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an orthogonal map associated with inner product
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Let $rleq n$ be a natural number and let $v_1,v_2,dots ,v_r$ and $w_1,w_2,dots ,w_r$ be two linearly independent subsets of $mathbbR^n$ such that $langle v_i,v_j rangle = langle w_i,w_j rangle forall 1leq i,j leq r$, where $langle ,rangle$ denotes the standard inner product on $mathbbR^n$. Prove that there exists an orthogonal operator $T$ on $mathbbR^n$ such that $T(v_i)=w_i$ for all $1leq i leq r$.
An orthogonal mapping preserves inner products, and it is also known that $T$ is orthogonal on $mathbbR^n$ if $langle T(alpha),T(beta)rangle = langle alpha ,beta rangle$ for all $alpha,beta in mathbbR^n$. But the question here demands to be proved that the existence of such an orthogonal map. So do I need to find an example of such map or is there another way to prove that in general? Any help will be appreciated.
linear-algebra
asked Jul 27 at 16:28
am_11235...
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Let $rleq n$ be a natural number and let $v_1,v_2,dots ,v_r$ and $w_1,w_2,dots ,w_r$ be two linearly independent subsets of $mathbbR^n$ such that $langle v_i,v_j rangle = langle w_i,w_j rangle forall 1leq i,j leq r$, where $langle ,rangle$ denotes the standard inner product on $mathbbR^n$. Prove that there exists an orthogonal operator $T$ on $mathbbR^n$ such that $T(v_i)=w_i$ for all $1leq i leq r$.
An orthogonal mapping preserves inner products, and it is also known that $T$ is orthogonal on $mathbbR^n$ if $langle T(alpha),T(beta)rangle = langle alpha ,beta rangle$ for all $alpha,beta in mathbbR^n$. But the question here demands to be proved that the existence of such an orthogonal map. So do I need to find an example of such map or is there another way to prove that in general? Any help will be appreciated.
linear-algebra
asked Jul 27 at 16:28
am_11235...
1797
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $rleq n$ be a natural number and let $v_1,v_2,dots ,v_r$ and $w_1,w_2,dots ,w_r$ be two linearly independent subsets of $mathbbR^n$ such that $langle v_i,v_j rangle = langle w_i,w_j rangle forall 1leq i,j leq r$, where $langle ,rangle$ denotes the standard inner product on $mathbbR^n$. Prove that there exists an orthogonal operator $T$ on $mathbbR^n$ such that $T(v_i)=w_i$ for all $1leq i leq r$.
An orthogonal mapping preserves inner products, and it is also known that $T$ is orthogonal on $mathbbR^n$ if $langle T(alpha),T(beta)rangle = langle alpha ,beta rangle$ for all $alpha,beta in mathbbR^n$. But the question here demands to be proved that the existence of such an orthogonal map. So do I need to find an example of such map or is there another way to prove that in general? Any help will be appreciated.
linear-algebra
asked Jul 27 at 16:28
am_11235...
1797
Let $rleq n$ be a natural number and let $v_1,v_2,dots ,v_r$ and $w_1,w_2,dots ,w_r$ be two linearly independent subsets of $mathbbR^n$ such that $langle v_i,v_j rangle = langle w_i,w_j rangle forall 1leq i,j leq r$, where $langle ,rangle$ denotes the standard inner product on $mathbbR^n$. Prove that there exists an orthogonal operator $T$ on $mathbbR^n$ such that $T(v_i)=w_i$ for all $1leq i leq r$.
An orthogonal mapping preserves inner products, and it is also known that $T$ is orthogonal on $mathbbR^n$ if $langle T(alpha),T(beta)rangle = langle alpha ,beta rangle$ for all $alpha,beta in mathbbR^n$. But the question here demands to be proved that the existence of such an orthogonal map. So do I need to find an example of such map or is there another way to prove that in general? Any help will be appreciated.
linear-algebra
asked Jul 27 at 16:28
am_11235...
1797
asked Jul 27 at 16:28
am_11235...
1797
asked Jul 27 at 16:28
am_11235...
1797
asked Jul 27 at 16:28
am_11235...
1797
1797
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add a comment |Â
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You define $Tv_j=w_j$ for all $j=1,ldots,r$, and since both sets are linearly independent, $T$ gets defined by linearity on their span.
