Find $text{card}(T_n( mathbb R ) cap O(n))$.












0












$begingroup$


Let $T_n( mathbb R )$ be the set of upper triangular matrices of size $n$.



Let $O(n)$ be the set of general orthogonal matrices and $SO(n)$ the set of special orthogonal matrices.



Find the cardinal of $T_n( mathbb R ) cap O(n)$ and $T_n( mathbb R ) cap SO(n)$.



For the second question, I found that case $n = 2$ is pretty easy, since
$forall M in SO(2), M$ can be written as $M = begin{bmatrix}
cos(theta) & -sin(theta) \
sin(theta) & cos(theta)
end{bmatrix}$, $theta in mathbb R$.



We can see that $M$ is upper triangular iff $theta = k pi$ with $k in mathbb Z$.



So that gives us $text{card}(T_2( mathbb R ) cap SO(2)) = 2$.



I have trouble generalizing this result.






linear-algebra matrices orthogonal-matrices







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Can you prove that the inverse of an upper triangular matrix is an upper triangular matrix?
    $endgroup$
    – Mindlack
    Jan 21 at 12:55
















0












$begingroup$


Let $T_n( mathbb R )$ be the set of upper triangular matrices of size $n$.



Let $O(n)$ be the set of general orthogonal matrices and $SO(n)$ the set of special orthogonal matrices.



Find the cardinal of $T_n( mathbb R ) cap O(n)$ and $T_n( mathbb R ) cap SO(n)$.



For the second question, I found that case $n = 2$ is pretty easy, since
$forall M in SO(2), M$ can be written as $M = begin{bmatrix}
cos(theta) & -sin(theta) \
sin(theta) & cos(theta)
end{bmatrix}$, $theta in mathbb R$.



We can see that $M$ is upper triangular iff $theta = k pi$ with $k in mathbb Z$.



So that gives us $text{card}(T_2( mathbb R ) cap SO(2)) = 2$.



I have trouble generalizing this result.






linear-algebra matrices orthogonal-matrices







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Can you prove that the inverse of an upper triangular matrix is an upper triangular matrix?
    $endgroup$
    – Mindlack
    Jan 21 at 12:55














0












0








0





$begingroup$


Let $T_n( mathbb R )$ be the set of upper triangular matrices of size $n$.



Let $O(n)$ be the set of general orthogonal matrices and $SO(n)$ the set of special orthogonal matrices.



Find the cardinal of $T_n( mathbb R ) cap O(n)$ and $T_n( mathbb R ) cap SO(n)$.



For the second question, I found that case $n = 2$ is pretty easy, since
$forall M in SO(2), M$ can be written as $M = begin{bmatrix}
cos(theta) & -sin(theta) \
sin(theta) & cos(theta)
end{bmatrix}$, $theta in mathbb R$.



We can see that $M$ is upper triangular iff $theta = k pi$ with $k in mathbb Z$.



So that gives us $text{card}(T_2( mathbb R ) cap SO(2)) = 2$.



I have trouble generalizing this result.






linear-algebra matrices orthogonal-matrices







share|cite|improve this question









$endgroup$




Let $T_n( mathbb R )$ be the set of upper triangular matrices of size $n$.



Let $O(n)$ be the set of general orthogonal matrices and $SO(n)$ the set of special orthogonal matrices.



Find the cardinal of $T_n( mathbb R ) cap O(n)$ and $T_n( mathbb R ) cap SO(n)$.



For the second question, I found that case $n = 2$ is pretty easy, since
$forall M in SO(2), M$ can be written as $M = begin{bmatrix}
cos(theta) & -sin(theta) \
sin(theta) & cos(theta)
end{bmatrix}$, $theta in mathbb R$.



We can see that $M$ is upper triangular iff $theta = k pi$ with $k in mathbb Z$.



So that gives us $text{card}(T_2( mathbb R ) cap SO(2)) = 2$.



I have trouble generalizing this result.






linear-algebra matrices orthogonal-matrices




linear-algebra matrices orthogonal-matrices






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 21 at 12:51









Euler PythagorasEuler Pythagoras

58111




58111








  • 2




    $begingroup$
    Can you prove that the inverse of an upper triangular matrix is an upper triangular matrix?
    $endgroup$
    – Mindlack
    Jan 21 at 12:55














  • 2




    $begingroup$
    Can you prove that the inverse of an upper triangular matrix is an upper triangular matrix?
    $endgroup$
    – Mindlack
    Jan 21 at 12:55








2




2




$begingroup$
Can you prove that the inverse of an upper triangular matrix is an upper triangular matrix?
$endgroup$
– Mindlack
Jan 21 at 12:55




