Find a matrix R α ∈ R 2 × 2 such that f ( x ) = f R α ( x ) for every x ∈ R 2
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I'm trying to solve this task.
Let α∈R be given. Consider the linear map f:R^2→R^2,
(x1 --> ( x1·cos(α)−x2·sin(α)
x2) x1.sin(α)+x2·cos(α)).
a) Find a matrix Rα∈R2×2 such that f(x) =f Rα(x) for everyx∈R^2.
b) Interpret the map f, i.e. state what it does to a given vector x∈R^2.
First I thought, I should use this rule
|A| = ad − bc , but I think, It's already used . should i gave α a number like zero and then I will have a matrix and that's it?
linear-algebra matrices linear-transformations
asked Nov 25 at 22:03
Amerov
65
add a comment |
up vote
0
down vote
favorite
I'm trying to solve this task.
Let α∈R be given. Consider the linear map f:R^2→R^2,
(x1 --> ( x1·cos(α)−x2·sin(α)
x2) x1.sin(α)+x2·cos(α)).
a) Find a matrix Rα∈R2×2 such that f(x) =f Rα(x) for everyx∈R^2.
b) Interpret the map f, i.e. state what it does to a given vector x∈R^2.
First I thought, I should use this rule
|A| = ad − bc , but I think, It's already used . should i gave α a number like zero and then I will have a matrix and that's it?
linear-algebra matrices linear-transformations
asked Nov 25 at 22:03
Amerov
65
Your question is very difficult to read! You really have to learn MathJax. See this link for a good tutorial.
– TonyK
Nov 25 at 22:08
You have a description there of the result of applying $f$ to an arbitrary column vector $(x_1,x_2)^T$. Can you write that result as a matrix times $(x_1,x_2)^T$ in any way?
– Arthur
Nov 25 at 22:08
@TonyK thanks..
– Amerov
Nov 25 at 22:25
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to solve this task.
Let α∈R be given. Consider the linear map f:R^2→R^2,
(x1 --> ( x1·cos(α)−x2·sin(α)
x2) x1.sin(α)+x2·cos(α)).
a) Find a matrix Rα∈R2×2 such that f(x) =f Rα(x) for everyx∈R^2.
b) Interpret the map f, i.e. state what it does to a given vector x∈R^2.
First I thought, I should use this rule
|A| = ad − bc , but I think, It's already used . should i gave α a number like zero and then I will have a matrix and that's it?
linear-algebra matrices linear-transformations
asked Nov 25 at 22:03
Amerov
65
I'm trying to solve this task.
Let α∈R be given. Consider the linear map f:R^2→R^2,
(x1 --> ( x1·cos(α)−x2·sin(α)
x2) x1.sin(α)+x2·cos(α)).
a) Find a matrix Rα∈R2×2 such that f(x) =f Rα(x) for everyx∈R^2.
b) Interpret the map f, i.e. state what it does to a given vector x∈R^2.
First I thought, I should use this rule
|A| = ad − bc , but I think, It's already used . should i gave α a number like zero and then I will have a matrix and that's it?
linear-algebra matrices linear-transformations
linear-algebra matrices linear-transformations
asked Nov 25 at 22:03
Amerov
65
asked Nov 25 at 22:03
Amerov
65
edited Nov 25 at 22:10
asked Nov 25 at 22:03
Amerov
65
asked Nov 25 at 22:03
Amerov
65
asked Nov 25 at 22:03
Amerov
65
65
Your question is very difficult to read! You really have to learn MathJax. See this link for a good tutorial.
– TonyK
Nov 25 at 22:08
You have a description there of the result of applying $f$ to an arbitrary column vector $(x_1,x_2)^T$. Can you write that result as a matrix times $(x_1,x_2)^T$ in any way?
– Arthur
Nov 25 at 22:08
@TonyK thanks..
– Amerov
Nov 25 at 22:25
add a comment |
Your question is very difficult to read! You really have to learn MathJax. See this link for a good tutorial.
– TonyK
Nov 25 at 22:08
You have a description there of the result of applying $f$ to an arbitrary column vector $(x_1,x_2)^T$. Can you write that result as a matrix times $(x_1,x_2)^T$ in any way?
– Arthur
Nov 25 at 22:08
@TonyK thanks..
– Amerov
Nov 25 at 22:25
Your question is very difficult to read! You really have to learn MathJax. See this link for a good tutorial.
– TonyK
Nov 25 at 22:08
Your question is very difficult to read! You really have to learn MathJax. See this link for a good tutorial.
– TonyK
Nov 25 at 22:08
You have a description there of the result of applying $f$ to an arbitrary column vector $(x_1,x_2)^T$. Can you write that result as a matrix times $(x_1,x_2)^T$ in any way?
– Arthur
Nov 25 at 22:08
You have a description there of the result of applying $f$ to an arbitrary column vector $(x_1,x_2)^T$. Can you write that result as a matrix times $(x_1,x_2)^T$ in any way?
