Mixing
Orthogonal matrix with single $0$ entry
$begingroup$
To my surprise there is an orthogonal matrix dimension $3 times 3$ with a single $0$ entry as it was shown in this answer.
Moreover it was possible to identify the pattern for matrix entries to satisfy this condition i.e.
$$Q=begin{bmatrix} 0 & -a & -b \ a & -b^2 & ab \ b & ab & -a^2end{bmatrix}$$
when $a,b$ are constrained by $a^2+b^2=1$.
For orthogonal matrix dimension $2 times 2$ single $0$ entry is not possible.
Here my new questions:
Is it possible a single $0$ entry for orthogonal matrix dimension $4
times 4$ ?Can we also provide for this case some more general formula? (to be more specific in a similar form as it was shown above for 3d)
Is the method listed above the only one for orthogonal matrices dimension $3
times 3$ or can we generate such matrix also with some other substantially different algorithm?
linear-algebra matrices orthogonal-matrices
asked Jan 30 at 16:07
WidawensenWidawensen
4,76331447
$endgroup$
add a comment |
$begingroup$
To my surprise there is an orthogonal matrix dimension $3 times 3$ with a single $0$ entry as it was shown in this answer.
Moreover it was possible to identify the pattern for matrix entries to satisfy this condition i.e.
$$Q=begin{bmatrix} 0 & -a & -b \ a & -b^2 & ab \ b & ab & -a^2end{bmatrix}$$
when $a,b$ are constrained by $a^2+b^2=1$.
For orthogonal matrix dimension $2 times 2$ single $0$ entry is not possible.
Here my new questions:
Is it possible a single $0$ entry for orthogonal matrix dimension $4
times 4$ ?Can we also provide for this case some more general formula? (to be more specific in a similar form as it was shown above for 3d)
Is the method listed above the only one for orthogonal matrices dimension $3
times 3$ or can we generate such matrix also with some other substantially different algorithm?
linear-algebra matrices orthogonal-matrices
asked Jan 30 at 16:07
WidawensenWidawensen
4,76331447
$endgroup$
add a comment |
$begingroup$
To my surprise there is an orthogonal matrix dimension $3 times 3$ with a single $0$ entry as it was shown in this answer.
Moreover it was possible to identify the pattern for matrix entries to satisfy this condition i.e.
$$Q=begin{bmatrix} 0 & -a & -b \ a & -b^2 & ab \ b & ab & -a^2end{bmatrix}$$
when $a,b$ are constrained by $a^2+b^2=1$.
For orthogonal matrix dimension $2 times 2$ single $0$ entry is not possible.
Here my new questions:
Is it possible a single $0$ entry for orthogonal matrix dimension $4
times 4$ ?Can we also provide for this case some more general formula? (to be more specific in a similar form as it was shown above for 3d)
Is the method listed above the only one for orthogonal matrices dimension $3
times 3$ or can we generate such matrix also with some other substantially different algorithm?
linear-algebra matrices orthogonal-matrices
asked Jan 30 at 16:07
WidawensenWidawensen
4,76331447
$endgroup$
To my surprise there is an orthogonal matrix dimension $3 times 3$ with a single $0$ entry as it was shown in this answer.
Moreover it was possible to identify the pattern for matrix entries to satisfy this condition i.e.
$$Q=begin{bmatrix} 0 & -a & -b \ a & -b^2 & ab \ b & ab & -a^2end{bmatrix}$$
when $a,b$ are constrained by $a^2+b^2=1$.
For orthogonal matrix dimension $2 times 2$ single $0$ entry is not possible.
Here my new questions:
Is it possible a single $0$ entry for orthogonal matrix dimension $4
times 4$ ?Can we also provide for this case some more general formula? (to be more specific in a similar form as it was shown above for 3d)
Is the method listed above the only one for orthogonal matrices dimension $3
times 3$ or can we generate such matrix also with some other substantially different algorithm?
linear-algebra matrices orthogonal-matrices
linear-algebra matrices orthogonal-matrices
asked Jan 30 at 16:07
WidawensenWidawensen
4,76331447
asked Jan 30 at 16:07
WidawensenWidawensen
4,76331447
edited Jan 31 at 9:46
Widawensen
asked Jan 30 at 16:07
WidawensenWidawensen
4,76331447
asked Jan 30 at 16:07
WidawensenWidawensen
4,76331447
asked Jan 30 at 16:07
WidawensenWidawensen
4,76331447
4,76331447
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I can answer your first question, but the other two seem less reasonable to answer concisely.