But now you have the problem that maybe $r<n$. Expanding both sets to bases does not work right away, because you have no way to preserve the inner product relations. So what you do is take $v_r+1,ldots,v_n^perp$ to be an orthonormal basis of $v_1,ldots,v_r^perp$, and $w_r+1,ldots,w_n$ an orthonormal basis of $w_1,ldots,w_r^perp$. Then define $Tv_j=w_j$ for all $j=1,ldots, n$. The inner product relations are preserved because ${v_j,v_krangle=delta_kj=langle w_j,w_krangle$ for $j,k>r$, and when one index is below $r$ and the other above, the inner product is zero by the orthogonality.
answered Jul 27 at 16:35
Martin Argerami
115k1071164
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Let the subspace
$V = textspan( v_1, v_2, ldots, v_r ) = textspan( w_1, w_2, ldots, w_r ); tag 1$
we define $T$ on $V$ by setting
$Tv_k = w_k, ; 1 le k le r, tag 2$
and extending $T$ by linearity to all of $V$, viz:
$Tleft ( displaystyle sum_1^r alpha_i v_i right ) = displaystylesum_1^r alpha_i Tv_i = sum_1^r alpha_i w_i; tag 3$
that such an extension is possible follows from the linear independence of the $v_i$ and the $w_i$ as is well-known.
If
$alpha, beta in V tag 4$
we may write
$alpha = displaystyle sum_1^r alpha_i v_i, ; beta = sum_1^r beta_j v_j, tag 5$
whence
$left langle Talpha, Tbeta right rangle = left langle T left (displaystyle sum_1^r alpha_i v_i right), T left (displaystyle sum_1^r beta_j v_j right) right rangle = left langle displaystyle sum_1^r alpha_i Tv_i, displaystyle sum_1^r beta_j Tv_j right rangle = displaystyle sum_i, j = 1^r alpha_i beta_j langle Tv_i, Tv_j rangle = sum_i, j = 1^r alpha_i beta_j langle w_i, w_j rangle = sum_i, j = 1^r alpha_i beta_j langle v_i, v_j rangle$
$= left langle displaystyle sum_1^r alpha_i v_i, displaystyle sum_1^r beta_j v_j right rangle = langle alpha, beta rangle, tag 6$
which proves that $T:V to V$ is orthogonal.
In the event that $r < n$, we extend $T$ from $V$ to all of $Bbb R^n$ as follows: let $W subset Bbb R^n$ be the orthogonal compliment of $V$, $W = V^bot$; then we have
$Bbb R^n = V oplus W; tag 7$
we observe that any $x in Bbb R^n$ may be represented uniquely as
$x = v + w, ; v in V, w in W; tag8$
existence of such a decomposition follows directly from (7), i.e. from the fact that $W = V^bot$; to see uniqueness, suppose
$x = v_1 + w_1 = v_2 + w_2, ; v_1, v_2 in V, w_1, w_2 in W; tag 9$
then
$v_1 - v_2 = w_2 - w_1, tag10$
which forces
$v_1 - v_2 = 0 = w_2 - w_1, tag11$
since
$v_1 - v_2 in V, ; w_1 - w_2 in W = V^bot, tag12$
and hence
$Vert v_1 - v_2 Vert^2 = langle v_1 - v_2, v_1 - v_2 rangle = langle v_1 - v_2, w_2 - w_1 rangle = 0, tag13$
implying $v_1 = v_2$, and thus $w_1 = w_2$ via (11); so the decomposition (8) is unique; this uniqueness means we may unabiguously define an extension $T_E$ of $T$ from $V$ to $V oplus W = Bbb R^n$ by
$T_E(v + w) = Tv + w, ; v in V, w in W; tag14$
that is,
$T_E = T oplus I: V oplus W to V oplus W; tag15$
we can formally show $T_E$ is orthogonal on all of $Bbb R^n$ if we observe that, for $v_1, v_2 in V$, $w_1, w_2 in W$,
$langle v_1 + w_1, v_2 + w_2 rangle$
$= langle v_1, v_2 rangle + langle v_1, w_2 rangle + langle w_1, v_2 rangle + langle w_1, w_2 rangle = langle v_1, v_2 rangle + langle w_1, w_2 rangle; tag16$
thus,
$langle T_E(v_1 + w_1), T_E(v_2 + w_2) rangle = langle (T oplus I)(v_1 + w_1), (T oplus I)(v_2 + w_2) rangle$
$= langle Tv_1 + w_1, Tv_2 + w_2 rangle = langle Tv_1, Tv_2 rangle + langle w_1, w_2 rangle$
$= langle v_1, v_2 rangle + langle w_1, w_2 rangle = langle v_1 + w_1, v_2 + w_2 rangle, tag17$
which shows that $T_E = T oplus I$ is orthogonal on $Bbb R^n$.
