on the matrix representation of a canonical linear map on the vector space of $4times 4$ skew-symmetric matrices
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For $t_1,t_2,...,t_6in mathbb R$, let $P_(t_1,t_2,...,t_6)=beginpmatrix 0&t_1&t_2&t_3\ -t_1&0&t_4&t_5\ -t_2&-t_4 &0&t_6\-t_3&-t_5&-t_6&0endpmatrix$.
Let $V=P_(t_1,t_2,...,t_6) : t_1,...,t_6in mathbb R$ i.e. $V$ is the vector space of all $4times 4$ skew-symmetric matrices with real entries .
Then $V$ is a $6$-dimensional vector space with a basis given by
$beta=P_(1,0,...,0), P_(0,1,0,...,0), P_(0,0,1,0,...,0),...,P_(0,...,0,1)$ .
Now given $Ain M(4,mathbb R)$, consider the map $phi_A:V to V$ given by $phi_A(P)=APA^t,forall Pin V$. Let $widehat A$ be the matrix of $phi_A$ w.r.t. the basis $beta $ of $V$ (so $widehat A in M(6,mathbb R)$ ). It is easy to see that if $A in GL(4,mathbb R)$ then $phi_A$ is an isomorphism, hence $A in GL(4,mathbb R)$ implies $widehat Ain GL(6,mathbb R) $. Now consider the map $psi :GL(4,mathbb R) to GL(6,mathbb R)$ given by $psi(A)=widehat A$ . Since $phi_AB=phi_A circ phi_B$, hence $widehat AB=widehat A . widehat B ,forall A,B in GL(4,mathbb R)$, thus $psi$ is a group homomorphism.
My question is: How can we describe the kernel of $psi$ i.e. how can we describe the set $A in GL(4,mathbb R) : widehat A=Id$ ?
My try: $widehat A=Id $ if and only if $phi_A$ is the identity map if and only if $APA^t=P$ for every $4times 4$ skew-symmetric matrix $P$. But I am unable to simplify this further.
Please help.
linear-algebra matrices vector-spaces multilinear-algebra exterior-algebra
asked Sep 7 at 2:22
user521337
298111
add a comment |Â
up vote
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For $t_1,t_2,...,t_6in mathbb R$, let $P_(t_1,t_2,...,t_6)=beginpmatrix 0&t_1&t_2&t_3\ -t_1&0&t_4&t_5\ -t_2&-t_4 &0&t_6\-t_3&-t_5&-t_6&0endpmatrix$.
Let $V=P_(t_1,t_2,...,t_6) : t_1,...,t_6in mathbb R$ i.e. $V$ is the vector space of all $4times 4$ skew-symmetric matrices with real entries .
Then $V$ is a $6$-dimensional vector space with a basis given by
$beta=P_(1,0,...,0), P_(0,1,0,...,0), P_(0,0,1,0,...,0),...,P_(0,...,0,1)$ .
Now given $Ain M(4,mathbb R)$, consider the map $phi_A:V to V$ given by $phi_A(P)=APA^t,forall Pin V$. Let $widehat A$ be the matrix of $phi_A$ w.r.t. the basis $beta $ of $V$ (so $widehat A in M(6,mathbb R)$ ). It is easy to see that if $A in GL(4,mathbb R)$ then $phi_A$ is an isomorphism, hence $A in GL(4,mathbb R)$ implies $widehat Ain GL(6,mathbb R) $. Now consider the map $psi :GL(4,mathbb R) to GL(6,mathbb R)$ given by $psi(A)=widehat A$ . Since $phi_AB=phi_A circ phi_B$, hence $widehat AB=widehat A . widehat B ,forall A,B in GL(4,mathbb R)$, thus $psi$ is a group homomorphism.
My question is: How can we describe the kernel of $psi$ i.e. how can we describe the set $A in GL(4,mathbb R) : widehat A=Id$ ?
