Two labelled sets of vectors in $mathbb{R^n}$ with same pairwise dots products
$begingroup$
Given ${x_i}_{1 leq i leq m}$ and ${y_i}_{1 leq i leq m}$ such that $x_i cdot x_j$ = $y_i cdot y_j, forall 1 leq i,j leq m$, what can I conclude about the two sets of vectors?
Clearly they need not be identical as $y_i = Ax_i$ for any orthogonal matrix $A$ is a valid solution. Intuitively it seems that all valid solutions may be of this form, but I am not certain that is the case, and would ideally like an algebraic proof if it is the case.
As a follow on question, I am also curious if the answer would change if instead we said $x_i cdot x_j$ = $y_i cdot y_j, forall 1 leq i neq j leq m$, so that we no longer had (explicitly anyway) that $lVert x_i rVert$ = $lVert y_i rVert$.
Any help would be much appreciated, thanks.
linear-algebra linear-transformations
asked Dec 6 '18 at 17:58
ludogludog
155
$endgroup$
|
show 1 more comment
$begingroup$
Given ${x_i}_{1 leq i leq m}$ and ${y_i}_{1 leq i leq m}$ such that $x_i cdot x_j$ = $y_i cdot y_j, forall 1 leq i,j leq m$, what can I conclude about the two sets of vectors?
Clearly they need not be identical as $y_i = Ax_i$ for any orthogonal matrix $A$ is a valid solution. Intuitively it seems that all valid solutions may be of this form, but I am not certain that is the case, and would ideally like an algebraic proof if it is the case.
As a follow on question, I am also curious if the answer would change if instead we said $x_i cdot x_j$ = $y_i cdot y_j, forall 1 leq i neq j leq m$, so that we no longer had (explicitly anyway) that $lVert x_i rVert$ = $lVert y_i rVert$.
Any help would be much appreciated, thanks.
linear-algebra linear-transformations
asked Dec 6 '18 at 17:58
ludogludog
155
$endgroup$
$begingroup$
Can you try by induction?
$endgroup$
– Federico
Dec 6 '18 at 18:03
$begingroup$
Another hint. Let's say that $c_1x_1+dots+c_mx_m=0$. Can you prove that $c_1y_1+dots+c_my_m=0$? Try taking the squared norm of both vectors
$endgroup$
– Federico
Dec 6 '18 at 18:05
$begingroup$
This should tell you that there exists a linear transformation such that $A x_i=y_i$. Then what can you say about $A$? Does it necessarily have to be orthogonal?
$endgroup$
– Federico
Dec 6 '18 at 18:06
$begingroup$
Ok so $c_1x_1 + dots + c_mx_m = 0 Rightarrow lVert c_1x_1 + dots + c_mx_m rVert^2 = 0 = sum_{1 leq i,j, leq m} c_ic_jx_i cdot x_j = sum_{1 leq i,j, leq m} c_ic_jy_i cdot y_j = lVert c_1y_1 + dots + c_my_m rVert^2 Rightarrow c_1y_1 + dots + c_my_m = 0$, but why does this tell me the transformation is linear?
$endgroup$
– ludog
Dec 6 '18 at 20:20
$begingroup$
The inverse statement is clear, but I can't see why $A(c_1x_1 + dots c_kx_k) = c_1A(x_1) + dots + c_kA(x_k)$
$endgroup$
– ludog
Dec 6 '18 at 20:51
|
show 1 more comment
$begingroup$
Given ${x_i}_{1 leq i leq m}$ and ${y_i}_{1 leq i leq m}$ such that $x_i cdot x_j$ = $y_i cdot y_j, forall 1 leq i,j leq m$, what can I conclude about the two sets of vectors?
Clearly they need not be identical as $y_i = Ax_i$ for any orthogonal matrix $A$ is a valid solution. Intuitively it seems that all valid solutions may be of this form, but I am not certain that is the case, and would ideally like an algebraic proof if it is the case.
As a follow on question, I am also curious if the answer would change if instead we said $x_i cdot x_j$ = $y_i cdot y_j, forall 1 leq i neq j leq m$, so that we no longer had (explicitly anyway) that $lVert x_i rVert$ = $lVert y_i rVert$.
Any help would be much appreciated, thanks.
linear-algebra linear-transformations
asked Dec 6 '18 at 17:58
ludogludog
155
$endgroup$
Given ${x_i}_{1 leq i leq m}$ and ${y_i}_{1 leq i leq m}$ such that $x_i cdot x_j$ = $y_i cdot y_j, forall 1 leq i,j leq m$, what can I conclude about the two sets of vectors?