It's possible to find such a matrix for any $nge 3$, and there are tons of them. Let's just choose the first column to be $w'=(0,1,dots,1)^T$ and $w = w'/||w'||$. We are thus looking to complete $w$ to an orthonormal basis $w,v_1,dots,v_{n-1}$ with no coordinate of any $v_i$ equal to $0$. The space of vectors orthogonal to $w$ is the space
$$W = {(x_1,dots,x_n)^T : x_2+dots+x_n = 0},$$
and we want to find an orthonormal basis of this space with no zero coordinates. That is, we want to find an orthonormal basis of this space which avoids the $n$ hyperplanes
$$A_i = {(x_1,dots,x_n)^T : x_i = 0}.$$
The key use of having $nge 3$ is so that $W cap A_i$ is $(n-2)$-dimensional for each $i$. When $n=2$, for example, $W cap A_2$ is $1$-dimensional. Now, for each $i$, let $B_i = W cap A_i$, an $(n-2)$-dimensional subspace of the $(n-1)$-dimensional space $W$. For each $i$, let $C_i = B_i^perp cap W$, a $1$-dimensional subspace of $W$ (and $1le n-2$, remember). Now, since all the $B_i$ and $C_i$ have dimension smaller than that of $W$, their union cannot be $W$. Pick a unit vector $v_1$ in $W$ but not any $B_i$ or $C_i$ (you can pick it with norm $1$ because if $v$ is any vector in $W$ not in their union, then $lambda v$ is also not in their union for $lambda ne 0$: if $lambda v$ were in their union, then it would be in one of them in particular, and then so would $v=(1/lambda)lambda v$). Now let $W_1 = Wcap {v_1}^perp$, which is $(n-2)$-dimensional. And for each $i$, let $B_i^1 = B_i cap W_1$, which is $(n-3)$-dimensional, since we picked $v_1$ not in any $C_i$. Now we're looking for $n-2$ unit-norm vectors in $W_1$ not in any of the $B_i^1$, and we can iterate.
answered Jan 30 at 17:59
cspruncsprun
2,795211
$endgroup$
$begingroup$
Thank you for the answer csprun, it is very good (I have upvoted it) because it generalizes the method for any matrix dimension ( I would call your approach geometrical), however it would be useful for me also to know a matrix in algebraic form (second part of a question), similar to manner when the solution for 3d was presented in the question. It could be interesting to know such matrix in the form for example (here I have to divide the comment - see part two)
$endgroup$
– Widawensen
Jan 31 at 9:40
$begingroup$
$Q=begin{bmatrix} 0 & f_{12}(a,b,c) & f_{13}(a,b,c) & f_{14}(a,b,c) \ a & f_{22}(a,b,c) & f_{23}(a,b,c) & f_{24}(a,b,c) \ b & f_{32}(a,b,c) & f_{33}(a,b,c) & f_{34}(a,b,c) \ c & f_{42}(a,b,c) & f_{43}(a,b,c) & f_{44}(a,b,c) end{bmatrix}$ where $a,b,c$ are constrained by $a^2+b^2+c^2=1$. It would give me an algorithm for easy generating such matrices.
$endgroup$
– Widawensen
Jan 31 at 9:40
add a comment |
$begingroup$
As so far there was no answer which could provide an algebraic form of orthogonal matrix with single $0$ entry.
However looking a few days at the 3d form I was able to find the missing pattern at the end yesterday.
The pattern is amazingly simple.
Let $ v=begin{bmatrix} a_1 & a_2 & dots & a_n end{bmatrix}^T $ be n-dimensional unit vector ($v^Tv=1$).