$begingroup$
Can you prove that the inverse of an upper triangular matrix is an upper triangular matrix?
$endgroup$
– Mindlack
Jan 21 at 12:55










1 Answer
1






active

oldest

votes


















0












$begingroup$

Take $Min T_n(mathbb{R})cap O(n)$. Then $M^{-1}=M^T$ (since $Min O(n)$) and $M^{-1}in T_n(mathbb{R})$ (since the inverse of an upper triangular matrix is also upper triangular. So, $M$ and $M^T$ are both upper triangular, which means that $M$ is, in fact, a diagonal matrix. Now, how many diagonal matrices are there in $O(n)$? And in $SO(n)$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for responding. Let $M = (m_{ij}). $Since $det(M) = 1$ and $forall (i,j) in [![ 1, n]!]^2, m_{ij} in [-1, 1]$, the only choices for the coefficients of $M$ is ${-1,1}$. So, $text{card}(T_n( mathbb R ) cap O(n)) = text{card}(T_n( mathbb R ) cap SO(n)) = 2^n$. Is that all right?
    $endgroup$
    – Euler Pythagoras
    Jan 21 at 18:34












  • $begingroup$
    No. I don't know how you jump from $M_{ij}in[-1,1]$ to $M_{ij}in{1,-1}$. Besides, $operatorname{card}(T_n(mathbb{R})cap SO(n))=2^{n-1}$.
    $endgroup$
    – José Carlos Santos
    Jan 21 at 18:36










  • $begingroup$
    We know that $M in O(n)$ so every column of $M$ has norm $1$. But why is $text{card}(T_n( mathbb R ) cap SO(n)) = 2^{n-1}$?
    $endgroup$
    – Euler Pythagoras
    Jan 21 at 18:41












  • $begingroup$
    Because, the number of $-1$'s in the main diagonal of a diagonal matrix in $SO(n)$ must be even, so that its determinant is $-1$.
    $endgroup$
    – José Carlos Santos
    Jan 21 at 18:47










  • $begingroup$
    Is that really? Because $begin{bmatrix} cos(theta) & -sin(theta) \ sin(theta) & cos(theta) end{bmatrix} in SO(2)$ and has determinant $1$.
    $endgroup$
    – Euler Pythagoras
    Jan 21 at 19:16











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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









0












$begingroup$

Take $Min T_n(mathbb{R})cap O(n)$. Then $M^{-1}=M^T$ (since $Min O(n)$) and $M^{-1}in T_n(mathbb{R})$ (since the inverse of an upper triangular matrix is also upper triangular. So, $M$ and $M^T$ are both upper triangular, which means that $M$ is, in fact, a diagonal matrix. Now, how many diagonal matrices are there in $O(n)$? And in $SO(n)$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for responding. Let $M = (m_{ij}). $Since $det(M) = 1$ and $forall (i,j) in [![ 1, n]!]^2, m_{ij} in [-1, 1]$, the only choices for the coefficients of $M$ is ${-1,1}$. So, $text{card}(T_n( mathbb R ) cap O(n)) = text{card}(T_n( mathbb R ) cap SO(n)) = 2^n$. Is that all right?
    $endgroup$
    – Euler Pythagoras
    Jan 21 at 18:34












  • $begingroup$
    No. I don't know how you jump from $M_{ij}in[-1,1]$ to $M_{ij}in{1,-1}$. Besides, $operatorname{card}(T_n(mathbb{R})cap SO(n))=2^{n-1}$.
    $endgroup$
    – José Carlos Santos
    Jan 21 at 18:36










  • $begingroup$
    We know that $M in O(n)$ so every column of $M$ has norm $1$. But why is $text{card}(T_n( mathbb R ) cap SO(n)) = 2^{n-1}$?
    $endgroup$
    – Euler Pythagoras
    Jan 21 at 18:41












  • $begingroup$
    Because, the number of $-1$'s in the main diagonal of a diagonal matrix in $SO(n)$ must be even, so that its determinant is $-1$.
    $endgroup$
    – José Carlos Santos
    Jan 21 at 18:47










  • $begingroup$
    Is that really? Because $begin{bmatrix} cos(theta) & -sin(theta) \ sin(theta) & cos(theta) end{bmatrix} in SO(2)$ and has determinant $1$.
    $endgroup$
    – Euler Pythagoras
    Jan 21 at 19:16
















