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The angle between the $x$-axis and the vector that reachs the $2$-norm of a matrix
The 2019 Stack Overflow Developer Survey Results Are InCalculating axis angle from matrix without wrapping at PIWhat is the rotation axis and rotation angle of the composition of two rotation matrix in $mathbbR^3$Let $theta=frac2 pi67$ consider the rotation matrix $A$. What is $A^2010$?Find axis of rotation for matrix BA given matrix ABFind a matrix for the linear transformation of reflection about a $theta$ line using the matrix for projection$nD$ rotation around a general $(n-2)$-dimensional subspaceRotation Matrix From Axis and AngleHow to rotate a vector by a given angle about a given axisWhile converting rotation matrix to angle-axis representation how to find axis of rotation when angle of rotation is Pi?Angle definition confusion in Rodrigues rotation matrix
$begingroup$
Suppose we have a matrix $Ain mathbbR^2times 2$. We have the $2$-norm defined as $left |Aright |_2=max_left left |Axright |_2$.
We know that in order to find a unit vector $v$ such that $left |Avright |_2=left |Aright |_2$, we have to find the eigenvalues of $A^TA$, take the greatest eigenvalue, find a corresponding eigenvector and finally normalize it.
Of course, this would lead us to two different solutions: a vector $v$ and its corresponding additive opposite. Nevertheless, we will take the one with non-negative value of $x$ (i.e. the one which is to the right of the $y$-axis).
Once we have this vector, it is easy to find the angle it forms with the $x$-axis. But I have the following question: Is there any better way to find this angle, providing that it is the only information we want?
This question arises from the following example: if $A=beginpmatrix
1 & 1\
0 & 1
endpmatrix$, the eigenvalues of $A^TA$ are $frac3pm sqrt52$, and the calculation becomes very difficult. Moreover, I found a solution (which I could not understand) that finds the angle without finding the vector. This is the solution:
Have to maximize the function $fleft (thetaright )=cos^2theta +2sinthetacostheta +2sin^2theta=1+sin2theta+sin^2theta$ and so setting the first derivative to be zero: $0=f'left (thetaright )=2cos2theta +2sinthetacostheta=2cos2theta+sin2theta$. That is $tan2theta=2$.
Maybe this can be generalized in order to obtain a general way to find this angle? In that case, I would like to have a better-explained solution. I do not understand where the function $f$ comes from.
matrices eigenvalues-eigenvectors norm normed-spaces
asked Mar 30 at 17:13
solomeo paredessolomeo paredes
485210
$endgroup$
add a comment |
$begingroup$
Suppose we have a matrix $Ain mathbbR^2times 2$. We have the $2$-norm defined as $left |Aright |_2=max_left left |Axright |_2$.
We know that in order to find a unit vector $v$ such that $left |Avright |_2=left |Aright |_2$, we have to find the eigenvalues of $A^TA$, take the greatest eigenvalue, find a corresponding eigenvector and finally normalize it.
Of course, this would lead us to two different solutions: a vector $v$ and its corresponding additive opposite. Nevertheless, we will take the one with non-negative value of $x$ (i.e. the one which is to the right of the $y$-axis).
Once we have this vector, it is easy to find the angle it forms with the $x$-axis. But I have the following question: Is there any better way to find this angle, providing that it is the only information we want?
This question arises from the following example: if $A=beginpmatrix
1 & 1\
0 & 1
endpmatrix$, the eigenvalues of $A^TA$ are $frac3pm sqrt52$, and the calculation becomes very difficult. Moreover, I found a solution (which I could not understand) that finds the angle without finding the vector. This is the solution:
Have to maximize the function $fleft (thetaright )=cos^2theta +2sinthetacostheta +2sin^2theta=1+sin2theta+sin^2theta$ and so setting the first derivative to be zero: $0=f'left (thetaright )=2cos2theta +2sinthetacostheta=2cos2theta+sin2theta$. That is $tan2theta=2$.
Maybe this can be generalized in order to obtain a general way to find this angle? In that case, I would like to have a better-explained solution. I do not understand where the function $f$ comes from.
matrices eigenvalues-eigenvectors norm normed-spaces
asked Mar 30 at 17:13
solomeo paredessolomeo paredes
485210
$endgroup$
add a comment |
$begingroup$
Suppose we have a matrix $Ain mathbbR^2times 2$. We have the $2$-norm defined as $left |Aright |_2=max_left left |Axright |_2$.