Now we can construct a block matrix dimension ${(n+1) times (n+1)}$ based only on $v$ vector:
$$Q_{(n+1) times (n+1)} =begin{bmatrix} 0_{1 times 1} & -v^T_{1 times n} \ v_{n times 1} & vv^T-I_{n times n} end{bmatrix}$$
For this matrix:
$$Q^T =begin{bmatrix} 0 & v^T \ -v & vv^T-I end{bmatrix}$$
One can check that really $QQ^T=I$, hence if only unit vector $v$ has non-zero components then we have orthogonal matrix with single zero entry.
answered Feb 1 at 13:19
WidawensenWidawensen
4,76331447
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
I can answer your first question, but the other two seem less reasonable to answer concisely.
It's possible to find such a matrix for any $nge 3$, and there are tons of them. Let's just choose the first column to be $w'=(0,1,dots,1)^T$ and $w = w'/||w'||$. We are thus looking to complete $w$ to an orthonormal basis $w,v_1,dots,v_{n-1}$ with no coordinate of any $v_i$ equal to $0$. The space of vectors orthogonal to $w$ is the space
$$W = {(x_1,dots,x_n)^T : x_2+dots+x_n = 0},$$
and we want to find an orthonormal basis of this space with no zero coordinates. That is, we want to find an orthonormal basis of this space which avoids the $n$ hyperplanes
$$A_i = {(x_1,dots,x_n)^T : x_i = 0}.$$
The key use of having $nge 3$ is so that $W cap A_i$ is $(n-2)$-dimensional for each $i$. When $n=2$, for example, $W cap A_2$ is $1$-dimensional. Now, for each $i$, let $B_i = W cap A_i$, an $(n-2)$-dimensional subspace of the $(n-1)$-dimensional space $W$. For each $i$, let $C_i = B_i^perp cap W$, a $1$-dimensional subspace of $W$ (and $1le n-2$, remember). Now, since all the $B_i$ and $C_i$ have dimension smaller than that of $W$, their union cannot be $W$. Pick a unit vector $v_1$ in $W$ but not any $B_i$ or $C_i$ (you can pick it with norm $1$ because if $v$ is any vector in $W$ not in their union, then $lambda v$ is also not in their union for $lambda ne 0$: if $lambda v$ were in their union, then it would be in one of them in particular, and then so would $v=(1/lambda)lambda v$). Now let $W_1 = Wcap {v_1}^perp$, which is $(n-2)$-dimensional. And for each $i$, let $B_i^1 = B_i cap W_1$, which is $(n-3)$-dimensional, since we picked $v_1$ not in any $C_i$. Now we're looking for $n-2$ unit-norm vectors in $W_1$ not in any of the $B_i^1$, and we can iterate.
answered Jan 30 at 17:59
cspruncsprun
2,795211
$endgroup$
$begingroup$
Thank you for the answer csprun, it is very good (I have upvoted it) because it generalizes the method for any matrix dimension ( I would call your approach geometrical), however it would be useful for me also to know a matrix in algebraic form (second part of a question), similar to manner when the solution for 3d was presented in the question. It could be interesting to know such matrix in the form for example (here I have to divide the comment - see part two)
$endgroup$
– Widawensen
Jan 31 at 9:40
$begingroup$
$Q=begin{bmatrix} 0 & f_{12}(a,b,c) & f_{13}(a,b,c) & f_{14}(a,b,c) \ a & f_{22}(a,b,c) & f_{23}(a,b,c) & f_{24}(a,b,c) \ b & f_{32}(a,b,c) & f_{33}(a,b,c) & f_{34}(a,b,c) \ c & f_{42}(a,b,c) & f_{43}(a,b,c) & f_{44}(a,b,c) end{bmatrix}$ where $a,b,c$ are constrained by $a^2+b^2+c^2=1$. It would give me an algorithm for easy generating such matrices.
$endgroup$
– Widawensen
Jan 31 at 9:40
add a comment |
$begingroup$
I can answer your first question, but the other two seem less reasonable to answer concisely.