Clearly they need not be identical as $y_i = Ax_i$ for any orthogonal matrix $A$ is a valid solution. Intuitively it seems that all valid solutions may be of this form, but I am not certain that is the case, and would ideally like an algebraic proof if it is the case.
As a follow on question, I am also curious if the answer would change if instead we said $x_i cdot x_j$ = $y_i cdot y_j, forall 1 leq i neq j leq m$, so that we no longer had (explicitly anyway) that $lVert x_i rVert$ = $lVert y_i rVert$.
Any help would be much appreciated, thanks.
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Dec 6 '18 at 17:58
ludogludog
155
asked Dec 6 '18 at 17:58
ludogludog
155
asked Dec 6 '18 at 17:58
ludogludog
155
asked Dec 6 '18 at 17:58
ludogludog
155
asked Dec 6 '18 at 17:58
ludogludog
155
155
$begingroup$
Can you try by induction?
$endgroup$
– Federico
Dec 6 '18 at 18:03
$begingroup$
Another hint. Let's say that $c_1x_1+dots+c_mx_m=0$. Can you prove that $c_1y_1+dots+c_my_m=0$? Try taking the squared norm of both vectors
$endgroup$
– Federico
Dec 6 '18 at 18:05
$begingroup$
This should tell you that there exists a linear transformation such that $A x_i=y_i$. Then what can you say about $A$? Does it necessarily have to be orthogonal?
$endgroup$
– Federico
Dec 6 '18 at 18:06
$begingroup$
Ok so $c_1x_1 + dots + c_mx_m = 0 Rightarrow lVert c_1x_1 + dots + c_mx_m rVert^2 = 0 = sum_{1 leq i,j, leq m} c_ic_jx_i cdot x_j = sum_{1 leq i,j, leq m} c_ic_jy_i cdot y_j = lVert c_1y_1 + dots + c_my_m rVert^2 Rightarrow c_1y_1 + dots + c_my_m = 0$, but why does this tell me the transformation is linear?
$endgroup$
– ludog
Dec 6 '18 at 20:20
$begingroup$
The inverse statement is clear, but I can't see why $A(c_1x_1 + dots c_kx_k) = c_1A(x_1) + dots + c_kA(x_k)$
$endgroup$
– ludog
Dec 6 '18 at 20:51
|
show 1 more comment
$begingroup$
Can you try by induction?
$endgroup$
– Federico
Dec 6 '18 at 18:03
$begingroup$
Another hint. Let's say that $c_1x_1+dots+c_mx_m=0$. Can you prove that $c_1y_1+dots+c_my_m=0$? Try taking the squared norm of both vectors
$endgroup$
– Federico
Dec 6 '18 at 18:05
$begingroup$
This should tell you that there exists a linear transformation such that $A x_i=y_i$. Then what can you say about $A$? Does it necessarily have to be orthogonal?
$endgroup$
– Federico
Dec 6 '18 at 18:06
$begingroup$
Ok so $c_1x_1 + dots + c_mx_m = 0 Rightarrow lVert c_1x_1 + dots + c_mx_m rVert^2 = 0 = sum_{1 leq i,j, leq m} c_ic_jx_i cdot x_j = sum_{1 leq i,j, leq m} c_ic_jy_i cdot y_j = lVert c_1y_1 + dots + c_my_m rVert^2 Rightarrow c_1y_1 + dots + c_my_m = 0$, but why does this tell me the transformation is linear?
$endgroup$
– ludog
Dec 6 '18 at 20:20
$begingroup$
The inverse statement is clear, but I can't see why $A(c_1x_1 + dots c_kx_k) = c_1A(x_1) + dots + c_kA(x_k)$
$endgroup$
– ludog
Dec 6 '18 at 20:51
$begingroup$
Can you try by induction?
$endgroup$
– Federico
Dec 6 '18 at 18:03
$begingroup$
Can you try by induction?
