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Higher powers of matrix A
Properties for a matrix being invariant under rotation?Prove the angle of $v in mathbb R^2 $ after rotation by $R_{theta}$ (rotation matrix) change $theta$-degrees.Matrix multiplication as combination of rotation and stretchingFind the inverse of a Trig MatrixMatrix representation for linear transformation on $mathbb{R}^{3}$Reduced row echelon form of matrix with trigonometric expressionsFind the rotation/reflection angle for orthogonal matrix ALooking for a conceptual understanding of a rotation matrix transformationProve fact using matrix multiplcationPowers of Rotation matrix
$begingroup$
Consider the matrix $$A= begin{bmatrix}cos(theta)&-sin(theta)\sin(theta)&cos(theta)end{bmatrix}$$
Calculate $A^9$ in the case of $theta = pi/27$.
Calculate $A^{2019}$ in the case of $theta = pi/6$.
$$$$
I'm fully clueless on this one. Don't know where to start.
linear-algebra
asked 1 hour ago
JayJay
183
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Consider the matrix $$A= begin{bmatrix}cos(theta)&-sin(theta)\sin(theta)&cos(theta)end{bmatrix}$$
Calculate $A^9$ in the case of $theta = pi/27$.
Calculate $A^{2019}$ in the case of $theta = pi/6$.
$$$$
I'm fully clueless on this one. Don't know where to start.
linear-algebra
asked 1 hour ago
JayJay
183
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Calculate the square and cube of the given matrix. Spot a pattern, make a claim and verify it via induction.
$endgroup$
– астон вілла олоф мэллбэрг
58 mins ago
$begingroup$
Maybe making an eigendecomposition? $A=PDP^{-1}$ which implies $A^n=PD^nP^{-1}$. I am just guessing
$endgroup$
– RScrlli
29 mins ago
add a comment |
$begingroup$
Consider the matrix $$A= begin{bmatrix}cos(theta)&-sin(theta)\sin(theta)&cos(theta)end{bmatrix}$$
Calculate $A^9$ in the case of $theta = pi/27$.
Calculate $A^{2019}$ in the case of $theta = pi/6$.
$$$$
I'm fully clueless on this one. Don't know where to start.
linear-algebra
asked 1 hour ago
JayJay
183
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Consider the matrix $$A= begin{bmatrix}cos(theta)&-sin(theta)\sin(theta)&cos(theta)end{bmatrix}$$
Calculate $A^9$ in the case of $theta = pi/27$.
Calculate $A^{2019}$ in the case of $theta = pi/6$.
$$$$
I'm fully clueless on this one. Don't know where to start.
linear-algebra
linear-algebra
asked 1 hour ago
JayJay
183
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
JayJay
183
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
JayJay
183
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
JayJay
183
asked 1 hour ago
JayJay
183
183
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Calculate the square and cube of the given matrix. Spot a pattern, make a claim and verify it via induction.
$endgroup$
– астон вілла олоф мэллбэрг
58 mins ago
$begingroup$
Maybe making an eigendecomposition? $A=PDP^{-1}$ which implies $A^n=PD^nP^{-1}$. I am just guessing
$endgroup$
– RScrlli
29 mins ago
add a comment |
1
$begingroup$
Calculate the square and cube of the given matrix. Spot a pattern, make a claim and verify it via induction.
$endgroup$
– астон вілла олоф мэллбэрг
58 mins ago
$begingroup$
Maybe making an eigendecomposition? $A=PDP^{-1}$ which implies $A^n=PD^nP^{-1}$. I am just guessing
$endgroup$
– RScrlli
29 mins ago
1
1
$begingroup$
Calculate the square and cube of the given matrix. Spot a pattern, make a claim and verify it via induction.
$endgroup$
– астон вілла олоф мэллбэрг
58 mins ago
$begingroup$
Calculate the square and cube of the given matrix. Spot a pattern, make a claim and verify it via induction.
