Mixing
For given column vectors $x$, $y$ how to find an orthogonal matrix $Q$ such that $Qx=y$
Problem: Given $x=[1,7,2,3,-1]^T $and $y=[-4,4,4,0,-4]^T$, find an orthogonal matrix $Q$ such that $Qx=y$.
My attempt: I know the definition of an orthogonal matrix. By definition, if $Q$ is an orthogonal matrix then $Q^{-1}=Q^T$. Further, I know if system of equations $Ax=b$ where $A$ is square matrix, has a solution if $b$ is in column space of $A$ i.e. if $rank(A:b)=rank(A)$.
I didn't able to solve above problem :-( please help me.
linear-algebra orthogonal-matrices
edited Jan 1 at 12:38
Saad
19.7k92352
asked Jan 1 at 6:45
Akash PatalwanshiAkash Patalwanshi
9821816
add a comment |
Problem: Given $x=[1,7,2,3,-1]^T $and $y=[-4,4,4,0,-4]^T$, find an orthogonal matrix $Q$ such that $Qx=y$.
My attempt: I know the definition of an orthogonal matrix. By definition, if $Q$ is an orthogonal matrix then $Q^{-1}=Q^T$. Further, I know if system of equations $Ax=b$ where $A$ is square matrix, has a solution if $b$ is in column space of $A$ i.e. if $rank(A:b)=rank(A)$.
I didn't able to solve above problem :-( please help me.
linear-algebra orthogonal-matrices
edited Jan 1 at 12:38
Saad
19.7k92352
asked Jan 1 at 6:45
Akash PatalwanshiAkash Patalwanshi
9821816
Look up Householder reflection.
– Chris Custer
Jan 1 at 7:04
Sir, I didn't know the notion of the Householder reflection. Can't we solve directly?
– Akash Patalwanshi
Jan 1 at 7:09
add a comment |
Problem: Given $x=[1,7,2,3,-1]^T $and $y=[-4,4,4,0,-4]^T$, find an orthogonal matrix $Q$ such that $Qx=y$.
My attempt: I know the definition of an orthogonal matrix. By definition, if $Q$ is an orthogonal matrix then $Q^{-1}=Q^T$. Further, I know if system of equations $Ax=b$ where $A$ is square matrix, has a solution if $b$ is in column space of $A$ i.e. if $rank(A:b)=rank(A)$.
I didn't able to solve above problem :-( please help me.
linear-algebra orthogonal-matrices
edited Jan 1 at 12:38
Saad
19.7k92352
asked Jan 1 at 6:45
Akash PatalwanshiAkash Patalwanshi
9821816
Problem: Given $x=[1,7,2,3,-1]^T $and $y=[-4,4,4,0,-4]^T$, find an orthogonal matrix $Q$ such that $Qx=y$.
My attempt: I know the definition of an orthogonal matrix. By definition, if $Q$ is an orthogonal matrix then $Q^{-1}=Q^T$. Further, I know if system of equations $Ax=b$ where $A$ is square matrix, has a solution if $b$ is in column space of $A$ i.e. if $rank(A:b)=rank(A)$.
I didn't able to solve above problem :-( please help me.
linear-algebra orthogonal-matrices
linear-algebra orthogonal-matrices
edited Jan 1 at 12:38
Saad
19.7k92352
asked Jan 1 at 6:45
Akash PatalwanshiAkash Patalwanshi
9821816
edited Jan 1 at 12:38
Saad
19.7k92352
asked Jan 1 at 6:45
Akash PatalwanshiAkash Patalwanshi
9821816
edited Jan 1 at 12:38
Saad
19.7k92352
edited Jan 1 at 12:38
Saad
19.7k92352
edited Jan 1 at 12:38
Saad
19.7k92352
19.7k92352
asked Jan 1 at 6:45
Akash PatalwanshiAkash Patalwanshi
9821816
asked Jan 1 at 6:45
Akash PatalwanshiAkash Patalwanshi
9821816
asked Jan 1 at 6:45
Akash PatalwanshiAkash Patalwanshi
9821816
9821816
Look up Householder reflection.
– Chris Custer
Jan 1 at 7:04
Sir, I didn't know the notion of the Householder reflection. Can't we solve directly?
– Akash Patalwanshi
Jan 1 at 7:09
add a comment |
Look up Householder reflection.
– Chris Custer
Jan 1 at 7:04
Sir, I didn't know the notion of the Householder reflection. Can't we solve directly?
– Akash Patalwanshi
Jan 1 at 7:09
Look up Householder reflection.
– Chris Custer
Jan 1 at 7:04
Look up Householder reflection.
– Chris Custer
Jan 1 at 7:04
Sir, I didn't know the notion of the Householder reflection. Can't we solve directly?