answered Jul 28 at 1:54
Robert Lewis
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2 Answers
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2 Answers
2
active
oldest
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active
oldest
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active
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up vote
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down vote
You define $Tv_j=w_j$ for all $j=1,ldots,r$, and since both sets are linearly independent, $T$ gets defined by linearity on their span.
But now you have the problem that maybe $r<n$. Expanding both sets to bases does not work right away, because you have no way to preserve the inner product relations. So what you do is take $v_r+1,ldots,v_n^perp$ to be an orthonormal basis of $v_1,ldots,v_r^perp$, and $w_r+1,ldots,w_n$ an orthonormal basis of $w_1,ldots,w_r^perp$. Then define $Tv_j=w_j$ for all $j=1,ldots, n$. The inner product relations are preserved because ${v_j,v_krangle=delta_kj=langle w_j,w_krangle$ for $j,k>r$, and when one index is below $r$ and the other above, the inner product is zero by the orthogonality.
answered Jul 27 at 16:35
Martin Argerami
115k1071164
add a comment |Â
up vote
0
down vote
You define $Tv_j=w_j$ for all $j=1,ldots,r$, and since both sets are linearly independent, $T$ gets defined by linearity on their span.
But now you have the problem that maybe $r<n$. Expanding both sets to bases does not work right away, because you have no way to preserve the inner product relations. So what you do is take $v_r+1,ldots,v_n^perp$ to be an orthonormal basis of $v_1,ldots,v_r^perp$, and $w_r+1,ldots,w_n$ an orthonormal basis of $w_1,ldots,w_r^perp$. Then define $Tv_j=w_j$ for all $j=1,ldots, n$. The inner product relations are preserved because ${v_j,v_krangle=delta_kj=langle w_j,w_krangle$ for $j,k>r$, and when one index is below $r$ and the other above, the inner product is zero by the orthogonality.
answered Jul 27 at 16:35
Martin Argerami
115k1071164
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You define $Tv_j=w_j$ for all $j=1,ldots,r$, and since both sets are linearly independent, $T$ gets defined by linearity on their span.
But now you have the problem that maybe $r<n$. Expanding both sets to bases does not work right away, because you have no way to preserve the inner product relations. So what you do is take $v_r+1,ldots,v_n^perp$ to be an orthonormal basis of $v_1,ldots,v_r^perp$, and $w_r+1,ldots,w_n$ an orthonormal basis of $w_1,ldots,w_r^perp$. Then define $Tv_j=w_j$ for all $j=1,ldots, n$. The inner product relations are preserved because ${v_j,v_krangle=delta_kj=langle w_j,w_krangle$ for $j,k>r$, and when one index is below $r$ and the other above, the inner product is zero by the orthogonality.
answered Jul 27 at 16:35
Martin Argerami
115k1071164
You define $Tv_j=w_j$ for all $j=1,ldots,r$, and since both sets are linearly independent, $T$ gets defined by linearity on their span.
But now you have the problem that maybe $r<n$. Expanding both sets to bases does not work right away, because you have no way to preserve the inner product relations. So what you do is take $v_r+1,ldots,v_n^perp$ to be an orthonormal basis of $v_1,ldots,v_r^perp$, and $w_r+1,ldots,w_n$ an orthonormal basis of $w_1,ldots,w_r^perp$. Then define $Tv_j=w_j$ for all $j=1,ldots, n$. The inner product relations are preserved because ${v_j,v_krangle=delta_kj=langle w_j,w_krangle$ for $j,k>r$, and when one index is below $r$ and the other above, the inner product is zero by the orthogonality.
