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Eigenvalue of loosely defined matrix
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Find $left|lambdaright|_2$.
$lambda(=a+bi)$ is eigenvalue of A. $;;left|lambdaright|_2=sqrta^2+b^2$ .
$ A=I-frac2u^Tuuu^T,;;;(non-zero);uin mathbbR^n, ;;;Ain mathbbR^ntimes n $
I'm not sure how to find A so that I can find the determinant of it.
All I could think was $;;u^Tu= left|uright|^2 ;;$ and $;;uu^T$ is symmetric.
and A would have a form like this.
$$ A=beginbmatrix
-1 & ? &? \
? & -1 & ? \
? & ? & -1
endbmatrix$$
Since it's symmetrical it'll have real eigenvalues.
Now, how should I proceed?
linear-algebra
asked Aug 2 at 13:09
nik
233
 |Â
show 12 more comments
up vote
0
down vote
favorite
Find $left|lambdaright|_2$.
$lambda(=a+bi)$ is eigenvalue of A. $;;left|lambdaright|_2=sqrta^2+b^2$ .
$ A=I-frac2u^Tuuu^T,;;;(non-zero);uin mathbbR^n, ;;;Ain mathbbR^ntimes n $
I'm not sure how to find A so that I can find the determinant of it.
All I could think was $;;u^Tu= left|uright|^2 ;;$ and $;;uu^T$ is symmetric.
and A would have a form like this.
$$ A=beginbmatrix
-1 & ? &? \
? & -1 & ? \
? & ? & -1
endbmatrix$$
Since it's symmetrical it'll have real eigenvalues.
Now, how should I proceed?
linear-algebra
asked Aug 2 at 13:09
nik
233
1st) Calculate $Au$; 2nd) Calculate $A^TA$ $(=A^2)$. That gives you the full eigen-structure.
– Tobias
Aug 2 at 13:18
um... I'm not sure what you mean. What do I get from Au and $A^TA$ ?
– nik
Aug 3 at 11:44
$A^TA=mathbf1$ says that $A$ is orthogonal. Therefore it has an orthogonal eigensystem. Actually you can use any base for the orthogonal complement of $operatornamespanu$ to complete the eigensystem of $A$ because $A$ acts as identity on the orthogonal complement of $operatornamespanu$ ($A$ is a (Householder) reflection). $u$ has eigenvalue -1 while the other eigenvectors correspond to the eigenvalue 1.
– Tobias
Aug 3 at 11:49
Many things in your comment confuse me. I'm bad at this haha;; let me ask you one by one. how is $A^TA$ equal to 1? when I imangine 3-by-3 identity matrix, I can't get 1. I get $sqrt3$
– nik
Aug 3 at 12:07
Didn't write 1, wrote $mathbf1$ standing for the identity matrix. Sorry, should have mentioned that.
– Tobias
Aug 3 at 12:22
 |Â
show 12 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find $left|lambdaright|_2$.
$lambda(=a+bi)$ is eigenvalue of A. $;;left|lambdaright|_2=sqrta^2+b^2$ .
$ A=I-frac2u^Tuuu^T,;;;(non-zero);uin mathbbR^n, ;;;Ain mathbbR^ntimes n $
I'm not sure how to find A so that I can find the determinant of it.
All I could think was $;;u^Tu= left|uright|^2 ;;$ and $;;uu^T$ is symmetric.
and A would have a form like this.
$$ A=beginbmatrix
-1 & ? &? \
? & -1 & ? \
? & ? & -1
endbmatrix$$
Since it's symmetrical it'll have real eigenvalues.
Now, how should I proceed?
linear-algebra
asked Aug 2 at 13:09
nik
233
Find $left|lambdaright|_2$.
$lambda(=a+bi)$ is eigenvalue of A. $;;left|lambdaright|_2=sqrta^2+b^2$ .