We know that in order to find a unit vector $v$ such that $left |Avright |_2=left |Aright |_2$, we have to find the eigenvalues of $A^TA$, take the greatest eigenvalue, find a corresponding eigenvector and finally normalize it.
Of course, this would lead us to two different solutions: a vector $v$ and its corresponding additive opposite. Nevertheless, we will take the one with non-negative value of $x$ (i.e. the one which is to the right of the $y$-axis).
Once we have this vector, it is easy to find the angle it forms with the $x$-axis. But I have the following question: Is there any better way to find this angle, providing that it is the only information we want?
This question arises from the following example: if $A=beginpmatrix
1 & 1\
0 & 1
endpmatrix$, the eigenvalues of $A^TA$ are $frac3pm sqrt52$, and the calculation becomes very difficult. Moreover, I found a solution (which I could not understand) that finds the angle without finding the vector. This is the solution:
Have to maximize the function $fleft (thetaright )=cos^2theta +2sinthetacostheta +2sin^2theta=1+sin2theta+sin^2theta$ and so setting the first derivative to be zero: $0=f'left (thetaright )=2cos2theta +2sinthetacostheta=2cos2theta+sin2theta$. That is $tan2theta=2$.
Maybe this can be generalized in order to obtain a general way to find this angle? In that case, I would like to have a better-explained solution. I do not understand where the function $f$ comes from.
matrices eigenvalues-eigenvectors norm normed-spaces
asked Mar 30 at 17:13
solomeo paredessolomeo paredes
485210
$endgroup$
Suppose we have a matrix $Ain mathbbR^2times 2$. We have the $2$-norm defined as $left |Aright |_2=max_left left |Axright |_2$.
We know that in order to find a unit vector $v$ such that $left |Avright |_2=left |Aright |_2$, we have to find the eigenvalues of $A^TA$, take the greatest eigenvalue, find a corresponding eigenvector and finally normalize it.
Of course, this would lead us to two different solutions: a vector $v$ and its corresponding additive opposite. Nevertheless, we will take the one with non-negative value of $x$ (i.e. the one which is to the right of the $y$-axis).
Once we have this vector, it is easy to find the angle it forms with the $x$-axis. But I have the following question: Is there any better way to find this angle, providing that it is the only information we want?
This question arises from the following example: if $A=beginpmatrix
1 & 1\
0 & 1
endpmatrix$, the eigenvalues of $A^TA$ are $frac3pm sqrt52$, and the calculation becomes very difficult. Moreover, I found a solution (which I could not understand) that finds the angle without finding the vector. This is the solution:
Have to maximize the function $fleft (thetaright )=cos^2theta +2sinthetacostheta +2sin^2theta=1+sin2theta+sin^2theta$ and so setting the first derivative to be zero: $0=f'left (thetaright )=2cos2theta +2sinthetacostheta=2cos2theta+sin2theta$. That is $tan2theta=2$.
Maybe this can be generalized in order to obtain a general way to find this angle? In that case, I would like to have a better-explained solution. I do not understand where the function $f$ comes from.
matrices eigenvalues-eigenvectors norm normed-spaces
matrices eigenvalues-eigenvectors norm normed-spaces
asked Mar 30 at 17:13
solomeo paredessolomeo paredes
485210
asked Mar 30 at 17:13
solomeo paredessolomeo paredes
485210
asked Mar 30 at 17:13
solomeo paredessolomeo paredes
485210
asked Mar 30 at 17:13
solomeo paredessolomeo paredes
485210
asked Mar 30 at 17:13
solomeo paredessolomeo paredes
485210
485210
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Because your setup is $2times2$, you can work with less theory. Since you only need to look at real vectors $(x,y)$ with $|(x,y)|=1$, you may assume that $x=costheta$, $y=sintheta$. So now (for your particular $A$) you are looking at
beginalign
|A(x,y)|^2=|(x+y,y)|^2
&=(costheta+sintheta)^2+sin^2theta=cos^2theta+sin^2theta+2costhetasintheta+sin^2theta\
&=1+2costhetasintheta+sin^2theta\
&=1+sin2theta+sin^2theta.
endalign
Now you are looking for $theta$ that maximizes this expression, so differentiate and equate to zero.