0












$begingroup$

Take $Min T_n(mathbb{R})cap O(n)$. Then $M^{-1}=M^T$ (since $Min O(n)$) and $M^{-1}in T_n(mathbb{R})$ (since the inverse of an upper triangular matrix is also upper triangular. So, $M$ and $M^T$ are both upper triangular, which means that $M$ is, in fact, a diagonal matrix. Now, how many diagonal matrices are there in $O(n)$? And in $SO(n)$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for responding. Let $M = (m_{ij}). $Since $det(M) = 1$ and $forall (i,j) in [![ 1, n]!]^2, m_{ij} in [-1, 1]$, the only choices for the coefficients of $M$ is ${-1,1}$. So, $text{card}(T_n( mathbb R ) cap O(n)) = text{card}(T_n( mathbb R ) cap SO(n)) = 2^n$. Is that all right?
    $endgroup$
    – Euler Pythagoras
    Jan 21 at 18:34












  • $begingroup$
    No. I don't know how you jump from $M_{ij}in[-1,1]$ to $M_{ij}in{1,-1}$. Besides, $operatorname{card}(T_n(mathbb{R})cap SO(n))=2^{n-1}$.
    $endgroup$
    – José Carlos Santos
    Jan 21 at 18:36










  • $begingroup$
    We know that $M in O(n)$ so every column of $M$ has norm $1$. But why is $text{card}(T_n( mathbb R ) cap SO(n)) = 2^{n-1}$?
    $endgroup$
    – Euler Pythagoras
    Jan 21 at 18:41












  • $begingroup$
    Because, the number of $-1$'s in the main diagonal of a diagonal matrix in $SO(n)$ must be even, so that its determinant is $-1$.
    $endgroup$
    – José Carlos Santos
    Jan 21 at 18:47










  • $begingroup$
    Is that really? Because $begin{bmatrix} cos(theta) & -sin(theta) \ sin(theta) & cos(theta) end{bmatrix} in SO(2)$ and has determinant $1$.
    $endgroup$
    – Euler Pythagoras
    Jan 21 at 19:16














0












0








0





$begingroup$

Take $Min T_n(mathbb{R})cap O(n)$. Then $M^{-1}=M^T$ (since $Min O(n)$) and $M^{-1}in T_n(mathbb{R})$ (since the inverse of an upper triangular matrix is also upper triangular. So, $M$ and $M^T$ are both upper triangular, which means that $M$ is, in fact, a diagonal matrix. Now, how many diagonal matrices are there in $O(n)$? And in $SO(n)$?






share|cite|improve this answer









$endgroup$



Take $Min T_n(mathbb{R})cap O(n)$. Then $M^{-1}=M^T$ (since $Min O(n)$) and $M^{-1}in T_n(mathbb{R})$ (since the inverse of an upper triangular matrix is also upper triangular. So, $M$ and $M^T$ are both upper triangular, which means that $M$ is, in fact, a diagonal matrix. Now, how many diagonal matrices are there in $O(n)$? And in $SO(n)$?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 21 at 13:07









José Carlos SantosJosé Carlos Santos

163k22131234




163k22131234












  • $begingroup$
    Thanks for responding. Let $M = (m_{ij}). $Since $det(M) = 1$ and $forall (i,j) in [![ 1, n]!]^2, m_{ij} in [-1, 1]$, the only choices for the coefficients of $M$ is ${-1,1}$. So, $text{card}(T_n( mathbb R ) cap O(n)) = text{card}(T_n( mathbb R ) cap SO(n)) = 2^n$. Is that all right?
    $endgroup$
    – Euler Pythagoras
    Jan 21 at 18:34












  • $begingroup$
    No. I don't know how you jump from $M_{ij}in[-1,1]$ to $M_{ij}in{1,-1}$. Besides, $operatorname{card}(T_n(mathbb{R})cap SO(n))=2^{n-1}$.
    $endgroup$
    – José Carlos Santos
    Jan 21 at 18:36










  • $begingroup$
    We know that $M in O(n)$ so every column of $M$ has norm $1$. But why is $text{card}(T_n( mathbb R ) cap SO(n)) = 2^{n-1}$?
    $endgroup$
    – Euler Pythagoras
    Jan 21 at 18:41












  • $begingroup$
    Because, the number of $-1$'s in the main diagonal of a diagonal matrix in $SO(n)$ must be even, so that its determinant is $-1$.
    $endgroup$
    – José Carlos Santos
    Jan 21 at 18:47










  • $begingroup$
    Is that really? Because $begin{bmatrix} cos(theta) & -sin(theta) \ sin(theta) & cos(theta) end{bmatrix} in SO(2)$ and has determinant $1$.
    $endgroup$
    – Euler Pythagoras
    Jan 21 at 19:16


