$ A=I-frac2u^Tuuu^T,;;;(non-zero);uin mathbbR^n, ;;;Ain mathbbR^ntimes n $
I'm not sure how to find A so that I can find the determinant of it.
All I could think was $;;u^Tu= left|uright|^2 ;;$ and $;;uu^T$ is symmetric.
and A would have a form like this.
$$ A=beginbmatrix
-1 & ? &? \
? & -1 & ? \
? & ? & -1
endbmatrix$$
Since it's symmetrical it'll have real eigenvalues.
Now, how should I proceed?
linear-algebra
asked Aug 2 at 13:09
nik
233
asked Aug 2 at 13:09
nik
233
asked Aug 2 at 13:09
nik
233
asked Aug 2 at 13:09
nik
233
233
1st) Calculate $Au$; 2nd) Calculate $A^TA$ $(=A^2)$. That gives you the full eigen-structure.
– Tobias
Aug 2 at 13:18
um... I'm not sure what you mean. What do I get from Au and $A^TA$ ?
– nik
Aug 3 at 11:44
$A^TA=mathbf1$ says that $A$ is orthogonal. Therefore it has an orthogonal eigensystem. Actually you can use any base for the orthogonal complement of $operatornamespanu$ to complete the eigensystem of $A$ because $A$ acts as identity on the orthogonal complement of $operatornamespanu$ ($A$ is a (Householder) reflection). $u$ has eigenvalue -1 while the other eigenvectors correspond to the eigenvalue 1.
– Tobias
Aug 3 at 11:49
Many things in your comment confuse me. I'm bad at this haha;; let me ask you one by one. how is $A^TA$ equal to 1? when I imangine 3-by-3 identity matrix, I can't get 1. I get $sqrt3$
– nik
Aug 3 at 12:07
Didn't write 1, wrote $mathbf1$ standing for the identity matrix. Sorry, should have mentioned that.
– Tobias
Aug 3 at 12:22
 |Â
show 12 more comments
1st) Calculate $Au$; 2nd) Calculate $A^TA$ $(=A^2)$. That gives you the full eigen-structure.
– Tobias
Aug 2 at 13:18
um... I'm not sure what you mean. What do I get from Au and $A^TA$ ?
– nik
Aug 3 at 11:44
$A^TA=mathbf1$ says that $A$ is orthogonal. Therefore it has an orthogonal eigensystem. Actually you can use any base for the orthogonal complement of $operatornamespanu$ to complete the eigensystem of $A$ because $A$ acts as identity on the orthogonal complement of $operatornamespanu$ ($A$ is a (Householder) reflection). $u$ has eigenvalue -1 while the other eigenvectors correspond to the eigenvalue 1.
– Tobias
Aug 3 at 11:49
Many things in your comment confuse me. I'm bad at this haha;; let me ask you one by one. how is $A^TA$ equal to 1? when I imangine 3-by-3 identity matrix, I can't get 1. I get $sqrt3$
– nik
Aug 3 at 12:07
Didn't write 1, wrote $mathbf1$ standing for the identity matrix. Sorry, should have mentioned that.
– Tobias
Aug 3 at 12:22
1st) Calculate $Au$; 2nd) Calculate $A^TA$ $(=A^2)$. That gives you the full eigen-structure.
– Tobias
Aug 2 at 13:18
1st) Calculate $Au$; 2nd) Calculate $A^TA$ $(=A^2)$. That gives you the full eigen-structure.
– Tobias
Aug 2 at 13:18
um... I'm not sure what you mean. What do I get from Au and $A^TA$ ?
– nik
Aug 3 at 11:44
um... I'm not sure what you mean. What do I get from Au and $A^TA$ ?
– nik
Aug 3 at 11:44
$A^TA=mathbf1$ says that $A$ is orthogonal. Therefore it has an orthogonal eigensystem. Actually you can use any base for the orthogonal complement of $operatornamespanu$ to complete the eigensystem of $A$ because $A$ acts as identity on the orthogonal complement of $operatornamespanu$ ($A$ is a (Householder) reflection). $u$ has eigenvalue -1 while the other eigenvectors correspond to the eigenvalue 1.