If you have $A=beginbmatrix a_11&a_12\ a_21&a_22endbmatrix$, now the function to be maximized will be
$$
(a_11costheta+a_12sintheta)^2+(a_21costheta+a_22sintheta)^2.
$$
answered Mar 31 at 4:58
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Because your setup is $2times2$, you can work with less theory. Since you only need to look at real vectors $(x,y)$ with $|(x,y)|=1$, you may assume that $x=costheta$, $y=sintheta$. So now (for your particular $A$) you are looking at
beginalign
|A(x,y)|^2=|(x+y,y)|^2
&=(costheta+sintheta)^2+sin^2theta=cos^2theta+sin^2theta+2costhetasintheta+sin^2theta\
&=1+2costhetasintheta+sin^2theta\
&=1+sin2theta+sin^2theta.
endalign
Now you are looking for $theta$ that maximizes this expression, so differentiate and equate to zero.
If you have $A=beginbmatrix a_11&a_12\ a_21&a_22endbmatrix$, now the function to be maximized will be
$$
(a_11costheta+a_12sintheta)^2+(a_21costheta+a_22sintheta)^2.
$$
answered Mar 31 at 4:58
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
Because your setup is $2times2$, you can work with less theory. Since you only need to look at real vectors $(x,y)$ with $|(x,y)|=1$, you may assume that $x=costheta$, $y=sintheta$. So now (for your particular $A$) you are looking at
beginalign
|A(x,y)|^2=|(x+y,y)|^2
&=(costheta+sintheta)^2+sin^2theta=cos^2theta+sin^2theta+2costhetasintheta+sin^2theta\
&=1+2costhetasintheta+sin^2theta\
&=1+sin2theta+sin^2theta.
endalign
Now you are looking for $theta$ that maximizes this expression, so differentiate and equate to zero.
If you have $A=beginbmatrix a_11&a_12\ a_21&a_22endbmatrix$, now the function to be maximized will be
$$
(a_11costheta+a_12sintheta)^2+(a_21costheta+a_22sintheta)^2.
$$
answered Mar 31 at 4:58
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
Because your setup is $2times2$, you can work with less theory. Since you only need to look at real vectors $(x,y)$ with $|(x,y)|=1$, you may assume that $x=costheta$, $y=sintheta$. So now (for your particular $A$) you are looking at
beginalign
|A(x,y)|^2=|(x+y,y)|^2
&=(costheta+sintheta)^2+sin^2theta=cos^2theta+sin^2theta+2costhetasintheta+sin^2theta\
&=1+2costhetasintheta+sin^2theta\
&=1+sin2theta+sin^2theta.
endalign
Now you are looking for $theta$ that maximizes this expression, so differentiate and equate to zero.
If you have $A=beginbmatrix a_11&a_12\ a_21&a_22endbmatrix$, now the function to be maximized will be
$$
(a_11costheta+a_12sintheta)^2+(a_21costheta+a_22sintheta)^2.
$$
answered Mar 31 at 4:58
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
Because your setup is $2times2$, you can work with less theory. Since you only need to look at real vectors $(x,y)$ with $|(x,y)|=1$, you may assume that $x=costheta$, $y=sintheta$. So now (for your particular $A$) you are looking at
beginalign
|A(x,y)|^2=|(x+y,y)|^2
&=(costheta+sintheta)^2+sin^2theta=cos^2theta+sin^2theta+2costhetasintheta+sin^2theta\
&=1+2costhetasintheta+sin^2theta\
&=1+sin2theta+sin^2theta.
endalign
Now you are looking for $theta$ that maximizes this expression, so differentiate and equate to zero.
If you have $A=beginbmatrix a_11&a_12\ a_21&a_22endbmatrix$, now the function to be maximized will be
$$
(a_11costheta+a_12sintheta)^2+(a_21costheta+a_22sintheta)^2.
$$
answered Mar 31 at 4:58
Martin ArgeramiMartin Argerami
129k1184185
answered Mar 31 at 4:58
Martin ArgeramiMartin Argerami
129k1184185
answered Mar 31 at 4:58
Martin ArgeramiMartin Argerami
129k1184185
answered Mar 31 at 4:58
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
add a comment |
add a comment |
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