My try: $widehat A=Id $ if and only if $phi_A$ is the identity map if and only if $APA^t=P$ for every $4times 4$ skew-symmetric matrix $P$. But I am unable to simplify this further.
Please help.
linear-algebra matrices vector-spaces multilinear-algebra exterior-algebra
asked Sep 7 at 2:22
user521337
298111
Put in all basis elements for $P$. For example, let $P = P_(1,0,0,0,0,0)$, set $J = (beginsmallmatrix0 & 1\-1 & 0endsmallmatrix)$, and let $A = (beginsmallmatrixB & C\D & Eendsmallmatrix)$ with $2times 2$ blocks $B,C,D,E$. Then you'll immediately see that $APA^T = P$ implies $BJB^T = J$ (which means $det B = 1$) and $DJB^T = 0$. As $B$ is invertible ($det B = 1$), this implies $D = 0$. Similarly, choosing $P = P_(0,0,0,0,0,1)$ gives $det E=1$ and $C = 0$. I guess this is not the end...
– amsmath
Sep 7 at 5:26
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For $t_1,t_2,...,t_6in mathbb R$, let $P_(t_1,t_2,...,t_6)=beginpmatrix 0&t_1&t_2&t_3\ -t_1&0&t_4&t_5\ -t_2&-t_4 &0&t_6\-t_3&-t_5&-t_6&0endpmatrix$.
Let $V=P_(t_1,t_2,...,t_6) : t_1,...,t_6in mathbb R$ i.e. $V$ is the vector space of all $4times 4$ skew-symmetric matrices with real entries .
Then $V$ is a $6$-dimensional vector space with a basis given by
$beta=P_(1,0,...,0), P_(0,1,0,...,0), P_(0,0,1,0,...,0),...,P_(0,...,0,1)$ .
Now given $Ain M(4,mathbb R)$, consider the map $phi_A:V to V$ given by $phi_A(P)=APA^t,forall Pin V$. Let $widehat A$ be the matrix of $phi_A$ w.r.t. the basis $beta $ of $V$ (so $widehat A in M(6,mathbb R)$ ). It is easy to see that if $A in GL(4,mathbb R)$ then $phi_A$ is an isomorphism, hence $A in GL(4,mathbb R)$ implies $widehat Ain GL(6,mathbb R) $. Now consider the map $psi :GL(4,mathbb R) to GL(6,mathbb R)$ given by $psi(A)=widehat A$ . Since $phi_AB=phi_A circ phi_B$, hence $widehat AB=widehat A . widehat B ,forall A,B in GL(4,mathbb R)$, thus $psi$ is a group homomorphism.
My question is: How can we describe the kernel of $psi$ i.e. how can we describe the set $A in GL(4,mathbb R) : widehat A=Id$ ?
My try: $widehat A=Id $ if and only if $phi_A$ is the identity map if and only if $APA^t=P$ for every $4times 4$ skew-symmetric matrix $P$. But I am unable to simplify this further.
Please help.
linear-algebra matrices vector-spaces multilinear-algebra exterior-algebra
asked Sep 7 at 2:22
user521337
298111
For $t_1,t_2,...,t_6in mathbb R$, let $P_(t_1,t_2,...,t_6)=beginpmatrix 0&t_1&t_2&t_3\ -t_1&0&t_4&t_5\ -t_2&-t_4 &0&t_6\-t_3&-t_5&-t_6&0endpmatrix$.
Let $V=P_(t_1,t_2,...,t_6) : t_1,...,t_6in mathbb R$ i.e. $V$ is the vector space of all $4times 4$ skew-symmetric matrices with real entries .
Then $V$ is a $6$-dimensional vector space with a basis given by
$beta=P_(1,0,...,0), P_(0,1,0,...,0), P_(0,0,1,0,...,0),...,P_(0,...,0,1)$ .