answered Jul 27 at 16:35
Martin Argerami
115k1071164
answered Jul 27 at 16:35
Martin Argerami
115k1071164
answered Jul 27 at 16:35
Martin Argerami
115k1071164
answered Jul 27 at 16:35
Martin Argerami
115k1071164
115k1071164
add a comment |Â
add a comment |Â
up vote
0
down vote
Let the subspace
$V = textspan( v_1, v_2, ldots, v_r ) = textspan( w_1, w_2, ldots, w_r ); tag 1$
we define $T$ on $V$ by setting
$Tv_k = w_k, ; 1 le k le r, tag 2$
and extending $T$ by linearity to all of $V$, viz:
$Tleft ( displaystyle sum_1^r alpha_i v_i right ) = displaystylesum_1^r alpha_i Tv_i = sum_1^r alpha_i w_i; tag 3$
that such an extension is possible follows from the linear independence of the $v_i$ and the $w_i$ as is well-known.
If
$alpha, beta in V tag 4$
we may write
$alpha = displaystyle sum_1^r alpha_i v_i, ; beta = sum_1^r beta_j v_j, tag 5$
whence
$left langle Talpha, Tbeta right rangle = left langle T left (displaystyle sum_1^r alpha_i v_i right), T left (displaystyle sum_1^r beta_j v_j right) right rangle = left langle displaystyle sum_1^r alpha_i Tv_i, displaystyle sum_1^r beta_j Tv_j right rangle = displaystyle sum_i, j = 1^r alpha_i beta_j langle Tv_i, Tv_j rangle = sum_i, j = 1^r alpha_i beta_j langle w_i, w_j rangle = sum_i, j = 1^r alpha_i beta_j langle v_i, v_j rangle$
$= left langle displaystyle sum_1^r alpha_i v_i, displaystyle sum_1^r beta_j v_j right rangle = langle alpha, beta rangle, tag 6$
which proves that $T:V to V$ is orthogonal.
In the event that $r < n$, we extend $T$ from $V$ to all of $Bbb R^n$ as follows: let $W subset Bbb R^n$ be the orthogonal compliment of $V$, $W = V^bot$; then we have
$Bbb R^n = V oplus W; tag 7$
we observe that any $x in Bbb R^n$ may be represented uniquely as
$x = v + w, ; v in V, w in W; tag8$
existence of such a decomposition follows directly from (7), i.e. from the fact that $W = V^bot$; to see uniqueness, suppose
$x = v_1 + w_1 = v_2 + w_2, ; v_1, v_2 in V, w_1, w_2 in W; tag 9$
then
$v_1 - v_2 = w_2 - w_1, tag10$
which forces
$v_1 - v_2 = 0 = w_2 - w_1, tag11$
since
$v_1 - v_2 in V, ; w_1 - w_2 in W = V^bot, tag12$
and hence
$Vert v_1 - v_2 Vert^2 = langle v_1 - v_2, v_1 - v_2 rangle = langle v_1 - v_2, w_2 - w_1 rangle = 0, tag13$
implying $v_1 = v_2$, and thus $w_1 = w_2$ via (11); so the decomposition (8) is unique; this uniqueness means we may unabiguously define an extension $T_E$ of $T$ from $V$ to $V oplus W = Bbb R^n$ by
$T_E(v + w) = Tv + w, ; v in V, w in W; tag14$
that is,
$T_E = T oplus I: V oplus W to V oplus W; tag15$
we can formally show $T_E$ is orthogonal on all of $Bbb R^n$ if we observe that, for $v_1, v_2 in V$, $w_1, w_2 in W$,
$langle v_1 + w_1, v_2 + w_2 rangle$
$= langle v_1, v_2 rangle + langle v_1, w_2 rangle + langle w_1, v_2 rangle + langle w_1, w_2 rangle = langle v_1, v_2 rangle + langle w_1, w_2 rangle; tag16$
thus,
$langle T_E(v_1 + w_1), T_E(v_2 + w_2) rangle = langle (T oplus I)(v_1 + w_1), (T oplus I)(v_2 + w_2) rangle$
$= langle Tv_1 + w_1, Tv_2 + w_2 rangle = langle Tv_1, Tv_2 rangle + langle w_1, w_2 rangle$
$= langle v_1, v_2 rangle + langle w_1, w_2 rangle = langle v_1 + w_1, v_2 + w_2 rangle, tag17$
which shows that $T_E = T oplus I$ is orthogonal on $Bbb R^n$.
