it a go at a later date)
answered 1 hour ago
Wen2400
257
answered 1 hour ago
Wen2400
257
answered 1 hour ago
Wen2400
257
answered 1 hour ago
Wen2400
257
257
add a comment |Â
add a comment |Â
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1
Try with the expression of $A^-1$ in term of its comatrix.
– AlexL
2 hours ago
Is the zero matrix also an upper triangular matrix?
– dan_fulea
2 hours ago
Yes, I believe so.
– AnotherJohnDoe
2 hours ago
Intuitively, in performing Gaussian elimination to convert the augmented matrix $beginbmatrix A & | & I endbmatrix$ to $beginbmatrix I & | & A^-1 endbmatrix$, you can restrict to operations which keep both halves of the augmented matrix upper triangular.
– Daniel Schepler
1 hour ago