$endgroup$
– Federico
Dec 6 '18 at 18:03
$begingroup$
Another hint. Let's say that $c_1x_1+dots+c_mx_m=0$. Can you prove that $c_1y_1+dots+c_my_m=0$? Try taking the squared norm of both vectors
$endgroup$
– Federico
Dec 6 '18 at 18:05
$begingroup$
Another hint. Let's say that $c_1x_1+dots+c_mx_m=0$. Can you prove that $c_1y_1+dots+c_my_m=0$? Try taking the squared norm of both vectors
$endgroup$
– Federico
Dec 6 '18 at 18:05
$begingroup$
This should tell you that there exists a linear transformation such that $A x_i=y_i$. Then what can you say about $A$? Does it necessarily have to be orthogonal?
$endgroup$
– Federico
Dec 6 '18 at 18:06
$begingroup$
This should tell you that there exists a linear transformation such that $A x_i=y_i$. Then what can you say about $A$? Does it necessarily have to be orthogonal?
$endgroup$
– Federico
Dec 6 '18 at 18:06
$begingroup$
Ok so $c_1x_1 + dots + c_mx_m = 0 Rightarrow lVert c_1x_1 + dots + c_mx_m rVert^2 = 0 = sum_{1 leq i,j, leq m} c_ic_jx_i cdot x_j = sum_{1 leq i,j, leq m} c_ic_jy_i cdot y_j = lVert c_1y_1 + dots + c_my_m rVert^2 Rightarrow c_1y_1 + dots + c_my_m = 0$, but why does this tell me the transformation is linear?
$endgroup$
– ludog
Dec 6 '18 at 20:20
$begingroup$
Ok so $c_1x_1 + dots + c_mx_m = 0 Rightarrow lVert c_1x_1 + dots + c_mx_m rVert^2 = 0 = sum_{1 leq i,j, leq m} c_ic_jx_i cdot x_j = sum_{1 leq i,j, leq m} c_ic_jy_i cdot y_j = lVert c_1y_1 + dots + c_my_m rVert^2 Rightarrow c_1y_1 + dots + c_my_m = 0$, but why does this tell me the transformation is linear?
$endgroup$
– ludog
Dec 6 '18 at 20:20
$begingroup$
The inverse statement is clear, but I can't see why $A(c_1x_1 + dots c_kx_k) = c_1A(x_1) + dots + c_kA(x_k)$
$endgroup$
– ludog
Dec 6 '18 at 20:51
$begingroup$
The inverse statement is clear, but I can't see why $A(c_1x_1 + dots c_kx_k) = c_1A(x_1) + dots + c_kA(x_k)$
$endgroup$
– ludog
Dec 6 '18 at 20:51
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
The following holds under the assumption of the problem: $x_icdot x_j=y_icdot y_j$ for all $1leq i,jleq m$.
Lemma. If $c_1x_1+dots+c_mx_m=0$, then also $c_1y_1+dots+c_my_m=0$.
Proof.
$$
begin{split}
|c_1y_1+dots+c_my_m|^2
&= sum_{1leq i,jleq m} c_ic_j y_icdot y_j \
&= sum_{1leq i,jleq m} c_ic_j x_icdot x_j
= |c_1x_1+dots+c_mx_m|^2 = 0 .
end{split}
$$
□
Now I define a map $A:mathrm{span}{x_1,dots,x_m}tomathrm{span}{y_1,dots,y_m}$.
For $x=a_1x_1+dots+a_mx_minmathrm{span}{x_1,dots,x_m}$, define
$$
Ax = a_1y_1+dots+a_my_m.
$$
The map is well defined because if
$$
x=a_1x_1+dots+a_mx_m=b_1x_1+dots+b_mx_m,
$$
then letting $c_i=a_i-b_i$ we have $c_1x_1+dots+c_mx_m = 0$, which by the previous lemma implies $c_1y_1+dots+c_my_m$, which is equivalent to
$$
a_1y_1+dots+a_my_m = b_1y_1+dots+b_my_m.
$$
The map is clearly linear.
Moreover $A$ preserves the norm, because, as we have already computed, if $x=a_1x_1+dots+a_mx_m$, then
$$
begin{split}
|Ax| &= |a_1y_1+dots+a_my_m|^2
= sum_{1leq i,jleq m} a_ia_j y_icdot y_j \
&= sum_{1leq i,jleq m} a_ia_j x_icdot x_j
= |a_1x_1+dots+a_mx_m|^2 = |x|^2 .
end{split}
$$
This means that $A$ is orthogonal on its domain of definition, which is $mathrm{span}{x_1,dots,x_m}$, and can therefore be completed to an orthogonal linear transformation $A:mathbb R^ntomathbb R^n$.