– Tobias
Aug 3 at 11:49
$A^TA=mathbf1$ says that $A$ is orthogonal. Therefore it has an orthogonal eigensystem. Actually you can use any base for the orthogonal complement of $operatornamespanu$ to complete the eigensystem of $A$ because $A$ acts as identity on the orthogonal complement of $operatornamespanu$ ($A$ is a (Householder) reflection). $u$ has eigenvalue -1 while the other eigenvectors correspond to the eigenvalue 1.
– Tobias
Aug 3 at 11:49
Many things in your comment confuse me. I'm bad at this haha;; let me ask you one by one. how is $A^TA$ equal to 1? when I imangine 3-by-3 identity matrix, I can't get 1. I get $sqrt3$
– nik
Aug 3 at 12:07
Many things in your comment confuse me. I'm bad at this haha;; let me ask you one by one. how is $A^TA$ equal to 1? when I imangine 3-by-3 identity matrix, I can't get 1. I get $sqrt3$
– nik
Aug 3 at 12:07
Didn't write 1, wrote $mathbf1$ standing for the identity matrix. Sorry, should have mentioned that.
– Tobias
Aug 3 at 12:22
Didn't write 1, wrote $mathbf1$ standing for the identity matrix. Sorry, should have mentioned that.
– Tobias
Aug 3 at 12:22
 |Â
show 12 more comments
1 Answer
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0
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I'll try to write what i've understood by the help of Tobias.
$A^TA=AA=left(I-frac2u^Tuuu^T right)left(I-frac2u^Tuuu^T right)=I-frac4u^Tuuu^T+frac4(u^Tu)^2uu^Tuu^T $
$= I-frac4leftuu^T+frac4^4uu^Tuu^T=I-frac4leftuu^T+frac4^4left|u^2right|uu^T=I-frac4leftuu^T+frac4leftuu^T=I$
So, A is orthogonal matrix. (I'm not sure why it matters though it means $det$ is $pm$1)
And,
$Au=left(I-frac2u^Tuuu^T right)u=u-2u=-u$
It means A is a "reflection about a plane" transformation matrix, thus for a v($perp u$), Av=v.
(I wonder If I need to figure out eigenvectors would be orthonormal from the fact that A is symmetrical to get to this conclusion.)
Eigenvalues of A is -1 and two 1.
answered Aug 4 at 7:50
nik
233
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I'll try to write what i've understood by the help of Tobias.
$A^TA=AA=left(I-frac2u^Tuuu^T right)left(I-frac2u^Tuuu^T right)=I-frac4u^Tuuu^T+frac4(u^Tu)^2uu^Tuu^T $
$= I-frac4leftuu^T+frac4^4uu^Tuu^T=I-frac4leftuu^T+frac4^4left|u^2right|uu^T=I-frac4leftuu^T+frac4leftuu^T=I$
So, A is orthogonal matrix. (I'm not sure why it matters though it means $det$ is $pm$1)
And,
$Au=left(I-frac2u^Tuuu^T right)u=u-2u=-u$
It means A is a "reflection about a plane" transformation matrix, thus for a v($perp u$), Av=v.
(I wonder If I need to figure out eigenvectors would be orthonormal from the fact that A is symmetrical to get to this conclusion.)
Eigenvalues of A is -1 and two 1.
answered Aug 4 at 7:50
nik
233
add a comment |Â
up vote
0
down vote
I'll try to write what i've understood by the help of Tobias.
