some confusion in upper traingular matrix?
up vote
1
down vote
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$1$. What is the dimension of the subspace of all real $n times n$ matrix of upper triangular matrix?
$2$. What is dimension of all real $n times n$ matrix of upper triangular ?
Is both option
$1)$
and
$2)$ having same dimension?
Any hints/solution
linear-algebra
edited Nov 27 at 15:35
amWhy
191k28223439
asked Nov 27 at 14:07
Messi fifa
50111
add a comment |
up vote
1
down vote
favorite
$1$. What is the dimension of the subspace of all real $n times n$ matrix of upper triangular matrix?
$2$. What is dimension of all real $n times n$ matrix of upper triangular ?
Is both option
$1)$
and
$2)$ having same dimension?
Any hints/solution
linear-algebra
edited Nov 27 at 15:35
amWhy
191k28223439
asked Nov 27 at 14:07
Messi fifa
50111
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$1$. What is the dimension of the subspace of all real $n times n$ matrix of upper triangular matrix?
$2$. What is dimension of all real $n times n$ matrix of upper triangular ?
Is both option
$1)$
and
$2)$ having same dimension?
Any hints/solution
linear-algebra
edited Nov 27 at 15:35
amWhy
191k28223439
asked Nov 27 at 14:07
Messi fifa
50111
$1$. What is the dimension of the subspace of all real $n times n$ matrix of upper triangular matrix?
$2$. What is dimension of all real $n times n$ matrix of upper triangular ?
Is both option
$1)$
and
$2)$ having same dimension?
Any hints/solution
linear-algebra
linear-algebra
edited Nov 27 at 15:35
amWhy
191k28223439
asked Nov 27 at 14:07
Messi fifa
50111
edited Nov 27 at 15:35
amWhy
191k28223439
asked Nov 27 at 14:07
Messi fifa
50111
edited Nov 27 at 15:35
amWhy
191k28223439
edited Nov 27 at 15:35
amWhy
191k28223439
edited Nov 27 at 15:35
amWhy
191k28223439
191k28223439
asked Nov 27 at 14:07
Messi fifa
50111
asked Nov 27 at 14:07
Messi fifa
50111
asked Nov 27 at 14:07
Messi fifa
50111
50111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
For 2), let be $S$ the space of upper triangular matrix $M_n(R)$. So the dimension is equal to the elements in that space that can be a base (every matrix with 1 in any single place) so the dimension will be dim($S$)$= n+(n-1)+(n-2)+...+3+2+1 = {n(n+1) over 2 }$; of course it depends on $n$.
For sure dim($S$) < $n*n$ in this case.
Every element of this kind can be a base remember!
For 1), is the same as 2), the "subspace" requirements seems identical to question 2).
answered Nov 27 at 14:18
Alessar
17213
.@alessar .is this method same for lower triangular ??
– Messi fifa
Nov 27 at 14:24
1
The matrix is $n*n$, so the dimension is the same; they are simply symmetrical respect the principal diagonal
– Alessar
Nov 27 at 14:26
1
thanks u got its
– Messi fifa
Nov 27 at 14:28
add a comment |
up vote
2
down vote
HINT
An upper triangular matrix is defined by $N=frac{n(n+1)}{2}$ elements.
answered Nov 27 at 14:19
gimusi
91.9k84495
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
For 2), let be $S$ the space of upper triangular matrix $M_n(R)$. So the dimension is equal to the elements in that space that can be a base (every matrix with 1 in any single place) so the dimension will be dim($S$)$= n+(n-1)+(n-2)+...+3+2+1 = {n(n+1) over 2 }$; of course it depends on $n$.
For sure dim($S$) < $n*n$ in this case.
Every element of this kind can be a base remember!
For 1), is the same as 2), the "subspace" requirements seems identical to question 2).
answered Nov 27 at 14:18
Alessar
17213
.@alessar .is this method same for lower triangular ??
– Messi fifa
Nov 27 at 14:24
1
The matrix is $n*n$, so the dimension is the same; they are simply symmetrical respect the principal diagonal
– Alessar
Nov 27 at 14:26
1
thanks u got its
– Messi fifa
Nov 27 at 14:28
add a comment |
up vote
2
down vote
accepted
For 2), let be $S$ the space of upper triangular matrix $M_n(R)$. So the dimension is equal to the elements in that space that can be a base (every matrix with 1 in any single place) so the dimension will be dim($S$)$= n+(n-1)+(n-2)+...+3+2+1 = {n(n+1) over 2 }$; of course it depends on $n$.