Notice that $A$ is uniquely determined only on $mathrm{span}{x_1,dots,x_m}$: its extension to the complement can be arbitrary, but in particular can be orthogonal.
Edit: proof that preserving the norm and being orthogonal are equivalent: Theorem 2.1.
answered Dec 7 '18 at 14:23
FedericoFederico
5,144514
$endgroup$
$begingroup$
Excellent, thank you. I know that orthogonal matrices preserve norms, but wasn't sure that the converse is necessarily true. After a bit of digging I found a proof in Thm 2.1 here that norm-preserving linear transformations are always orthogonal math.lsa.umich.edu/~rauch/555/Conformal_Matrices.pdf, (Thm 2.1). (worth editing your answer to include link?) So does this mean there may exist multiple square matrices $A$ with $Ax_i = y_i$, but that at least one such matrix will be orthogonal?
$endgroup$
– ludog
Dec 7 '18 at 16:02
$begingroup$
Regarding orthogonality, I thought it was a pretty well known fact that the two notions are equivalent. I'll add the link nevertheless
$endgroup$
– Federico
Dec 7 '18 at 16:07
$begingroup$
"So does this mean there may exist multiple square matrices $A$ with $Ax_i=y_i$, but that at least one such matrix will be orthogonal?" Exactly. The matrix is uniquely determined on $mathrm{span}{x_i}$. On the complement you can do whatever you want. But you can extend it to an orthogonal matrix. Notice that also this extension is not unique. You can extend it to two different orthogonal matrices. Because you just take an arbitrary basis of $mathrm{span}{x_i}^perp$ and $mathrm{span}{y_i}^perp$.
$endgroup$
– Federico
Dec 7 '18 at 16:11
$begingroup$
Hmm ok interesting. Thanks very much for your help :)
$endgroup$
– ludog
Dec 7 '18 at 16:16
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The following holds under the assumption of the problem: $x_icdot x_j=y_icdot y_j$ for all $1leq i,jleq m$.
Lemma. If $c_1x_1+dots+c_mx_m=0$, then also $c_1y_1+dots+c_my_m=0$.
Proof.
$$
begin{split}
|c_1y_1+dots+c_my_m|^2
&= sum_{1leq i,jleq m} c_ic_j y_icdot y_j \
&= sum_{1leq i,jleq m} c_ic_j x_icdot x_j
= |c_1x_1+dots+c_mx_m|^2 = 0 .
end{split}
$$
□
Now I define a map $A:mathrm{span}{x_1,dots,x_m}tomathrm{span}{y_1,dots,y_m}$.
For $x=a_1x_1+dots+a_mx_minmathrm{span}{x_1,dots,x_m}$, define
$$
Ax = a_1y_1+dots+a_my_m.
$$
The map is well defined because if
$$
x=a_1x_1+dots+a_mx_m=b_1x_1+dots+b_mx_m,
$$
then letting $c_i=a_i-b_i$ we have $c_1x_1+dots+c_mx_m = 0$, which by the previous lemma implies $c_1y_1+dots+c_my_m$, which is equivalent to
$$
a_1y_1+dots+a_my_m = b_1y_1+dots+b_my_m.
$$
The map is clearly linear.
Moreover $A$ preserves the norm, because, as we have already computed, if $x=a_1x_1+dots+a_mx_m$, then
$$
begin{split}
|Ax| &= |a_1y_1+dots+a_my_m|^2
= sum_{1leq i,jleq m} a_ia_j y_icdot y_j \
&= sum_{1leq i,jleq m} a_ia_j x_icdot x_j
= |a_1x_1+dots+a_mx_m|^2 = |x|^2 .
end{split}
$$
This means that $A$ is orthogonal on its domain of definition, which is $mathrm{span}{x_1,dots,x_m}$, and can therefore be completed to an orthogonal linear transformation $A:mathbb R^ntomathbb R^n$.
Notice that $A$ is uniquely determined only on $mathrm{span}{x_1,dots,x_m}$: its extension to the complement can be arbitrary, but in particular can be orthogonal.
Edit: proof that preserving the norm and being orthogonal are equivalent: Theorem 2.1.
answered Dec 7 '18 at 14:23
FedericoFederico
5,144514
$endgroup$
$begingroup$
Excellent, thank you. I know that orthogonal matrices preserve norms, but wasn't sure that the converse is necessarily true. After a bit of digging I found a proof in Thm 2.1 here that norm-preserving linear transformations are always orthogonal math.lsa.umich.edu/~rauch/555/Conformal_Matrices.pdf, (Thm 2.1). (worth editing your answer to include link?) So does this mean there may exist multiple square matrices $A$ with $Ax_i = y_i$, but that at least one such matrix will be orthogonal?