– Akash Patalwanshi
Jan 1 at 7:09
Sir, I didn't know the notion of the Householder reflection. Can't we solve directly?
– Akash Patalwanshi
Jan 1 at 7:09
add a comment |
1 Answer
1
active
oldest
votes
If you let $v=frac{x-y}{midmid y-xmidmid}$.
$A=I-2vv^t$ should do the trick.
This is called the Householder reflection.
answered Jan 1 at 7:23
Chris CusterChris Custer
11.1k3824
Sir thanks for your answer. Sir your answer shows such matrix exists. but, Why such a matrix exists? and how we know when it doesn't exist?
– Akash Patalwanshi
Jan 1 at 7:35
3
Note that the two vectors have the same norm.
– Chris Custer
Jan 1 at 7:38
1
This shows that it works:math.stackexchange.com/a/1903322
– Chris Custer
Jan 1 at 7:44
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you let $v=frac{x-y}{midmid y-xmidmid}$.
$A=I-2vv^t$ should do the trick.
This is called the Householder reflection.
answered Jan 1 at 7:23
Chris CusterChris Custer
11.1k3824
Sir thanks for your answer. Sir your answer shows such matrix exists. but, Why such a matrix exists? and how we know when it doesn't exist?
– Akash Patalwanshi
Jan 1 at 7:35
3
Note that the two vectors have the same norm.
– Chris Custer
Jan 1 at 7:38
1
This shows that it works:math.stackexchange.com/a/1903322
– Chris Custer
Jan 1 at 7:44
add a comment |
If you let $v=frac{x-y}{midmid y-xmidmid}$.
$A=I-2vv^t$ should do the trick.
This is called the Householder reflection.
answered Jan 1 at 7:23
Chris CusterChris Custer
11.1k3824
Sir thanks for your answer. Sir your answer shows such matrix exists. but, Why such a matrix exists? and how we know when it doesn't exist?
– Akash Patalwanshi
Jan 1 at 7:35
3
Note that the two vectors have the same norm.
– Chris Custer
Jan 1 at 7:38
1
This shows that it works:math.stackexchange.com/a/1903322
– Chris Custer
Jan 1 at 7:44
add a comment |
If you let $v=frac{x-y}{midmid y-xmidmid}$.
$A=I-2vv^t$ should do the trick.
This is called the Householder reflection.
answered Jan 1 at 7:23
Chris CusterChris Custer
11.1k3824
If you let $v=frac{x-y}{midmid y-xmidmid}$.
$A=I-2vv^t$ should do the trick.
This is called the Householder reflection.
answered Jan 1 at 7:23
Chris CusterChris Custer
11.1k3824
edited Jan 1 at 7:36
answered Jan 1 at 7:23
Chris CusterChris Custer
11.1k3824
answered Jan 1 at 7:23
Chris CusterChris Custer
11.1k3824
answered Jan 1 at 7:23
Chris CusterChris Custer
11.1k3824
11.1k3824
Sir thanks for your answer. Sir your answer shows such matrix exists. but, Why such a matrix exists? and how we know when it doesn't exist?
– Akash Patalwanshi
Jan 1 at 7:35
3
Note that the two vectors have the same norm.
– Chris Custer
Jan 1 at 7:38
1
This shows that it works:math.stackexchange.com/a/1903322
– Chris Custer
Jan 1 at 7:44
add a comment |
Sir thanks for your answer. Sir your answer shows such matrix exists. but, Why such a matrix exists? and how we know when it doesn't exist?
– Akash Patalwanshi
Jan 1 at 7:35
3
Note that the two vectors have the same norm.
– Chris Custer
Jan 1 at 7:38
1
This shows that it works:math.stackexchange.com/a/1903322
– Chris Custer
Jan 1 at 7:44
Sir thanks for your answer. Sir your answer shows such matrix exists. but, Why such a matrix exists? and how we know when it doesn't exist?
– Akash Patalwanshi
Jan 1 at 7:35
Sir thanks for your answer. Sir your answer shows such matrix exists. but, Why such a matrix exists? and how we know when it doesn't exist?
– Akash Patalwanshi
Jan 1 at 7:35
3
3
Note that the two vectors have the same norm.
– Chris Custer
Jan 1 at 7:38
Note that the two vectors have the same norm.
– Chris Custer
Jan 1 at 7:38
1
1
This shows that it works:math.stackexchange.com/a/1903322
– Chris Custer
Jan 1 at 7:44
This shows that it works:math.stackexchange.com/a/1903322
– Chris Custer
Jan 1 at 7:44
add a comment |
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Look up Householder reflection.
– Chris Custer
Jan 1 at 7:04
Sir, I didn't know the notion of the Householder reflection. Can't we solve directly?
– Akash Patalwanshi
Jan 1 at 7:09