$endgroup$
– астон вілла олоф мэллбэрг
58 mins ago
$begingroup$
Maybe making an eigendecomposition? $A=PDP^{-1}$ which implies $A^n=PD^nP^{-1}$. I am just guessing
$endgroup$
– RScrlli
29 mins ago
$begingroup$
Maybe making an eigendecomposition? $A=PDP^{-1}$ which implies $A^n=PD^nP^{-1}$. I am just guessing
$endgroup$
– RScrlli
29 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You may use the result that $$A^{n}= begin{pmatrix}cos(ntheta)&-sin(ntheta)\sin(ntheta)&cos(ntheta)end{pmatrix} $$
Or prove it by induction before using it.
answered 54 mins ago
Jade PangJade Pang
964
$endgroup$
add a comment |
$begingroup$
Hint $:$ The given matrix is the matrix of rotation by an angle $theta.$ Note that $text {Rot} (theta_1) circ text {Rot} (theta_2) = text {Rot} (theta_1 + theta_2).$
answered 54 mins ago
Dbchatto67Dbchatto67
2,193320
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
You may use the result that $$A^{n}= begin{pmatrix}cos(ntheta)&-sin(ntheta)\sin(ntheta)&cos(ntheta)end{pmatrix} $$
Or prove it by induction before using it.
answered 54 mins ago
Jade PangJade Pang
964
$endgroup$
add a comment |
$begingroup$
You may use the result that $$A^{n}= begin{pmatrix}cos(ntheta)&-sin(ntheta)\sin(ntheta)&cos(ntheta)end{pmatrix} $$
Or prove it by induction before using it.
answered 54 mins ago
Jade PangJade Pang
964
$endgroup$
add a comment |
$begingroup$
You may use the result that $$A^{n}= begin{pmatrix}cos(ntheta)&-sin(ntheta)\sin(ntheta)&cos(ntheta)end{pmatrix} $$
Or prove it by induction before using it.
answered 54 mins ago
Jade PangJade Pang
964
$endgroup$
You may use the result that $$A^{n}= begin{pmatrix}cos(ntheta)&-sin(ntheta)\sin(ntheta)&cos(ntheta)end{pmatrix} $$
Or prove it by induction before using it.
answered 54 mins ago
Jade PangJade Pang
964
answered 54 mins ago
Jade PangJade Pang
964
answered 54 mins ago
Jade PangJade Pang
964
answered 54 mins ago
Jade PangJade Pang
964
964
add a comment |
add a comment |
$begingroup$
Hint $:$ The given matrix is the matrix of rotation by an angle $theta.$ Note that $text {Rot} (theta_1) circ text {Rot} (theta_2) = text {Rot} (theta_1 + theta_2).$
answered 54 mins ago
Dbchatto67Dbchatto67
2,193320
$endgroup$
add a comment |
$begingroup$
Hint $:$ The given matrix is the matrix of rotation by an angle $theta.$ Note that $text {Rot} (theta_1) circ text {Rot} (theta_2) = text {Rot} (theta_1 + theta_2).$
answered 54 mins ago
Dbchatto67Dbchatto67
2,193320
$endgroup$
add a comment |
$begingroup$
Hint $:$ The given matrix is the matrix of rotation by an angle $theta.$ Note that $text {Rot} (theta_1) circ text {Rot} (theta_2) = text {Rot} (theta_1 + theta_2).$
answered 54 mins ago
Dbchatto67Dbchatto67
2,193320
$endgroup$
Hint $:$ The given matrix is the matrix of rotation by an angle $theta.$ Note that $text {Rot} (theta_1) circ text {Rot} (theta_2) = text {Rot} (theta_1 + theta_2).$
answered 54 mins ago
Dbchatto67Dbchatto67
2,193320
answered 54 mins ago
Dbchatto67Dbchatto67
2,193320
answered 54 mins ago
Dbchatto67Dbchatto67
2,193320
answered 54 mins ago
Dbchatto67Dbchatto67
2,193320
2,193320
add a comment |
add a comment |
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Jay is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
Calculate the square and cube of the given matrix. Spot a pattern, make a claim and verify it via induction.
$endgroup$
– астон вілла олоф мэллбэрг
58 mins ago
$begingroup$
Maybe making an eigendecomposition? $A=PDP^{-1}$ which implies $A^n=PD^nP^{-1}$. I am just guessing
$endgroup$
– RScrlli
29 mins ago