answered Jul 28 at 1:54
Robert Lewis
36.8k22155
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up vote
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Let the subspace
$V = textspan( v_1, v_2, ldots, v_r ) = textspan( w_1, w_2, ldots, w_r ); tag 1$
we define $T$ on $V$ by setting
$Tv_k = w_k, ; 1 le k le r, tag 2$
and extending $T$ by linearity to all of $V$, viz:
$Tleft ( displaystyle sum_1^r alpha_i v_i right ) = displaystylesum_1^r alpha_i Tv_i = sum_1^r alpha_i w_i; tag 3$
that such an extension is possible follows from the linear independence of the $v_i$ and the $w_i$ as is well-known.
If
$alpha, beta in V tag 4$
we may write
$alpha = displaystyle sum_1^r alpha_i v_i, ; beta = sum_1^r beta_j v_j, tag 5$
whence
$left langle Talpha, Tbeta right rangle = left langle T left (displaystyle sum_1^r alpha_i v_i right), T left (displaystyle sum_1^r beta_j v_j right) right rangle = left langle displaystyle sum_1^r alpha_i Tv_i, displaystyle sum_1^r beta_j Tv_j right rangle = displaystyle sum_i, j = 1^r alpha_i beta_j langle Tv_i, Tv_j rangle = sum_i, j = 1^r alpha_i beta_j langle w_i, w_j rangle = sum_i, j = 1^r alpha_i beta_j langle v_i, v_j rangle$
$= left langle displaystyle sum_1^r alpha_i v_i, displaystyle sum_1^r beta_j v_j right rangle = langle alpha, beta rangle, tag 6$
which proves that $T:V to V$ is orthogonal.
In the event that $r < n$, we extend $T$ from $V$ to all of $Bbb R^n$ as follows: let $W subset Bbb R^n$ be the orthogonal compliment of $V$, $W = V^bot$; then we have
$Bbb R^n = V oplus W; tag 7$
we observe that any $x in Bbb R^n$ may be represented uniquely as
$x = v + w, ; v in V, w in W; tag8$
existence of such a decomposition follows directly from (7), i.e. from the fact that $W = V^bot$; to see uniqueness, suppose
$x = v_1 + w_1 = v_2 + w_2, ; v_1, v_2 in V, w_1, w_2 in W; tag 9$
then
$v_1 - v_2 = w_2 - w_1, tag10$
which forces
$v_1 - v_2 = 0 = w_2 - w_1, tag11$
since
$v_1 - v_2 in V, ; w_1 - w_2 in W = V^bot, tag12$
and hence
$Vert v_1 - v_2 Vert^2 = langle v_1 - v_2, v_1 - v_2 rangle = langle v_1 - v_2, w_2 - w_1 rangle = 0, tag13$
implying $v_1 = v_2$, and thus $w_1 = w_2$ via (11); so the decomposition (8) is unique; this uniqueness means we may unabiguously define an extension $T_E$ of $T$ from $V$ to $V oplus W = Bbb R^n$ by
$T_E(v + w) = Tv + w, ; v in V, w in W; tag14$
that is,
$T_E = T oplus I: V oplus W to V oplus W; tag15$
we can formally show $T_E$ is orthogonal on all of $Bbb R^n$ if we observe that, for $v_1, v_2 in V$, $w_1, w_2 in W$,
$langle v_1 + w_1, v_2 + w_2 rangle$
$= langle v_1, v_2 rangle + langle v_1, w_2 rangle + langle w_1, v_2 rangle + langle w_1, w_2 rangle = langle v_1, v_2 rangle + langle w_1, w_2 rangle; tag16$
thus,
$langle T_E(v_1 + w_1), T_E(v_2 + w_2) rangle = langle (T oplus I)(v_1 + w_1), (T oplus I)(v_2 + w_2) rangle$
$= langle Tv_1 + w_1, Tv_2 + w_2 rangle = langle Tv_1, Tv_2 rangle + langle w_1, w_2 rangle$
$= langle v_1, v_2 rangle + langle w_1, w_2 rangle = langle v_1 + w_1, v_2 + w_2 rangle, tag17$
which shows that $T_E = T oplus I$ is orthogonal on $Bbb R^n$.
answered Jul 28 at 1:54
Robert Lewis
36.8k22155
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Let the subspace
$V = textspan( v_1, v_2, ldots, v_r ) = textspan( w_1, w_2, ldots, w_r ); tag 1$
we define $T$ on $V$ by setting
$Tv_k = w_k, ; 1 le k le r, tag 2$
and extending $T$ by linearity to all of $V$, viz:
$Tleft ( displaystyle sum_1^r alpha_i v_i right ) = displaystylesum_1^r alpha_i Tv_i = sum_1^r alpha_i w_i; tag 3$
that such an extension is possible follows from the linear independence of the $v_i$ and the $w_i$ as is well-known.