For sure dim($S$) < $n*n$ in this case.
Every element of this kind can be a base remember!
For 1), is the same as 2), the "subspace" requirements seems identical to question 2).
answered Nov 27 at 14:18
Alessar
17213
.@alessar .is this method same for lower triangular ??
– Messi fifa
Nov 27 at 14:24
1
The matrix is $n*n$, so the dimension is the same; they are simply symmetrical respect the principal diagonal
– Alessar
Nov 27 at 14:26
1
thanks u got its
– Messi fifa
Nov 27 at 14:28
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
For 2), let be $S$ the space of upper triangular matrix $M_n(R)$. So the dimension is equal to the elements in that space that can be a base (every matrix with 1 in any single place) so the dimension will be dim($S$)$= n+(n-1)+(n-2)+...+3+2+1 = {n(n+1) over 2 }$; of course it depends on $n$.
For sure dim($S$) < $n*n$ in this case.
Every element of this kind can be a base remember!
For 1), is the same as 2), the "subspace" requirements seems identical to question 2).
answered Nov 27 at 14:18
Alessar
17213
For 2), let be $S$ the space of upper triangular matrix $M_n(R)$. So the dimension is equal to the elements in that space that can be a base (every matrix with 1 in any single place) so the dimension will be dim($S$)$= n+(n-1)+(n-2)+...+3+2+1 = {n(n+1) over 2 }$; of course it depends on $n$.
For sure dim($S$) < $n*n$ in this case.
Every element of this kind can be a base remember!
For 1), is the same as 2), the "subspace" requirements seems identical to question 2).
answered Nov 27 at 14:18
Alessar
17213
answered Nov 27 at 14:18
Alessar
17213
answered Nov 27 at 14:18
Alessar
17213
answered Nov 27 at 14:18
Alessar
17213
17213
.@alessar .is this method same for lower triangular ??
– Messi fifa
Nov 27 at 14:24
1
The matrix is $n*n$, so the dimension is the same; they are simply symmetrical respect the principal diagonal
– Alessar
Nov 27 at 14:26
1
thanks u got its
– Messi fifa
Nov 27 at 14:28
add a comment |
.@alessar .is this method same for lower triangular ??
– Messi fifa
Nov 27 at 14:24
1
The matrix is $n*n$, so the dimension is the same; they are simply symmetrical respect the principal diagonal
– Alessar
Nov 27 at 14:26
1
thanks u got its
– Messi fifa
Nov 27 at 14:28
.@alessar .is this method same for lower triangular ??
– Messi fifa
Nov 27 at 14:24
.@alessar .is this method same for lower triangular ??
– Messi fifa
Nov 27 at 14:24
1
1
The matrix is $n*n$, so the dimension is the same; they are simply symmetrical respect the principal diagonal
– Alessar
Nov 27 at 14:26
The matrix is $n*n$, so the dimension is the same; they are simply symmetrical respect the principal diagonal
– Alessar
Nov 27 at 14:26
1
1
thanks u got its
– Messi fifa
Nov 27 at 14:28
thanks u got its
– Messi fifa
Nov 27 at 14:28
add a comment |
up vote
2
down vote
HINT
An upper triangular matrix is defined by $N=frac{n(n+1)}{2}$ elements.
answered Nov 27 at 14:19
gimusi
91.9k84495
add a comment |
up vote
2
down vote
HINT
An upper triangular matrix is defined by $N=frac{n(n+1)}{2}$ elements.
answered Nov 27 at 14:19
gimusi
91.9k84495
add a comment |
up vote
2
down vote
up vote
2
down vote
HINT
An upper triangular matrix is defined by $N=frac{n(n+1)}{2}$ elements.
answered Nov 27 at 14:19
gimusi
91.9k84495
HINT
An upper triangular matrix is defined by $N=frac{n(n+1)}{2}$ elements.
answered Nov 27 at 14:19
gimusi
91.9k84495
answered Nov 27 at 14:19
gimusi
91.9k84495
answered Nov 27 at 14:19
gimusi
91.9k84495
answered Nov 27 at 14:19
gimusi
91.9k84495
91.9k84495
add a comment |
add a comment |
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