  • $begingroup$
    Thanks for responding. Let $M = (m_{ij}). $Since $det(M) = 1$ and $forall (i,j) in [![ 1, n]!]^2, m_{ij} in [-1, 1]$, the only choices for the coefficients of $M$ is ${-1,1}$. So, $text{card}(T_n( mathbb R ) cap O(n)) = text{card}(T_n( mathbb R ) cap SO(n)) = 2^n$. Is that all right?
    $endgroup$
    – Euler Pythagoras
    Jan 21 at 18:34












  • $begingroup$
    No. I don't know how you jump from $M_{ij}in[-1,1]$ to $M_{ij}in{1,-1}$. Besides, $operatorname{card}(T_n(mathbb{R})cap SO(n))=2^{n-1}$.
    $endgroup$
    – José Carlos Santos
    Jan 21 at 18:36










  • $begingroup$
    We know that $M in O(n)$ so every column of $M$ has norm $1$. But why is $text{card}(T_n( mathbb R ) cap SO(n)) = 2^{n-1}$?
    $endgroup$
    – Euler Pythagoras
    Jan 21 at 18:41












  • $begingroup$
    Because, the number of $-1$'s in the main diagonal of a diagonal matrix in $SO(n)$ must be even, so that its determinant is $-1$.
    $endgroup$
    – José Carlos Santos
    Jan 21 at 18:47










  • $begingroup$
    Is that really? Because $begin{bmatrix} cos(theta) & -sin(theta) \ sin(theta) & cos(theta) end{bmatrix} in SO(2)$ and has determinant $1$.
    $endgroup$
    – Euler Pythagoras
    Jan 21 at 19:16
















$begingroup$
Thanks for responding. Let $M = (m_{ij}). $Since $det(M) = 1$ and $forall (i,j) in [![ 1, n]!]^2, m_{ij} in [-1, 1]$, the only choices for the coefficients of $M$ is ${-1,1}$. So, $text{card}(T_n( mathbb R ) cap O(n)) = text{card}(T_n( mathbb R ) cap SO(n)) = 2^n$. Is that all right?
$endgroup$
– Euler Pythagoras
Jan 21 at 18:34






$begingroup$
Thanks for responding. Let $M = (m_{ij}). $Since $det(M) = 1$ and $forall (i,j) in [![ 1, n]!]^2, m_{ij} in [-1, 1]$, the only choices for the coefficients of $M$ is ${-1,1}$. So, $text{card}(T_n( mathbb R ) cap O(n)) = text{card}(T_n( mathbb R ) cap SO(n)) = 2^n$. Is that all right?
$endgroup$
– Euler Pythagoras
Jan 21 at 18:34














$begingroup$
No. I don't know how you jump from $M_{ij}in[-1,1]$ to $M_{ij}in{1,-1}$. Besides, $operatorname{card}(T_n(mathbb{R})cap SO(n))=2^{n-1}$.
$endgroup$
– José Carlos Santos
Jan 21 at 18:36




$begingroup$
No. I don't know how you jump from $M_{ij}in[-1,1]$ to $M_{ij}in{1,-1}$. Besides, $operatorname{card}(T_n(mathbb{R})cap SO(n))=2^{n-1}$.
$endgroup$
– José Carlos Santos
Jan 21 at 18:36












$begingroup$
We know that $M in O(n)$ so every column of $M$ has norm $1$. But why is $text{card}(T_n( mathbb R ) cap SO(n)) = 2^{n-1}$?
$endgroup$
– Euler Pythagoras
Jan 21 at 18:41






$begingroup$
We know that $M in O(n)$ so every column of $M$ has norm $1$. But why is $text{card}(T_n( mathbb R ) cap SO(n)) = 2^{n-1}$?
$endgroup$
– Euler Pythagoras
Jan 21 at 18:41














$begingroup$
Because, the number of $-1$'s in the main diagonal of a diagonal matrix in $SO(n)$ must be even, so that its determinant is $-1$.
$endgroup$
– José Carlos Santos
Jan 21 at 18:47




$begingroup$
Because, the number of $-1$'s in the main diagonal of a diagonal matrix in $SO(n)$ must be even, so that its determinant is $-1$.
$endgroup$
– José Carlos Santos
Jan 21 at 18:47












$begingroup$
Is that really? Because $begin{bmatrix} cos(theta) & -sin(theta) \ sin(theta) & cos(theta) end{bmatrix} in SO(2)$ and has determinant $1$.
$endgroup$
– Euler Pythagoras
Jan 21 at 19:16




$begingroup$
Is that really? Because $begin{bmatrix} cos(theta) & -sin(theta) \ sin(theta) & cos(theta) end{bmatrix} in SO(2)$ and has determinant $1$.
$endgroup$
– Euler Pythagoras
Jan 21 at 19:16


















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