Rotation of Linear Transformation in R2 The 2019 Stack Overflow Developer Survey Results Are InGiven this transformation matrix, how do I decompose it into translation, rotation and scale matrices?3D rotation groupEuler angles to rotation matrix. Rotation directionFinding a rotation matrixWhy is rotation about the y axis in $mathbbR^3$ different from rotation about the x and y axis.Linear Transformation Rotation, reflection, and projectionFor which point is the transformation linear?Let T1 be the linear transformation corresponding to a counterclockwise rotation of 120 degreesAngle definition confusion in Rodrigues rotation matrixLinear operator and rotation matrices
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Rotation of Linear Transformation in R2
The 2019 Stack Overflow Developer Survey Results Are InGiven this transformation matrix, how do I decompose it into translation, rotation and scale matrices?3D rotation groupEuler angles to rotation matrix. Rotation directionFinding a rotation matrixWhy is rotation about the y axis in $mathbbR^3$ different from rotation about the x and y axis.Linear Transformation Rotation, reflection, and projectionFor which point is the transformation linear?Let T1 be the linear transformation corresponding to a counterclockwise rotation of 120 degreesAngle definition confusion in Rodrigues rotation matrixLinear operator and rotation matrices
$begingroup$
I have a matrix which represents a 45 degrees rotation counterclockwise: $$beginpmatrix
frac1sqrt2 & -frac1sqrt2 \
frac1sqrt2 & frac1sqrt2
endpmatrix$$.
To find the matrices of Rot^4, why is this not the same matrix?
Thanks
linear-algebra matrices
edited Mar 23 at 22:18
Yuval Gat
9641213
asked Mar 23 at 18:00
MCCMCC
368
$endgroup$
add a comment |
$begingroup$
I have a matrix which represents a 45 degrees rotation counterclockwise: $$beginpmatrix
frac1sqrt2 & -frac1sqrt2 \
frac1sqrt2 & frac1sqrt2
endpmatrix$$.
To find the matrices of Rot^4, why is this not the same matrix?
Thanks
linear-algebra matrices
edited Mar 23 at 22:18
Yuval Gat
9641213
asked Mar 23 at 18:00
MCCMCC
368
$endgroup$
2
$begingroup$
Because $8*45=360$
$endgroup$
– John Douma
Mar 23 at 18:04
add a comment |
$begingroup$
I have a matrix which represents a 45 degrees rotation counterclockwise: $$beginpmatrix
frac1sqrt2 & -frac1sqrt2 \
frac1sqrt2 & frac1sqrt2
endpmatrix$$.
To find the matrices of Rot^4, why is this not the same matrix?
Thanks
linear-algebra matrices
edited Mar 23 at 22:18
Yuval Gat
9641213
asked Mar 23 at 18:00
MCCMCC
368
$endgroup$
I have a matrix which represents a 45 degrees rotation counterclockwise: $$beginpmatrix
frac1sqrt2 & -frac1sqrt2 \
frac1sqrt2 & frac1sqrt2
endpmatrix$$.
To find the matrices of Rot^4, why is this not the same matrix?
Thanks
linear-algebra matrices
linear-algebra matrices
edited Mar 23 at 22:18
Yuval Gat
9641213
asked Mar 23 at 18:00
MCCMCC
368
edited Mar 23 at 22:18
Yuval Gat
9641213
asked Mar 23 at 18:00
MCCMCC
368
edited Mar 23 at 22:18
Yuval Gat
9641213
edited Mar 23 at 22:18
Yuval Gat
9641213
edited Mar 23 at 22:18
Yuval Gat
9641213
9641213
asked Mar 23 at 18:00
MCCMCC
368
asked Mar 23 at 18:00
MCCMCC
368
asked Mar 23 at 18:00
MCCMCC
368
368
2
$begingroup$
Because $8*45=360$
$endgroup$
– John Douma
Mar 23 at 18:04
add a comment |
2
$begingroup$
Because $8*45=360$
$endgroup$
– John Douma
Mar 23 at 18:04
2
2
$begingroup$
Because $8*45=360$
$endgroup$
– John Douma
Mar 23 at 18:04
$begingroup$
Because $8*45=360$
$endgroup$
– John Douma
Mar 23 at 18:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You matrix is
$$
A=beginbmatrix costfracpi4 & -sintfracpi4 \ sintfracpi4& costfracpi4
endbmatrix.