$endgroup$
– ludog
Dec 7 '18 at 16:02
$begingroup$
Regarding orthogonality, I thought it was a pretty well known fact that the two notions are equivalent. I'll add the link nevertheless
$endgroup$
– Federico
Dec 7 '18 at 16:07
$begingroup$
"So does this mean there may exist multiple square matrices $A$ with $Ax_i=y_i$, but that at least one such matrix will be orthogonal?" Exactly. The matrix is uniquely determined on $mathrm{span}{x_i}$. On the complement you can do whatever you want. But you can extend it to an orthogonal matrix. Notice that also this extension is not unique. You can extend it to two different orthogonal matrices. Because you just take an arbitrary basis of $mathrm{span}{x_i}^perp$ and $mathrm{span}{y_i}^perp$.
$endgroup$
– Federico
Dec 7 '18 at 16:11
$begingroup$
Hmm ok interesting. Thanks very much for your help :)
$endgroup$
– ludog
Dec 7 '18 at 16:16
add a comment |
$begingroup$
The following holds under the assumption of the problem: $x_icdot x_j=y_icdot y_j$ for all $1leq i,jleq m$.
Lemma. If $c_1x_1+dots+c_mx_m=0$, then also $c_1y_1+dots+c_my_m=0$.
Proof.
$$
begin{split}
|c_1y_1+dots+c_my_m|^2
&= sum_{1leq i,jleq m} c_ic_j y_icdot y_j \
&= sum_{1leq i,jleq m} c_ic_j x_icdot x_j
= |c_1x_1+dots+c_mx_m|^2 = 0 .
end{split}
$$
□
Now I define a map $A:mathrm{span}{x_1,dots,x_m}tomathrm{span}{y_1,dots,y_m}$.
For $x=a_1x_1+dots+a_mx_minmathrm{span}{x_1,dots,x_m}$, define
$$
Ax = a_1y_1+dots+a_my_m.
$$
The map is well defined because if
$$
x=a_1x_1+dots+a_mx_m=b_1x_1+dots+b_mx_m,
$$
then letting $c_i=a_i-b_i$ we have $c_1x_1+dots+c_mx_m = 0$, which by the previous lemma implies $c_1y_1+dots+c_my_m$, which is equivalent to
$$
a_1y_1+dots+a_my_m = b_1y_1+dots+b_my_m.
$$
The map is clearly linear.
Moreover $A$ preserves the norm, because, as we have already computed, if $x=a_1x_1+dots+a_mx_m$, then
$$
begin{split}
|Ax| &= |a_1y_1+dots+a_my_m|^2
= sum_{1leq i,jleq m} a_ia_j y_icdot y_j \
&= sum_{1leq i,jleq m} a_ia_j x_icdot x_j
= |a_1x_1+dots+a_mx_m|^2 = |x|^2 .
end{split}
$$
This means that $A$ is orthogonal on its domain of definition, which is $mathrm{span}{x_1,dots,x_m}$, and can therefore be completed to an orthogonal linear transformation $A:mathbb R^ntomathbb R^n$.
Notice that $A$ is uniquely determined only on $mathrm{span}{x_1,dots,x_m}$: its extension to the complement can be arbitrary, but in particular can be orthogonal.
Edit: proof that preserving the norm and being orthogonal are equivalent: Theorem 2.1.
answered Dec 7 '18 at 14:23
FedericoFederico
5,144514
$endgroup$
$begingroup$
Excellent, thank you. I know that orthogonal matrices preserve norms, but wasn't sure that the converse is necessarily true. After a bit of digging I found a proof in Thm 2.1 here that norm-preserving linear transformations are always orthogonal math.lsa.umich.edu/~rauch/555/Conformal_Matrices.pdf, (Thm 2.1). (worth editing your answer to include link?) So does this mean there may exist multiple square matrices $A$ with $Ax_i = y_i$, but that at least one such matrix will be orthogonal?
