Can a linear isometry always be expressed in terms of an orthogonal matrix?
$begingroup$
Is the following true?
Let $S: mathbb{R}^n to mathbb{R}^n$ be a linear transformation such that $||S(v)|| = ||v|| text{for all} v in mathbb{R}^n$, where $||cdot||$ denotes the Euclidean norm. Then, for some $A in O(n)$ and for all $v in mathbb{R}^n$, we have $S(v) = Av$. Here $O(n)$ denotes the set of all orthogonal $n times n$ matrices.
If so, how can one prove it?
linear-algebra linear-transformations orthogonal-matrices
asked Dec 6 '18 at 16:10
E-muE-mu
786417
$endgroup$
add a comment |
$begingroup$
Is the following true?
Let $S: mathbb{R}^n to mathbb{R}^n$ be a linear transformation such that $||S(v)|| = ||v|| text{for all} v in mathbb{R}^n$, where $||cdot||$ denotes the Euclidean norm. Then, for some $A in O(n)$ and for all $v in mathbb{R}^n$, we have $S(v) = Av$. Here $O(n)$ denotes the set of all orthogonal $n times n$ matrices.
If so, how can one prove it?
linear-algebra linear-transformations orthogonal-matrices
asked Dec 6 '18 at 16:10
E-muE-mu
786417
$endgroup$
add a comment |
$begingroup$
Is the following true?
Let $S: mathbb{R}^n to mathbb{R}^n$ be a linear transformation such that $||S(v)|| = ||v|| text{for all} v in mathbb{R}^n$, where $||cdot||$ denotes the Euclidean norm. Then, for some $A in O(n)$ and for all $v in mathbb{R}^n$, we have $S(v) = Av$. Here $O(n)$ denotes the set of all orthogonal $n times n$ matrices.
If so, how can one prove it?
linear-algebra linear-transformations orthogonal-matrices
asked Dec 6 '18 at 16:10
E-muE-mu
786417
$endgroup$
Is the following true?
Let $S: mathbb{R}^n to mathbb{R}^n$ be a linear transformation such that $||S(v)|| = ||v|| text{for all} v in mathbb{R}^n$, where $||cdot||$ denotes the Euclidean norm. Then, for some $A in O(n)$ and for all $v in mathbb{R}^n$, we have $S(v) = Av$. Here $O(n)$ denotes the set of all orthogonal $n times n$ matrices.
If so, how can one prove it?
linear-algebra linear-transformations orthogonal-matrices
linear-algebra linear-transformations orthogonal-matrices
asked Dec 6 '18 at 16:10
E-muE-mu
786417
asked Dec 6 '18 at 16:10
E-muE-mu
786417
asked Dec 6 '18 at 16:10
E-muE-mu
786417
asked Dec 6 '18 at 16:10
E-muE-mu
786417
asked Dec 6 '18 at 16:10
E-muE-mu
786417
786417
add a comment |
add a comment |
1 Answer
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$begingroup$
Let $A$ be the matrix of $S$. Since $langle u,vrangle=frac14lVert u+vrVert^2-frac14lVert u-vrVert^2$, we have that $langle Au,Avrangle=langle u,vrangle$ for all $u,v$. Id est, $u^t(A^tA)v=(Au)^t(Av)=u^tv=u^tIv$ for all $u,v$. Since $e_i^tXe_j=X_{ij}$, that identity implies $A^tA=I$.
answered Dec 6 '18 at 16:19
Saucy O'PathSaucy O'Path
5,8641626
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $A$ be the matrix of $S$. Since $langle u,vrangle=frac14lVert u+vrVert^2-frac14lVert u-vrVert^2$, we have that $langle Au,Avrangle=langle u,vrangle$ for all $u,v$. Id est, $u^t(A^tA)v=(Au)^t(Av)=u^tv=u^tIv$ for all $u,v$. Since $e_i^tXe_j=X_{ij}$, that identity implies $A^tA=I$.
answered Dec 6 '18 at 16:19
Saucy O'PathSaucy O'Path
5,8641626
$endgroup$
add a comment |
$begingroup$
Let $A$ be the matrix of $S$. Since $langle u,vrangle=frac14lVert u+vrVert^2-frac14lVert u-vrVert^2$, we have that $langle Au,Avrangle=langle u,vrangle$ for all $u,v$. Id est, $u^t(A^tA)v=(Au)^t(Av)=u^tv=u^tIv$ for all $u,v$. Since $e_i^tXe_j=X_{ij}$, that identity implies $A^tA=I$.
answered Dec 6 '18 at 16:19
Saucy O'PathSaucy O'Path
5,8641626
$endgroup$
add a comment |
$begingroup$
Let $A$ be the matrix of $S$. Since $langle u,vrangle=frac14lVert u+vrVert^2-frac14lVert u-vrVert^2$, we have that $langle Au,Avrangle=langle u,vrangle$ for all $u,v$. Id est, $u^t(A^tA)v=(Au)^t(Av)=u^tv=u^tIv$ for all $u,v$. Since $e_i^tXe_j=X_{ij}$, that identity implies $A^tA=I$.
answered Dec 6 '18 at 16:19
Saucy O'PathSaucy O'Path
5,8641626
$endgroup$
Let $A$ be the matrix of $S$. Since $langle u,vrangle=frac14lVert u+vrVert^2-frac14lVert u-vrVert^2$, we have that $langle Au,Avrangle=langle u,vrangle$ for all $u,v$. Id est, $u^t(A^tA)v=(Au)^t(Av)=u^tv=u^tIv$ for all $u,v$. Since $e_i^tXe_j=X_{ij}$, that identity implies $A^tA=I$.
answered Dec 6 '18 at 16:19
Saucy O'PathSaucy O'Path
5,8641626
answered Dec 6 '18 at 16:19
Saucy O'PathSaucy O'Path
5,8641626
answered Dec 6 '18 at 16:19
Saucy O'PathSaucy O'Path
5,8641626
answered Dec 6 '18 at 16:19
Saucy O'PathSaucy O'Path
5,8641626
5,8641626
add a comment |
add a comment |
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