If
$alpha, beta in V tag 4$
we may write
$alpha = displaystyle sum_1^r alpha_i v_i, ; beta = sum_1^r beta_j v_j, tag 5$
whence
$left langle Talpha, Tbeta right rangle = left langle T left (displaystyle sum_1^r alpha_i v_i right), T left (displaystyle sum_1^r beta_j v_j right) right rangle = left langle displaystyle sum_1^r alpha_i Tv_i, displaystyle sum_1^r beta_j Tv_j right rangle = displaystyle sum_i, j = 1^r alpha_i beta_j langle Tv_i, Tv_j rangle = sum_i, j = 1^r alpha_i beta_j langle w_i, w_j rangle = sum_i, j = 1^r alpha_i beta_j langle v_i, v_j rangle$
$= left langle displaystyle sum_1^r alpha_i v_i, displaystyle sum_1^r beta_j v_j right rangle = langle alpha, beta rangle, tag 6$
which proves that $T:V to V$ is orthogonal.
In the event that $r < n$, we extend $T$ from $V$ to all of $Bbb R^n$ as follows: let $W subset Bbb R^n$ be the orthogonal compliment of $V$, $W = V^bot$; then we have
$Bbb R^n = V oplus W; tag 7$
we observe that any $x in Bbb R^n$ may be represented uniquely as
$x = v + w, ; v in V, w in W; tag8$
existence of such a decomposition follows directly from (7), i.e. from the fact that $W = V^bot$; to see uniqueness, suppose
$x = v_1 + w_1 = v_2 + w_2, ; v_1, v_2 in V, w_1, w_2 in W; tag 9$
then
$v_1 - v_2 = w_2 - w_1, tag10$
which forces
$v_1 - v_2 = 0 = w_2 - w_1, tag11$
since
$v_1 - v_2 in V, ; w_1 - w_2 in W = V^bot, tag12$
and hence
$Vert v_1 - v_2 Vert^2 = langle v_1 - v_2, v_1 - v_2 rangle = langle v_1 - v_2, w_2 - w_1 rangle = 0, tag13$
implying $v_1 = v_2$, and thus $w_1 = w_2$ via (11); so the decomposition (8) is unique; this uniqueness means we may unabiguously define an extension $T_E$ of $T$ from $V$ to $V oplus W = Bbb R^n$ by
$T_E(v + w) = Tv + w, ; v in V, w in W; tag14$
that is,
$T_E = T oplus I: V oplus W to V oplus W; tag15$
we can formally show $T_E$ is orthogonal on all of $Bbb R^n$ if we observe that, for $v_1, v_2 in V$, $w_1, w_2 in W$,
$langle v_1 + w_1, v_2 + w_2 rangle$
$= langle v_1, v_2 rangle + langle v_1, w_2 rangle + langle w_1, v_2 rangle + langle w_1, w_2 rangle = langle v_1, v_2 rangle + langle w_1, w_2 rangle; tag16$
thus,
$langle T_E(v_1 + w_1), T_E(v_2 + w_2) rangle = langle (T oplus I)(v_1 + w_1), (T oplus I)(v_2 + w_2) rangle$
$= langle Tv_1 + w_1, Tv_2 + w_2 rangle = langle Tv_1, Tv_2 rangle + langle w_1, w_2 rangle$
$= langle v_1, v_2 rangle + langle w_1, w_2 rangle = langle v_1 + w_1, v_2 + w_2 rangle, tag17$
which shows that $T_E = T oplus I$ is orthogonal on $Bbb R^n$.
answered Jul 28 at 1:54
Robert Lewis
36.8k22155
Let the subspace
$V = textspan( v_1, v_2, ldots, v_r ) = textspan( w_1, w_2, ldots, w_r ); tag 1$
we define $T$ on $V$ by setting
$Tv_k = w_k, ; 1 le k le r, tag 2$
and extending $T$ by linearity to all of $V$, viz:
$Tleft ( displaystyle sum_1^r alpha_i v_i right ) = displaystylesum_1^r alpha_i Tv_i = sum_1^r alpha_i w_i; tag 3$
that such an extension is possible follows from the linear independence of the $v_i$ and the $w_i$ as is well-known.