$endgroup$
– ludog
Dec 7 '18 at 16:02
$begingroup$
Regarding orthogonality, I thought it was a pretty well known fact that the two notions are equivalent. I'll add the link nevertheless
$endgroup$
– Federico
Dec 7 '18 at 16:07
$begingroup$
"So does this mean there may exist multiple square matrices $A$ with $Ax_i=y_i$, but that at least one such matrix will be orthogonal?" Exactly. The matrix is uniquely determined on $mathrm{span}{x_i}$. On the complement you can do whatever you want. But you can extend it to an orthogonal matrix. Notice that also this extension is not unique. You can extend it to two different orthogonal matrices. Because you just take an arbitrary basis of $mathrm{span}{x_i}^perp$ and $mathrm{span}{y_i}^perp$.
$endgroup$
– Federico
Dec 7 '18 at 16:11
$begingroup$
Hmm ok interesting. Thanks very much for your help :)
$endgroup$
– ludog
Dec 7 '18 at 16:16
add a comment |
$begingroup$
The following holds under the assumption of the problem: $x_icdot x_j=y_icdot y_j$ for all $1leq i,jleq m$.
Lemma. If $c_1x_1+dots+c_mx_m=0$, then also $c_1y_1+dots+c_my_m=0$.
Proof.
$$
begin{split}
|c_1y_1+dots+c_my_m|^2
&= sum_{1leq i,jleq m} c_ic_j y_icdot y_j \
&= sum_{1leq i,jleq m} c_ic_j x_icdot x_j
= |c_1x_1+dots+c_mx_m|^2 = 0 .
end{split}
$$
□
Now I define a map $A:mathrm{span}{x_1,dots,x_m}tomathrm{span}{y_1,dots,y_m}$.
For $x=a_1x_1+dots+a_mx_minmathrm{span}{x_1,dots,x_m}$, define
$$
Ax = a_1y_1+dots+a_my_m.
$$
The map is well defined because if
$$
x=a_1x_1+dots+a_mx_m=b_1x_1+dots+b_mx_m,
$$
then letting $c_i=a_i-b_i$ we have $c_1x_1+dots+c_mx_m = 0$, which by the previous lemma implies $c_1y_1+dots+c_my_m$, which is equivalent to
$$
a_1y_1+dots+a_my_m = b_1y_1+dots+b_my_m.
$$
The map is clearly linear.
Moreover $A$ preserves the norm, because, as we have already computed, if $x=a_1x_1+dots+a_mx_m$, then
$$
begin{split}
|Ax| &= |a_1y_1+dots+a_my_m|^2
= sum_{1leq i,jleq m} a_ia_j y_icdot y_j \
&= sum_{1leq i,jleq m} a_ia_j x_icdot x_j
= |a_1x_1+dots+a_mx_m|^2 = |x|^2 .
end{split}
$$
This means that $A$ is orthogonal on its domain of definition, which is $mathrm{span}{x_1,dots,x_m}$, and can therefore be completed to an orthogonal linear transformation $A:mathbb R^ntomathbb R^n$.
Notice that $A$ is uniquely determined only on $mathrm{span}{x_1,dots,x_m}$: its extension to the complement can be arbitrary, but in particular can be orthogonal.
Edit: proof that preserving the norm and being orthogonal are equivalent: Theorem 2.1.
answered Dec 7 '18 at 14:23
FedericoFederico
5,144514
$endgroup$
The following holds under the assumption of the problem: $x_icdot x_j=y_icdot y_j$ for all $1leq i,jleq m$.
Lemma. If $c_1x_1+dots+c_mx_m=0$, then also $c_1y_1+dots+c_my_m=0$.
Proof.
$$
begin{split}
|c_1y_1+dots+c_my_m|^2
&= sum_{1leq i,jleq m} c_ic_j y_icdot y_j \
&= sum_{1leq i,jleq m} c_ic_j x_icdot x_j
= |c_1x_1+dots+c_mx_m|^2 = 0 .
end{split}
$$
□
Now I define a map $A:mathrm{span}{x_1,dots,x_m}tomathrm{span}{y_1,dots,y_m}$.
For $x=a_1x_1+dots+a_mx_minmathrm{span}{x_1,dots,x_m}$, define
$$
Ax = a_1y_1+dots+a_my_m.
$$
The map is well defined because if
$$
x=a_1x_1+dots+a_mx_m=b_1x_1+dots+b_mx_m,
$$
then letting $c_i=a_i-b_i$ we have $c_1x_1+dots+c_mx_m = 0$, which by the previous lemma implies $c_1y_1+dots+c_my_m$, which is equivalent to
$$
a_1y_1+dots+a_my_m = b_1y_1+dots+b_my_m.
$$
The map is clearly linear.