$$
Then one can show (using induction and the cosime of the sum formula) that
$$
A^k=beginbmatrix costfrackpi4 & -sintfrackpi4 \ sintfrackpi4& costfrackpi4
endbmatrix.
$$
So to get $A^k=A$ we need $kpi/4 = 2mpi + pi/4$, or $k=8m+1$. That is, you need to turn eight times to get $A$ again, so the first time you'll get $A$ again is $A^9$.
On a more intuitive way, your matrix is a rotation of $pi/4$, and you need 8 more turns to get to the same point.
answered Mar 24 at 1:20
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
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
$begingroup$
You matrix is
$$
A=beginbmatrix costfracpi4 & -sintfracpi4 \ sintfracpi4& costfracpi4
endbmatrix.
$$
Then one can show (using induction and the cosime of the sum formula) that
$$
A^k=beginbmatrix costfrackpi4 & -sintfrackpi4 \ sintfrackpi4& costfrackpi4
endbmatrix.
$$
So to get $A^k=A$ we need $kpi/4 = 2mpi + pi/4$, or $k=8m+1$. That is, you need to turn eight times to get $A$ again, so the first time you'll get $A$ again is $A^9$.
On a more intuitive way, your matrix is a rotation of $pi/4$, and you need 8 more turns to get to the same point.
answered Mar 24 at 1:20
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
You matrix is
$$
A=beginbmatrix costfracpi4 & -sintfracpi4 \ sintfracpi4& costfracpi4
endbmatrix.
$$
Then one can show (using induction and the cosime of the sum formula) that
$$
A^k=beginbmatrix costfrackpi4 & -sintfrackpi4 \ sintfrackpi4& costfrackpi4
endbmatrix.
$$
So to get $A^k=A$ we need $kpi/4 = 2mpi + pi/4$, or $k=8m+1$. That is, you need to turn eight times to get $A$ again, so the first time you'll get $A$ again is $A^9$.
On a more intuitive way, your matrix is a rotation of $pi/4$, and you need 8 more turns to get to the same point.
answered Mar 24 at 1:20
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
You matrix is
$$
A=beginbmatrix costfracpi4 & -sintfracpi4 \ sintfracpi4& costfracpi4
endbmatrix.
$$
Then one can show (using induction and the cosime of the sum formula) that
$$
A^k=beginbmatrix costfrackpi4 & -sintfrackpi4 \ sintfrackpi4& costfrackpi4
endbmatrix.
$$
So to get $A^k=A$ we need $kpi/4 = 2mpi + pi/4$, or $k=8m+1$. That is, you need to turn eight times to get $A$ again, so the first time you'll get $A$ again is $A^9$.
On a more intuitive way, your matrix is a rotation of $pi/4$, and you need 8 more turns to get to the same point.
answered Mar 24 at 1:20
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
You matrix is
$$
A=beginbmatrix costfracpi4 & -sintfracpi4 \ sintfracpi4& costfracpi4
endbmatrix.
$$
Then one can show (using induction and the cosime of the sum formula) that
$$
A^k=beginbmatrix costfrackpi4 & -sintfrackpi4 \ sintfrackpi4& costfrackpi4
endbmatrix.
$$
So to get $A^k=A$ we need $kpi/4 = 2mpi + pi/4$, or $k=8m+1$. That is, you need to turn eight times to get $A$ again, so the first time you'll get $A$ again is $A^9$.
On a more intuitive way, your matrix is a rotation of $pi/4$, and you need 8 more turns to get to the same point.
answered Mar 24 at 1:20
Martin ArgeramiMartin Argerami
129k1184185
answered Mar 24 at 1:20
Martin ArgeramiMartin Argerami
129k1184185
answered Mar 24 at 1:20
Martin ArgeramiMartin Argerami
129k1184185
answered Mar 24 at 1:20
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
add a comment |
add a comment |
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$begingroup$
Because $8*45=360$
$endgroup$
– John Douma
Mar 23 at 18:04