If
$alpha, beta in V tag 4$
we may write
$alpha = displaystyle sum_1^r alpha_i v_i, ; beta = sum_1^r beta_j v_j, tag 5$
whence
$left langle Talpha, Tbeta right rangle = left langle T left (displaystyle sum_1^r alpha_i v_i right), T left (displaystyle sum_1^r beta_j v_j right) right rangle = left langle displaystyle sum_1^r alpha_i Tv_i, displaystyle sum_1^r beta_j Tv_j right rangle = displaystyle sum_i, j = 1^r alpha_i beta_j langle Tv_i, Tv_j rangle = sum_i, j = 1^r alpha_i beta_j langle w_i, w_j rangle = sum_i, j = 1^r alpha_i beta_j langle v_i, v_j rangle$
$= left langle displaystyle sum_1^r alpha_i v_i, displaystyle sum_1^r beta_j v_j right rangle = langle alpha, beta rangle, tag 6$
which proves that $T:V to V$ is orthogonal.
In the event that $r < n$, we extend $T$ from $V$ to all of $Bbb R^n$ as follows: let $W subset Bbb R^n$ be the orthogonal compliment of $V$, $W = V^bot$; then we have
$Bbb R^n = V oplus W; tag 7$
we observe that any $x in Bbb R^n$ may be represented uniquely as
$x = v + w, ; v in V, w in W; tag8$
existence of such a decomposition follows directly from (7), i.e. from the fact that $W = V^bot$; to see uniqueness, suppose
$x = v_1 + w_1 = v_2 + w_2, ; v_1, v_2 in V, w_1, w_2 in W; tag 9$
then
$v_1 - v_2 = w_2 - w_1, tag10$
which forces
$v_1 - v_2 = 0 = w_2 - w_1, tag11$
since
$v_1 - v_2 in V, ; w_1 - w_2 in W = V^bot, tag12$
and hence
$Vert v_1 - v_2 Vert^2 = langle v_1 - v_2, v_1 - v_2 rangle = langle v_1 - v_2, w_2 - w_1 rangle = 0, tag13$
implying $v_1 = v_2$, and thus $w_1 = w_2$ via (11); so the decomposition (8) is unique; this uniqueness means we may unabiguously define an extension $T_E$ of $T$ from $V$ to $V oplus W = Bbb R^n$ by
$T_E(v + w) = Tv + w, ; v in V, w in W; tag14$
that is,
$T_E = T oplus I: V oplus W to V oplus W; tag15$
we can formally show $T_E$ is orthogonal on all of $Bbb R^n$ if we observe that, for $v_1, v_2 in V$, $w_1, w_2 in W$,
$langle v_1 + w_1, v_2 + w_2 rangle$
$= langle v_1, v_2 rangle + langle v_1, w_2 rangle + langle w_1, v_2 rangle + langle w_1, w_2 rangle = langle v_1, v_2 rangle + langle w_1, w_2 rangle; tag16$
thus,
$langle T_E(v_1 + w_1), T_E(v_2 + w_2) rangle = langle (T oplus I)(v_1 + w_1), (T oplus I)(v_2 + w_2) rangle$
$= langle Tv_1 + w_1, Tv_2 + w_2 rangle = langle Tv_1, Tv_2 rangle + langle w_1, w_2 rangle$
$= langle v_1, v_2 rangle + langle w_1, w_2 rangle = langle v_1 + w_1, v_2 + w_2 rangle, tag17$
which shows that $T_E = T oplus I$ is orthogonal on $Bbb R^n$.
answered Jul 28 at 1:54
Robert Lewis
36.8k22155
answered Jul 28 at 1:54
Robert Lewis
36.8k22155
answered Jul 28 at 1:54
Robert Lewis
36.8k22155
answered Jul 28 at 1:54
Robert Lewis
36.8k22155
36.8k22155
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