Moreover $A$ preserves the norm, because, as we have already computed, if $x=a_1x_1+dots+a_mx_m$, then
$$
begin{split}
|Ax| &= |a_1y_1+dots+a_my_m|^2
= sum_{1leq i,jleq m} a_ia_j y_icdot y_j \
&= sum_{1leq i,jleq m} a_ia_j x_icdot x_j
= |a_1x_1+dots+a_mx_m|^2 = |x|^2 .
end{split}
$$
This means that $A$ is orthogonal on its domain of definition, which is $mathrm{span}{x_1,dots,x_m}$, and can therefore be completed to an orthogonal linear transformation $A:mathbb R^ntomathbb R^n$.
Notice that $A$ is uniquely determined only on $mathrm{span}{x_1,dots,x_m}$: its extension to the complement can be arbitrary, but in particular can be orthogonal.
Edit: proof that preserving the norm and being orthogonal are equivalent: Theorem 2.1.
answered Dec 7 '18 at 14:23
FedericoFederico
5,144514
edited Dec 7 '18 at 16:08
answered Dec 7 '18 at 14:23
FedericoFederico
5,144514
answered Dec 7 '18 at 14:23
FedericoFederico
5,144514
answered Dec 7 '18 at 14:23
FedericoFederico
5,144514
5,144514
$begingroup$
Excellent, thank you. I know that orthogonal matrices preserve norms, but wasn't sure that the converse is necessarily true. After a bit of digging I found a proof in Thm 2.1 here that norm-preserving linear transformations are always orthogonal math.lsa.umich.edu/~rauch/555/Conformal_Matrices.pdf, (Thm 2.1). (worth editing your answer to include link?) So does this mean there may exist multiple square matrices $A$ with $Ax_i = y_i$, but that at least one such matrix will be orthogonal?
$endgroup$
– ludog
Dec 7 '18 at 16:02
$begingroup$
Regarding orthogonality, I thought it was a pretty well known fact that the two notions are equivalent. I'll add the link nevertheless
$endgroup$
– Federico
Dec 7 '18 at 16:07
$begingroup$
"So does this mean there may exist multiple square matrices $A$ with $Ax_i=y_i$, but that at least one such matrix will be orthogonal?" Exactly. The matrix is uniquely determined on $mathrm{span}{x_i}$. On the complement you can do whatever you want. But you can extend it to an orthogonal matrix. Notice that also this extension is not unique. You can extend it to two different orthogonal matrices. Because you just take an arbitrary basis of $mathrm{span}{x_i}^perp$ and $mathrm{span}{y_i}^perp$.
$endgroup$
– Federico
Dec 7 '18 at 16:11
$begingroup$
Hmm ok interesting. Thanks very much for your help :)
$endgroup$
– ludog
Dec 7 '18 at 16:16
add a comment |
$begingroup$
Excellent, thank you. I know that orthogonal matrices preserve norms, but wasn't sure that the converse is necessarily true. After a bit of digging I found a proof in Thm 2.1 here that norm-preserving linear transformations are always orthogonal math.lsa.umich.edu/~rauch/555/Conformal_Matrices.pdf, (Thm 2.1). (worth editing your answer to include link?) So does this mean there may exist multiple square matrices $A$ with $Ax_i = y_i$, but that at least one such matrix will be orthogonal?
$endgroup$
– ludog
Dec 7 '18 at 16:02
$begingroup$
Regarding orthogonality, I thought it was a pretty well known fact that the two notions are equivalent. I'll add the link nevertheless
$endgroup$
– Federico
Dec 7 '18 at 16:07
$begingroup$
"So does this mean there may exist multiple square matrices $A$ with $Ax_i=y_i$, but that at least one such matrix will be orthogonal?" Exactly. The matrix is uniquely determined on $mathrm{span}{x_i}$. On the complement you can do whatever you want. But you can extend it to an orthogonal matrix. Notice that also this extension is not unique. You can extend it to two different orthogonal matrices. Because you just take an arbitrary basis of $mathrm{span}{x_i}^perp$ and $mathrm{span}{y_i}^perp$.