It's possible to find such a matrix for any $nge 3$, and there are tons of them. Let's just choose the first column to be $w'=(0,1,dots,1)^T$ and $w = w'/||w'||$. We are thus looking to complete $w$ to an orthonormal basis $w,v_1,dots,v_{n-1}$ with no coordinate of any $v_i$ equal to $0$. The space of vectors orthogonal to $w$ is the space
$$W = {(x_1,dots,x_n)^T : x_2+dots+x_n = 0},$$
and we want to find an orthonormal basis of this space with no zero coordinates. That is, we want to find an orthonormal basis of this space which avoids the $n$ hyperplanes
$$A_i = {(x_1,dots,x_n)^T : x_i = 0}.$$
The key use of having $nge 3$ is so that $W cap A_i$ is $(n-2)$-dimensional for each $i$. When $n=2$, for example, $W cap A_2$ is $1$-dimensional. Now, for each $i$, let $B_i = W cap A_i$, an $(n-2)$-dimensional subspace of the $(n-1)$-dimensional space $W$. For each $i$, let $C_i = B_i^perp cap W$, a $1$-dimensional subspace of $W$ (and $1le n-2$, remember). Now, since all the $B_i$ and $C_i$ have dimension smaller than that of $W$, their union cannot be $W$. Pick a unit vector $v_1$ in $W$ but not any $B_i$ or $C_i$ (you can pick it with norm $1$ because if $v$ is any vector in $W$ not in their union, then $lambda v$ is also not in their union for $lambda ne 0$: if $lambda v$ were in their union, then it would be in one of them in particular, and then so would $v=(1/lambda)lambda v$). Now let $W_1 = Wcap {v_1}^perp$, which is $(n-2)$-dimensional. And for each $i$, let $B_i^1 = B_i cap W_1$, which is $(n-3)$-dimensional, since we picked $v_1$ not in any $C_i$. Now we're looking for $n-2$ unit-norm vectors in $W_1$ not in any of the $B_i^1$, and we can iterate.
answered Jan 30 at 17:59
cspruncsprun
2,795211
$endgroup$
$begingroup$
Thank you for the answer csprun, it is very good (I have upvoted it) because it generalizes the method for any matrix dimension ( I would call your approach geometrical), however it would be useful for me also to know a matrix in algebraic form (second part of a question), similar to manner when the solution for 3d was presented in the question. It could be interesting to know such matrix in the form for example (here I have to divide the comment - see part two)
$endgroup$
– Widawensen
Jan 31 at 9:40
$begingroup$
$Q=begin{bmatrix} 0 & f_{12}(a,b,c) & f_{13}(a,b,c) & f_{14}(a,b,c) \ a & f_{22}(a,b,c) & f_{23}(a,b,c) & f_{24}(a,b,c) \ b & f_{32}(a,b,c) & f_{33}(a,b,c) & f_{34}(a,b,c) \ c & f_{42}(a,b,c) & f_{43}(a,b,c) & f_{44}(a,b,c) end{bmatrix}$ where $a,b,c$ are constrained by $a^2+b^2+c^2=1$. It would give me an algorithm for easy generating such matrices.
$endgroup$
– Widawensen
Jan 31 at 9:40
add a comment |
$begingroup$
I can answer your first question, but the other two seem less reasonable to answer concisely.
It's possible to find such a matrix for any $nge 3$, and there are tons of them. Let's just choose the first column to be $w'=(0,1,dots,1)^T$ and $w = w'/||w'||$. We are thus looking to complete $w$ to an orthonormal basis $w,v_1,dots,v_{n-1}$ with no coordinate of any $v_i$ equal to $0$. The space of vectors orthogonal to $w$ is the space
$$W = {(x_1,dots,x_n)^T : x_2+dots+x_n = 0},$$
and we want to find an orthonormal basis of this space with no zero coordinates. That is, we want to find an orthonormal basis of this space which avoids the $n$ hyperplanes
$$A_i = {(x_1,dots,x_n)^T : x_i = 0}.$$
The key use of having $nge 3$ is so that $W cap A_i$ is $(n-2)$-dimensional for each $i$. When $n=2$, for example, $W cap A_2$ is $1$-dimensional. Now, for each $i$, let $B_i = W cap A_i$, an $(n-2)$-dimensional subspace of the $(n-1)$-dimensional space $W$. For each $i$, let $C_i = B_i^perp cap W$, a $1$-dimensional subspace of $W$ (and $1le n-2$, remember). Now, since all the $B_i$ and $C_i$ have dimension smaller than that of $W$, their union cannot be $W$. Pick a unit vector $v_1$ in $W$ but not any $B_i$ or $C_i$ (you can pick it with norm $1$ because if $v$ is any vector in $W$ not in their union, then $lambda v$ is also not in their union for $lambda ne 0$: if $lambda v$ were in their union, then it would be in one of them in particular, and then so would $v=(1/lambda)lambda v$). Now let $W_1 = Wcap {v_1}^perp$, which is $(n-2)$-dimensional. And for each $i$, let $B_i^1 = B_i cap W_1$, which is $(n-3)$-dimensional, since we picked $v_1$ not in any $C_i$. Now we're looking for $n-2$ unit-norm vectors in $W_1$ not in any of the $B_i^1$, and we can iterate.