Now given $Ain M(4,mathbb R)$, consider the map $phi_A:V to V$ given by $phi_A(P)=APA^t,forall Pin V$. Let $widehat A$ be the matrix of $phi_A$ w.r.t. the basis $beta $ of $V$ (so $widehat A in M(6,mathbb R)$ ). It is easy to see that if $A in GL(4,mathbb R)$ then $phi_A$ is an isomorphism, hence $A in GL(4,mathbb R)$ implies $widehat Ain GL(6,mathbb R) $. Now consider the map $psi :GL(4,mathbb R) to GL(6,mathbb R)$ given by $psi(A)=widehat A$ . Since $phi_AB=phi_A circ phi_B$, hence $widehat AB=widehat A . widehat B ,forall A,B in GL(4,mathbb R)$, thus $psi$ is a group homomorphism.
My question is: How can we describe the kernel of $psi$ i.e. how can we describe the set $A in GL(4,mathbb R) : widehat A=Id$ ?
My try: $widehat A=Id $ if and only if $phi_A$ is the identity map if and only if $APA^t=P$ for every $4times 4$ skew-symmetric matrix $P$. But I am unable to simplify this further.
Please help.
linear-algebra matrices vector-spaces multilinear-algebra exterior-algebra
linear-algebra matrices vector-spaces multilinear-algebra exterior-algebra
asked Sep 7 at 2:22
user521337
298111
asked Sep 7 at 2:22
user521337
298111
edited Sep 7 at 2:37
asked Sep 7 at 2:22
user521337
298111
asked Sep 7 at 2:22
user521337
298111
asked Sep 7 at 2:22
user521337
298111
298111
Put in all basis elements for $P$. For example, let $P = P_(1,0,0,0,0,0)$, set $J = (beginsmallmatrix0 & 1\-1 & 0endsmallmatrix)$, and let $A = (beginsmallmatrixB & C\D & Eendsmallmatrix)$ with $2times 2$ blocks $B,C,D,E$. Then you'll immediately see that $APA^T = P$ implies $BJB^T = J$ (which means $det B = 1$) and $DJB^T = 0$. As $B$ is invertible ($det B = 1$), this implies $D = 0$. Similarly, choosing $P = P_(0,0,0,0,0,1)$ gives $det E=1$ and $C = 0$. I guess this is not the end...
– amsmath
Sep 7 at 5:26
add a comment |Â
Put in all basis elements for $P$. For example, let $P = P_(1,0,0,0,0,0)$, set $J = (beginsmallmatrix0 & 1\-1 & 0endsmallmatrix)$, and let $A = (beginsmallmatrixB & C\D & Eendsmallmatrix)$ with $2times 2$ blocks $B,C,D,E$. Then you'll immediately see that $APA^T = P$ implies $BJB^T = J$ (which means $det B = 1$) and $DJB^T = 0$. As $B$ is invertible ($det B = 1$), this implies $D = 0$. Similarly, choosing $P = P_(0,0,0,0,0,1)$ gives $det E=1$ and $C = 0$. I guess this is not the end...
– amsmath
Sep 7 at 5:26
Put in all basis elements for $P$. For example, let $P = P_(1,0,0,0,0,0)$, set $J = (beginsmallmatrix0 & 1\-1 & 0endsmallmatrix)$, and let $A = (beginsmallmatrixB & C\D & Eendsmallmatrix)$ with $2times 2$ blocks $B,C,D,E$. Then you'll immediately see that $APA^T = P$ implies $BJB^T = J$ (which means $det B = 1$) and $DJB^T = 0$. As $B$ is invertible ($det B = 1$), this implies $D = 0$. Similarly, choosing $P = P_(0,0,0,0,0,1)$ gives $det E=1$ and $C = 0$. I guess this is not the end...