$endgroup$
– Federico
Dec 7 '18 at 16:11
$begingroup$
Hmm ok interesting. Thanks very much for your help :)
$endgroup$
– ludog
Dec 7 '18 at 16:16
$begingroup$
Excellent, thank you. I know that orthogonal matrices preserve norms, but wasn't sure that the converse is necessarily true. After a bit of digging I found a proof in Thm 2.1 here that norm-preserving linear transformations are always orthogonal math.lsa.umich.edu/~rauch/555/Conformal_Matrices.pdf, (Thm 2.1). (worth editing your answer to include link?) So does this mean there may exist multiple square matrices $A$ with $Ax_i = y_i$, but that at least one such matrix will be orthogonal?
$endgroup$
– ludog
Dec 7 '18 at 16:02
$begingroup$
Excellent, thank you. I know that orthogonal matrices preserve norms, but wasn't sure that the converse is necessarily true. After a bit of digging I found a proof in Thm 2.1 here that norm-preserving linear transformations are always orthogonal math.lsa.umich.edu/~rauch/555/Conformal_Matrices.pdf, (Thm 2.1). (worth editing your answer to include link?) So does this mean there may exist multiple square matrices $A$ with $Ax_i = y_i$, but that at least one such matrix will be orthogonal?
$endgroup$
– ludog
Dec 7 '18 at 16:02
$begingroup$
Regarding orthogonality, I thought it was a pretty well known fact that the two notions are equivalent. I'll add the link nevertheless
$endgroup$
– Federico
Dec 7 '18 at 16:07
$begingroup$
Regarding orthogonality, I thought it was a pretty well known fact that the two notions are equivalent. I'll add the link nevertheless
$endgroup$
– Federico
Dec 7 '18 at 16:07
$begingroup$
"So does this mean there may exist multiple square matrices $A$ with $Ax_i=y_i$, but that at least one such matrix will be orthogonal?" Exactly. The matrix is uniquely determined on $mathrm{span}{x_i}$. On the complement you can do whatever you want. But you can extend it to an orthogonal matrix. Notice that also this extension is not unique. You can extend it to two different orthogonal matrices. Because you just take an arbitrary basis of $mathrm{span}{x_i}^perp$ and $mathrm{span}{y_i}^perp$.
$endgroup$
– Federico
Dec 7 '18 at 16:11
$begingroup$
"So does this mean there may exist multiple square matrices $A$ with $Ax_i=y_i$, but that at least one such matrix will be orthogonal?" Exactly. The matrix is uniquely determined on $mathrm{span}{x_i}$. On the complement you can do whatever you want. But you can extend it to an orthogonal matrix. Notice that also this extension is not unique. You can extend it to two different orthogonal matrices. Because you just take an arbitrary basis of $mathrm{span}{x_i}^perp$ and $mathrm{span}{y_i}^perp$.
$endgroup$
– Federico
Dec 7 '18 at 16:11
$begingroup$
Hmm ok interesting. Thanks very much for your help :)
$endgroup$
– ludog
Dec 7 '18 at 16:16
$begingroup$
Hmm ok interesting. Thanks very much for your help :)
$endgroup$
– ludog
Dec 7 '18 at 16:16
add a comment |
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$begingroup$
Can you try by induction?
$endgroup$
– Federico
Dec 6 '18 at 18:03
$begingroup$
Another hint. Let's say that $c_1x_1+dots+c_mx_m=0$. Can you prove that $c_1y_1+dots+c_my_m=0$? Try taking the squared norm of both vectors
$endgroup$
– Federico
Dec 6 '18 at 18:05
$begingroup$
This should tell you that there exists a linear transformation such that $A x_i=y_i$. Then what can you say about $A$? Does it necessarily have to be orthogonal?
$endgroup$
– Federico
Dec 6 '18 at 18:06
$begingroup$
Ok so $c_1x_1 + dots + c_mx_m = 0 Rightarrow lVert c_1x_1 + dots + c_mx_m rVert^2 = 0 = sum_{1 leq i,j, leq m} c_ic_jx_i cdot x_j = sum_{1 leq i,j, leq m} c_ic_jy_i cdot y_j = lVert c_1y_1 + dots + c_my_m rVert^2 Rightarrow c_1y_1 + dots + c_my_m = 0$, but why does this tell me the transformation is linear?
$endgroup$
– ludog
Dec 6 '18 at 20:20
$begingroup$
The inverse statement is clear, but I can't see why $A(c_1x_1 + dots c_kx_k) = c_1A(x_1) + dots + c_kA(x_k)$
$endgroup$
– ludog
Dec 6 '18 at 20:51