answered Jan 30 at 17:59
cspruncsprun
2,795211
$endgroup$
I can answer your first question, but the other two seem less reasonable to answer concisely.
It's possible to find such a matrix for any $nge 3$, and there are tons of them. Let's just choose the first column to be $w'=(0,1,dots,1)^T$ and $w = w'/||w'||$. We are thus looking to complete $w$ to an orthonormal basis $w,v_1,dots,v_{n-1}$ with no coordinate of any $v_i$ equal to $0$. The space of vectors orthogonal to $w$ is the space
$$W = {(x_1,dots,x_n)^T : x_2+dots+x_n = 0},$$
and we want to find an orthonormal basis of this space with no zero coordinates. That is, we want to find an orthonormal basis of this space which avoids the $n$ hyperplanes
$$A_i = {(x_1,dots,x_n)^T : x_i = 0}.$$
The key use of having $nge 3$ is so that $W cap A_i$ is $(n-2)$-dimensional for each $i$. When $n=2$, for example, $W cap A_2$ is $1$-dimensional. Now, for each $i$, let $B_i = W cap A_i$, an $(n-2)$-dimensional subspace of the $(n-1)$-dimensional space $W$. For each $i$, let $C_i = B_i^perp cap W$, a $1$-dimensional subspace of $W$ (and $1le n-2$, remember). Now, since all the $B_i$ and $C_i$ have dimension smaller than that of $W$, their union cannot be $W$. Pick a unit vector $v_1$ in $W$ but not any $B_i$ or $C_i$ (you can pick it with norm $1$ because if $v$ is any vector in $W$ not in their union, then $lambda v$ is also not in their union for $lambda ne 0$: if $lambda v$ were in their union, then it would be in one of them in particular, and then so would $v=(1/lambda)lambda v$). Now let $W_1 = Wcap {v_1}^perp$, which is $(n-2)$-dimensional. And for each $i$, let $B_i^1 = B_i cap W_1$, which is $(n-3)$-dimensional, since we picked $v_1$ not in any $C_i$. Now we're looking for $n-2$ unit-norm vectors in $W_1$ not in any of the $B_i^1$, and we can iterate.
answered Jan 30 at 17:59
cspruncsprun
2,795211
edited Jan 30 at 18:04
answered Jan 30 at 17:59
cspruncsprun
2,795211
answered Jan 30 at 17:59
cspruncsprun
2,795211
answered Jan 30 at 17:59
cspruncsprun
2,795211
2,795211
$begingroup$
Thank you for the answer csprun, it is very good (I have upvoted it) because it generalizes the method for any matrix dimension ( I would call your approach geometrical), however it would be useful for me also to know a matrix in algebraic form (second part of a question), similar to manner when the solution for 3d was presented in the question. It could be interesting to know such matrix in the form for example (here I have to divide the comment - see part two)
$endgroup$
– Widawensen
Jan 31 at 9:40
$begingroup$
$Q=begin{bmatrix} 0 & f_{12}(a,b,c) & f_{13}(a,b,c) & f_{14}(a,b,c) \ a & f_{22}(a,b,c) & f_{23}(a,b,c) & f_{24}(a,b,c) \ b & f_{32}(a,b,c) & f_{33}(a,b,c) & f_{34}(a,b,c) \ c & f_{42}(a,b,c) & f_{43}(a,b,c) & f_{44}(a,b,c) end{bmatrix}$ where $a,b,c$ are constrained by $a^2+b^2+c^2=1$. It would give me an algorithm for easy generating such matrices.