– amsmath
Sep 7 at 5:26
Put in all basis elements for $P$. For example, let $P = P_(1,0,0,0,0,0)$, set $J = (beginsmallmatrix0 & 1\-1 & 0endsmallmatrix)$, and let $A = (beginsmallmatrixB & C\D & Eendsmallmatrix)$ with $2times 2$ blocks $B,C,D,E$. Then you'll immediately see that $APA^T = P$ implies $BJB^T = J$ (which means $det B = 1$) and $DJB^T = 0$. As $B$ is invertible ($det B = 1$), this implies $D = 0$. Similarly, choosing $P = P_(0,0,0,0,0,1)$ gives $det E=1$ and $C = 0$. I guess this is not the end...
– amsmath
Sep 7 at 5:26
add a comment |Â
1 Answer
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By my comment above, $A = (beginsmallmatrixB & 0\0 & Eendsmallmatrix)$, where $det B = det E = 1$. Now, let $X$ be any $2times 2$ matrix and set $P = (beginsmallmatrix0 & X\-X^T & 0endsmallmatrix)$. Then $P$ is skew-symmetric and
beginalign*
beginpmatrix0 & X\-X^T & 0endpmatrix
&= P = APA^T = beginpmatrixB & 0\0 & Eendpmatrixbeginpmatrix0 & X\-X^T & 0endpmatrixbeginpmatrixB^T & 0\0 & E^Tendpmatrix\
&= beginpmatrix0 & BXE^T\-EX^TB^T & 0endpmatrix.
endalign*
Hence, $BXE^T = X$ for all $XinBbb R^2times 2$. Setting $X = I_2$ gives $E^T = B^-1$ and thus $BX = XB$ for all $XinBbb R^2times 2$. This implies $B = bI_2$ with $binBbb R$. But $det B = 1$, so $b^2 = 1$, i.e., $B = pm I$. Finally,
$$
A = beginpmatrixpm I_2 & 0\0 & pm I_2endpmatrix = pm I_4.
$$
answered Sep 7 at 6:34
amsmath
2,670114
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
By my comment above, $A = (beginsmallmatrixB & 0\0 & Eendsmallmatrix)$, where $det B = det E = 1$. Now, let $X$ be any $2times 2$ matrix and set $P = (beginsmallmatrix0 & X\-X^T & 0endsmallmatrix)$. Then $P$ is skew-symmetric and
beginalign*
beginpmatrix0 & X\-X^T & 0endpmatrix
&= P = APA^T = beginpmatrixB & 0\0 & Eendpmatrixbeginpmatrix0 & X\-X^T & 0endpmatrixbeginpmatrixB^T & 0\0 & E^Tendpmatrix\
&= beginpmatrix0 & BXE^T\-EX^TB^T & 0endpmatrix.
endalign*
Hence, $BXE^T = X$ for all $XinBbb R^2times 2$. Setting $X = I_2$ gives $E^T = B^-1$ and thus $BX = XB$ for all $XinBbb R^2times 2$. This implies $B = bI_2$ with $binBbb R$. But $det B = 1$, so $b^2 = 1$, i.e., $B = pm I$. Finally,
$$
A = beginpmatrixpm I_2 & 0\0 & pm I_2endpmatrix = pm I_4.
$$
answered Sep 7 at 6:34
amsmath
2,670114
add a comment |Â
up vote
0
down vote
accepted
By my comment above, $A = (beginsmallmatrixB & 0\0 & Eendsmallmatrix)$, where $det B = det E = 1$. Now, let $X$ be any $2times 2$ matrix and set $P = (beginsmallmatrix0 & X\-X^T & 0endsmallmatrix)$. Then $P$ is skew-symmetric and
beginalign*
beginpmatrix0 & X\-X^T & 0endpmatrix
&= P = APA^T = beginpmatrixB & 0\0 & Eendpmatrixbeginpmatrix0 & X\-X^T & 0endpmatrixbeginpmatrixB^T & 0\0 & E^Tendpmatrix\
&= beginpmatrix0 & BXE^T\-EX^TB^T & 0endpmatrix.