$A^TA=AA=left(I-frac2u^Tuuu^T right)left(I-frac2u^Tuuu^T right)=I-frac4u^Tuuu^T+frac4(u^Tu)^2uu^Tuu^T $
$= I-frac4leftuu^T+frac4^4uu^Tuu^T=I-frac4leftuu^T+frac4^4left|u^2right|uu^T=I-frac4leftuu^T+frac4leftuu^T=I$
So, A is orthogonal matrix. (I'm not sure why it matters though it means $det$ is $pm$1)
And,
$Au=left(I-frac2u^Tuuu^T right)u=u-2u=-u$
It means A is a "reflection about a plane" transformation matrix, thus for a v($perp u$), Av=v.
(I wonder If I need to figure out eigenvectors would be orthonormal from the fact that A is symmetrical to get to this conclusion.)
Eigenvalues of A is -1 and two 1.
answered Aug 4 at 7:50
nik
233
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I'll try to write what i've understood by the help of Tobias.
$A^TA=AA=left(I-frac2u^Tuuu^T right)left(I-frac2u^Tuuu^T right)=I-frac4u^Tuuu^T+frac4(u^Tu)^2uu^Tuu^T $
$= I-frac4leftuu^T+frac4^4uu^Tuu^T=I-frac4leftuu^T+frac4^4left|u^2right|uu^T=I-frac4leftuu^T+frac4leftuu^T=I$
So, A is orthogonal matrix. (I'm not sure why it matters though it means $det$ is $pm$1)
And,
$Au=left(I-frac2u^Tuuu^T right)u=u-2u=-u$
It means A is a "reflection about a plane" transformation matrix, thus for a v($perp u$), Av=v.
(I wonder If I need to figure out eigenvectors would be orthonormal from the fact that A is symmetrical to get to this conclusion.)
Eigenvalues of A is -1 and two 1.
answered Aug 4 at 7:50
nik
233
I'll try to write what i've understood by the help of Tobias.
$A^TA=AA=left(I-frac2u^Tuuu^T right)left(I-frac2u^Tuuu^T right)=I-frac4u^Tuuu^T+frac4(u^Tu)^2uu^Tuu^T $
$= I-frac4leftuu^T+frac4^4uu^Tuu^T=I-frac4leftuu^T+frac4^4left|u^2right|uu^T=I-frac4leftuu^T+frac4leftuu^T=I$
So, A is orthogonal matrix. (I'm not sure why it matters though it means $det$ is $pm$1)
And,
$Au=left(I-frac2u^Tuuu^T right)u=u-2u=-u$
It means A is a "reflection about a plane" transformation matrix, thus for a v($perp u$), Av=v.
(I wonder If I need to figure out eigenvectors would be orthonormal from the fact that A is symmetrical to get to this conclusion.)
Eigenvalues of A is -1 and two 1.
answered Aug 4 at 7:50
nik
233
answered Aug 4 at 7:50
nik
233
answered Aug 4 at 7:50
nik
233
answered Aug 4 at 7:50
nik
233
233
add a comment |Â
add a comment |Â
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1st) Calculate $Au$; 2nd) Calculate $A^TA$ $(=A^2)$. That gives you the full eigen-structure.
– Tobias
Aug 2 at 13:18
um... I'm not sure what you mean. What do I get from Au and $A^TA$ ?
– nik
Aug 3 at 11:44
$A^TA=mathbf1$ says that $A$ is orthogonal. Therefore it has an orthogonal eigensystem. Actually you can use any base for the orthogonal complement of $operatornamespanu$ to complete the eigensystem of $A$ because $A$ acts as identity on the orthogonal complement of $operatornamespanu$ ($A$ is a (Householder) reflection). $u$ has eigenvalue -1 while the other eigenvectors correspond to the eigenvalue 1.
– Tobias
Aug 3 at 11:49
Many things in your comment confuse me. I'm bad at this haha;; let me ask you one by one. how is $A^TA$ equal to 1? when I imangine 3-by-3 identity matrix, I can't get 1. I get $sqrt3$
– nik
Aug 3 at 12:07
Didn't write 1, wrote $mathbf1$ standing for the identity matrix. Sorry, should have mentioned that.
– Tobias
Aug 3 at 12:22