$endgroup$
– Widawensen
Jan 31 at 9:40
add a comment |
$begingroup$
Thank you for the answer csprun, it is very good (I have upvoted it) because it generalizes the method for any matrix dimension ( I would call your approach geometrical), however it would be useful for me also to know a matrix in algebraic form (second part of a question), similar to manner when the solution for 3d was presented in the question. It could be interesting to know such matrix in the form for example (here I have to divide the comment - see part two)
$endgroup$
– Widawensen
Jan 31 at 9:40
$begingroup$
$Q=begin{bmatrix} 0 & f_{12}(a,b,c) & f_{13}(a,b,c) & f_{14}(a,b,c) \ a & f_{22}(a,b,c) & f_{23}(a,b,c) & f_{24}(a,b,c) \ b & f_{32}(a,b,c) & f_{33}(a,b,c) & f_{34}(a,b,c) \ c & f_{42}(a,b,c) & f_{43}(a,b,c) & f_{44}(a,b,c) end{bmatrix}$ where $a,b,c$ are constrained by $a^2+b^2+c^2=1$. It would give me an algorithm for easy generating such matrices.
$endgroup$
– Widawensen
Jan 31 at 9:40
$begingroup$
Thank you for the answer csprun, it is very good (I have upvoted it) because it generalizes the method for any matrix dimension ( I would call your approach geometrical), however it would be useful for me also to know a matrix in algebraic form (second part of a question), similar to manner when the solution for 3d was presented in the question. It could be interesting to know such matrix in the form for example (here I have to divide the comment - see part two)
$endgroup$
– Widawensen
Jan 31 at 9:40
$begingroup$
Thank you for the answer csprun, it is very good (I have upvoted it) because it generalizes the method for any matrix dimension ( I would call your approach geometrical), however it would be useful for me also to know a matrix in algebraic form (second part of a question), similar to manner when the solution for 3d was presented in the question. It could be interesting to know such matrix in the form for example (here I have to divide the comment - see part two)
$endgroup$
– Widawensen
Jan 31 at 9:40
$begingroup$
$Q=begin{bmatrix} 0 & f_{12}(a,b,c) & f_{13}(a,b,c) & f_{14}(a,b,c) \ a & f_{22}(a,b,c) & f_{23}(a,b,c) & f_{24}(a,b,c) \ b & f_{32}(a,b,c) & f_{33}(a,b,c) & f_{34}(a,b,c) \ c & f_{42}(a,b,c) & f_{43}(a,b,c) & f_{44}(a,b,c) end{bmatrix}$ where $a,b,c$ are constrained by $a^2+b^2+c^2=1$. It would give me an algorithm for easy generating such matrices.
$endgroup$
– Widawensen
Jan 31 at 9:40
$begingroup$
$Q=begin{bmatrix} 0 & f_{12}(a,b,c) & f_{13}(a,b,c) & f_{14}(a,b,c) \ a & f_{22}(a,b,c) & f_{23}(a,b,c) & f_{24}(a,b,c) \ b & f_{32}(a,b,c) & f_{33}(a,b,c) & f_{34}(a,b,c) \ c & f_{42}(a,b,c) & f_{43}(a,b,c) & f_{44}(a,b,c) end{bmatrix}$ where $a,b,c$ are constrained by $a^2+b^2+c^2=1$. It would give me an algorithm for easy generating such matrices.
$endgroup$
– Widawensen
Jan 31 at 9:40
add a comment |
$begingroup$
As so far there was no answer which could provide an algebraic form of orthogonal matrix with single $0$ entry.
However looking a few days at the 3d form I was able to find the missing pattern at the end yesterday.
The pattern is amazingly simple.
Let $ v=begin{bmatrix} a_1 & a_2 & dots & a_n end{bmatrix}^T $ be n-dimensional unit vector ($v^Tv=1$).
