Mixing
How to identify what kind of isometry the given matrix is?
$begingroup$
Let $E=mathbb{R}^3$ an euclidean space and let the following matrix :
$$dfrac{1}{9}cdotbegin{pmatrix}8&1&4\-4&4&7\1&8&-4end{pmatrix}$$
Since $AA^T=I_3$ I conclude that $Ain O(E)$, $det A$ gives $-1$, so $Ain O(E)backslash SO(E)$
It seems to be a combination of one rotation and one symmetry, but I don't know how to find the rotation axe.
In general if $r$ is a rotation and $s$ is a symmetry, I don't know how to study $rcirc s$
linear-algebra linear-transformations
edited Jan 15 at 21:08
Bernard
121k740116
asked Jan 15 at 21:06
StuStu
1,1881414
$endgroup$
add a comment |
$begingroup$
Let $E=mathbb{R}^3$ an euclidean space and let the following matrix :
$$dfrac{1}{9}cdotbegin{pmatrix}8&1&4\-4&4&7\1&8&-4end{pmatrix}$$
Since $AA^T=I_3$ I conclude that $Ain O(E)$, $det A$ gives $-1$, so $Ain O(E)backslash SO(E)$
It seems to be a combination of one rotation and one symmetry, but I don't know how to find the rotation axe.
In general if $r$ is a rotation and $s$ is a symmetry, I don't know how to study $rcirc s$
linear-algebra linear-transformations
edited Jan 15 at 21:08
Bernard
121k740116
asked Jan 15 at 21:06
StuStu
1,1881414
$endgroup$
add a comment |
$begingroup$
Let $E=mathbb{R}^3$ an euclidean space and let the following matrix :
$$dfrac{1}{9}cdotbegin{pmatrix}8&1&4\-4&4&7\1&8&-4end{pmatrix}$$
Since $AA^T=I_3$ I conclude that $Ain O(E)$, $det A$ gives $-1$, so $Ain O(E)backslash SO(E)$
It seems to be a combination of one rotation and one symmetry, but I don't know how to find the rotation axe.
In general if $r$ is a rotation and $s$ is a symmetry, I don't know how to study $rcirc s$
linear-algebra linear-transformations
edited Jan 15 at 21:08
Bernard
121k740116
asked Jan 15 at 21:06
StuStu
1,1881414
$endgroup$
Let $E=mathbb{R}^3$ an euclidean space and let the following matrix :
$$dfrac{1}{9}cdotbegin{pmatrix}8&1&4\-4&4&7\1&8&-4end{pmatrix}$$
Since $AA^T=I_3$ I conclude that $Ain O(E)$, $det A$ gives $-1$, so $Ain O(E)backslash SO(E)$
It seems to be a combination of one rotation and one symmetry, but I don't know how to find the rotation axe.
In general if $r$ is a rotation and $s$ is a symmetry, I don't know how to study $rcirc s$
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Jan 15 at 21:08
Bernard
121k740116
asked Jan 15 at 21:06
StuStu
1,1881414
edited Jan 15 at 21:08
Bernard
121k740116
asked Jan 15 at 21:06
StuStu
1,1881414
edited Jan 15 at 21:08
Bernard
121k740116
edited Jan 15 at 21:08
Bernard
121k740116
edited Jan 15 at 21:08
Bernard
121k740116
121k740116
asked Jan 15 at 21:06
StuStu
1,1881414
asked Jan 15 at 21:06
StuStu
1,1881414
asked Jan 15 at 21:06
StuStu
1,1881414
1,1881414
add a comment |
add a comment |
1 Answer
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$begingroup$
Your matrix has one real eigenvalue $-1$ and two non-real complex conjugate eigenvalues. An eigenvector $v$ for eigenvalue $-1$, i.e. a vector in the null space of $A + I$, is transformed to $-v$ by $A$, while the plane orthogonal to that vector is rotated by $A$. Thus $A$ consists of reflection
across this plane together with a rotation around the axis $v$.
answered Jan 15 at 21:18
Robert IsraelRobert Israel
324k23214468
$endgroup$
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
$begingroup$
Your matrix has one real eigenvalue $-1$ and two non-real complex conjugate eigenvalues. An eigenvector $v$ for eigenvalue $-1$, i.e. a vector in the null space of $A + I$, is transformed to $-v$ by $A$, while the plane orthogonal to that vector is rotated by $A$. Thus $A$ consists of reflection
across this plane together with a rotation around the axis $v$.
answered Jan 15 at 21:18
Robert IsraelRobert Israel
324k23214468
$endgroup$
add a comment |
$begingroup$
Your matrix has one real eigenvalue $-1$ and two non-real complex conjugate eigenvalues. An eigenvector $v$ for eigenvalue $-1$, i.e. a vector in the null space of $A + I$, is transformed to $-v$ by $A$, while the plane orthogonal to that vector is rotated by $A$. Thus $A$ consists of reflection
across this plane together with a rotation around the axis $v$.
answered Jan 15 at 21:18
Robert IsraelRobert Israel
324k23214468
$endgroup$
add a comment |
$begingroup$
Your matrix has one real eigenvalue $-1$ and two non-real complex conjugate eigenvalues. An eigenvector $v$ for eigenvalue $-1$, i.e. a vector in the null space of $A + I$, is transformed to $-v$ by $A$, while the plane orthogonal to that vector is rotated by $A$. Thus $A$ consists of reflection
across this plane together with a rotation around the axis $v$.
answered Jan 15 at 21:18
Robert IsraelRobert Israel
324k23214468
$endgroup$
Your matrix has one real eigenvalue $-1$ and two non-real complex conjugate eigenvalues. An eigenvector $v$ for eigenvalue $-1$, i.e. a vector in the null space of $A + I$, is transformed to $-v$ by $A$, while the plane orthogonal to that vector is rotated by $A$. Thus $A$ consists of reflection
across this plane together with a rotation around the axis $v$.
answered Jan 15 at 21:18
Robert IsraelRobert Israel
324k23214468
edited Jan 15 at 21:25
answered Jan 15 at 21:18
Robert IsraelRobert Israel
324k23214468
answered Jan 15 at 21:18
Robert IsraelRobert Israel
324k23214468
answered Jan 15 at 21:18
Robert IsraelRobert Israel
324k23214468
324k23214468
add a comment |
add a comment |
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