endalign*
Hence, $BXE^T = X$ for all $XinBbb R^2times 2$. Setting $X = I_2$ gives $E^T = B^-1$ and thus $BX = XB$ for all $XinBbb R^2times 2$. This implies $B = bI_2$ with $binBbb R$. But $det B = 1$, so $b^2 = 1$, i.e., $B = pm I$. Finally,
$$
A = beginpmatrixpm I_2 & 0\0 & pm I_2endpmatrix = pm I_4.
$$
answered Sep 7 at 6:34
amsmath
2,670114
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
By my comment above, $A = (beginsmallmatrixB & 0\0 & Eendsmallmatrix)$, where $det B = det E = 1$. Now, let $X$ be any $2times 2$ matrix and set $P = (beginsmallmatrix0 & X\-X^T & 0endsmallmatrix)$. Then $P$ is skew-symmetric and
beginalign*
beginpmatrix0 & X\-X^T & 0endpmatrix
&= P = APA^T = beginpmatrixB & 0\0 & Eendpmatrixbeginpmatrix0 & X\-X^T & 0endpmatrixbeginpmatrixB^T & 0\0 & E^Tendpmatrix\
&= beginpmatrix0 & BXE^T\-EX^TB^T & 0endpmatrix.
endalign*
Hence, $BXE^T = X$ for all $XinBbb R^2times 2$. Setting $X = I_2$ gives $E^T = B^-1$ and thus $BX = XB$ for all $XinBbb R^2times 2$. This implies $B = bI_2$ with $binBbb R$. But $det B = 1$, so $b^2 = 1$, i.e., $B = pm I$. Finally,
$$
A = beginpmatrixpm I_2 & 0\0 & pm I_2endpmatrix = pm I_4.
$$
answered Sep 7 at 6:34
amsmath
2,670114
By my comment above, $A = (beginsmallmatrixB & 0\0 & Eendsmallmatrix)$, where $det B = det E = 1$. Now, let $X$ be any $2times 2$ matrix and set $P = (beginsmallmatrix0 & X\-X^T & 0endsmallmatrix)$. Then $P$ is skew-symmetric and
beginalign*
beginpmatrix0 & X\-X^T & 0endpmatrix
&= P = APA^T = beginpmatrixB & 0\0 & Eendpmatrixbeginpmatrix0 & X\-X^T & 0endpmatrixbeginpmatrixB^T & 0\0 & E^Tendpmatrix\
&= beginpmatrix0 & BXE^T\-EX^TB^T & 0endpmatrix.
endalign*
Hence, $BXE^T = X$ for all $XinBbb R^2times 2$. Setting $X = I_2$ gives $E^T = B^-1$ and thus $BX = XB$ for all $XinBbb R^2times 2$. This implies $B = bI_2$ with $binBbb R$. But $det B = 1$, so $b^2 = 1$, i.e., $B = pm I$. Finally,
$$
A = beginpmatrixpm I_2 & 0\0 & pm I_2endpmatrix = pm I_4.
$$
answered Sep 7 at 6:34
amsmath
2,670114
answered Sep 7 at 6:34
amsmath
2,670114
answered Sep 7 at 6:34
amsmath
2,670114
answered Sep 7 at 6:34
amsmath
2,670114
2,670114
add a comment |Â
add a comment |Â
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Put in all basis elements for $P$. For example, let $P = P_(1,0,0,0,0,0)$, set $J = (beginsmallmatrix0 & 1\-1 & 0endsmallmatrix)$, and let $A = (beginsmallmatrixB & C\D & Eendsmallmatrix)$ with $2times 2$ blocks $B,C,D,E$. Then you'll immediately see that $APA^T = P$ implies $BJB^T = J$ (which means $det B = 1$) and $DJB^T = 0$. As $B$ is invertible ($det B = 1$), this implies $D = 0$. Similarly, choosing $P = P_(0,0,0,0,0,1)$ gives $det E=1$ and $C = 0$. I guess this is not the end...
– amsmath
Sep 7 at 5:26