Now we can construct a block matrix dimension ${(n+1) times (n+1)}$ based only on $v$ vector:
$$Q_{(n+1) times (n+1)} =begin{bmatrix} 0_{1 times 1} & -v^T_{1 times n} \ v_{n times 1} & vv^T-I_{n times n} end{bmatrix}$$
For this matrix:
$$Q^T =begin{bmatrix} 0 & v^T \ -v & vv^T-I end{bmatrix}$$
One can check that really $QQ^T=I$, hence if only unit vector $v$ has non-zero components then we have orthogonal matrix with single zero entry.
answered Feb 1 at 13:19
WidawensenWidawensen
4,76331447
$endgroup$
add a comment |
$begingroup$
As so far there was no answer which could provide an algebraic form of orthogonal matrix with single $0$ entry.
However looking a few days at the 3d form I was able to find the missing pattern at the end yesterday.
The pattern is amazingly simple.
Let $ v=begin{bmatrix} a_1 & a_2 & dots & a_n end{bmatrix}^T $ be n-dimensional unit vector ($v^Tv=1$).
Now we can construct a block matrix dimension ${(n+1) times (n+1)}$ based only on $v$ vector:
$$Q_{(n+1) times (n+1)} =begin{bmatrix} 0_{1 times 1} & -v^T_{1 times n} \ v_{n times 1} & vv^T-I_{n times n} end{bmatrix}$$
For this matrix:
$$Q^T =begin{bmatrix} 0 & v^T \ -v & vv^T-I end{bmatrix}$$
One can check that really $QQ^T=I$, hence if only unit vector $v$ has non-zero components then we have orthogonal matrix with single zero entry.
answered Feb 1 at 13:19
WidawensenWidawensen
4,76331447
$endgroup$
add a comment |
$begingroup$
As so far there was no answer which could provide an algebraic form of orthogonal matrix with single $0$ entry.
However looking a few days at the 3d form I was able to find the missing pattern at the end yesterday.
The pattern is amazingly simple.
Let $ v=begin{bmatrix} a_1 & a_2 & dots & a_n end{bmatrix}^T $ be n-dimensional unit vector ($v^Tv=1$).
Now we can construct a block matrix dimension ${(n+1) times (n+1)}$ based only on $v$ vector:
$$Q_{(n+1) times (n+1)} =begin{bmatrix} 0_{1 times 1} & -v^T_{1 times n} \ v_{n times 1} & vv^T-I_{n times n} end{bmatrix}$$
For this matrix:
$$Q^T =begin{bmatrix} 0 & v^T \ -v & vv^T-I end{bmatrix}$$
One can check that really $QQ^T=I$, hence if only unit vector $v$ has non-zero components then we have orthogonal matrix with single zero entry.
answered Feb 1 at 13:19
WidawensenWidawensen
4,76331447
$endgroup$
As so far there was no answer which could provide an algebraic form of orthogonal matrix with single $0$ entry.
However looking a few days at the 3d form I was able to find the missing pattern at the end yesterday.
The pattern is amazingly simple.
Let $ v=begin{bmatrix} a_1 & a_2 & dots & a_n end{bmatrix}^T $ be n-dimensional unit vector ($v^Tv=1$).
Now we can construct a block matrix dimension ${(n+1) times (n+1)}$ based only on $v$ vector:
$$Q_{(n+1) times (n+1)} =begin{bmatrix} 0_{1 times 1} & -v^T_{1 times n} \ v_{n times 1} & vv^T-I_{n times n} end{bmatrix}$$
For this matrix:
$$Q^T =begin{bmatrix} 0 & v^T \ -v & vv^T-I end{bmatrix}$$
One can check that really $QQ^T=I$, hence if only unit vector $v$ has non-zero components then we have orthogonal matrix with single zero entry.
answered Feb 1 at 13:19
WidawensenWidawensen
4,76331447
edited Feb 1 at 13:27
answered Feb 1 at 13:19
WidawensenWidawensen
4,76331447
answered Feb 1 at 13:19
WidawensenWidawensen
4,76331447
answered Feb 1 at 13:19
WidawensenWidawensen
4,76331447
4,76331447
add a comment |
add a comment |
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