– Arthur
Nov 25 at 22:08
@TonyK thanks..
– Amerov
Nov 25 at 22:25
@TonyK thanks..
– Amerov
Nov 25 at 22:25
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
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accepted
Hint: The columns of the matrix are $f(1,0)$ and $f(0,1)$.
For the second part, look at what happens to $(1,0)$ and $(0,1)$ and apply a little trigonometry.
answered Nov 25 at 22:13
Chris Custer
9,9093724
How did you know that the columns of the m. are f(1,0)and f(0,1)?
– Amerov
Nov 25 at 22:24
This is always true.
– Chris Custer
Nov 25 at 22:27
can you give me a simple example to know how the result should look like ?
– Amerov
Nov 25 at 22:33
For the first column, $f(1,0)=(costheta,sintheta)$. So we have $begin{pmatrix}costheta &*\sintheta &*end{pmatrix}$, so far. You try the second column.
– Chris Custer
Nov 25 at 22:39
it should be (x1,x2) if it's wrong please don't blame me XD XD XD
– Amerov
Nov 25 at 22:46
|
show 23 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Hint: The columns of the matrix are $f(1,0)$ and $f(0,1)$.
For the second part, look at what happens to $(1,0)$ and $(0,1)$ and apply a little trigonometry.
answered Nov 25 at 22:13
Chris Custer
9,9093724
How did you know that the columns of the m. are f(1,0)and f(0,1)?
– Amerov
Nov 25 at 22:24
This is always true.
– Chris Custer
Nov 25 at 22:27
can you give me a simple example to know how the result should look like ?
– Amerov
Nov 25 at 22:33
For the first column, $f(1,0)=(costheta,sintheta)$. So we have $begin{pmatrix}costheta &*\sintheta &*end{pmatrix}$, so far. You try the second column.
– Chris Custer
Nov 25 at 22:39
it should be (x1,x2) if it's wrong please don't blame me XD XD XD
– Amerov
Nov 25 at 22:46
|
show 23 more comments
up vote
0
down vote
accepted
Hint: The columns of the matrix are $f(1,0)$ and $f(0,1)$.
For the second part, look at what happens to $(1,0)$ and $(0,1)$ and apply a little trigonometry.
answered Nov 25 at 22:13
Chris Custer
9,9093724
How did you know that the columns of the m. are f(1,0)and f(0,1)?
– Amerov
Nov 25 at 22:24
This is always true.
– Chris Custer
Nov 25 at 22:27
can you give me a simple example to know how the result should look like ?
– Amerov
Nov 25 at 22:33
For the first column, $f(1,0)=(costheta,sintheta)$. So we have $begin{pmatrix}costheta &*\sintheta &*end{pmatrix}$, so far. You try the second column.
– Chris Custer
Nov 25 at 22:39
it should be (x1,x2) if it's wrong please don't blame me XD XD XD
– Amerov
Nov 25 at 22:46
|
show 23 more comments
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Hint: The columns of the matrix are $f(1,0)$ and $f(0,1)$.
For the second part, look at what happens to $(1,0)$ and $(0,1)$ and apply a little trigonometry.
answered Nov 25 at 22:13
Chris Custer
9,9093724
Hint: The columns of the matrix are $f(1,0)$ and $f(0,1)$.
For the second part, look at what happens to $(1,0)$ and $(0,1)$ and apply a little trigonometry.
answered Nov 25 at 22:13
Chris Custer
9,9093724
answered Nov 25 at 22:13
Chris Custer
9,9093724
answered Nov 25 at 22:13
Chris Custer
9,9093724
answered Nov 25 at 22:13
Chris Custer
9,9093724
9,9093724
How did you know that the columns of the m. are f(1,0)and f(0,1)?
– Amerov
Nov 25 at 22:24
This is always true.
– Chris Custer
Nov 25 at 22:27
can you give me a simple example to know how the result should look like ?
– Amerov
Nov 25 at 22:33
For the first column, $f(1,0)=(costheta,sintheta)$. So we have $begin{pmatrix}costheta &*\sintheta &*end{pmatrix}$, so far. You try the second column.
– Chris Custer
Nov 25 at 22:39
it should be (x1,x2) if it's wrong please don't blame me XD XD XD
– Amerov
Nov 25 at 22:46
|
show 23 more comments
How did you know that the columns of the m. are f(1,0)and f(0,1)?
– Amerov
Nov 25 at 22:24
This is always true.
– Chris Custer
Nov 25 at 22:27
can you give me a simple example to know how the result should look like ?
– Amerov
Nov 25 at 22:33
For the first column, $f(1,0)=(costheta,sintheta)$. So we have $begin{pmatrix}costheta &*\sintheta &*end{pmatrix}$, so far. You try the second column.
– Chris Custer
Nov 25 at 22:39
it should be (x1,x2) if it's wrong please don't blame me XD XD XD
– Amerov
Nov 25 at 22:46
How did you know that the columns of the m. are f(1,0)and f(0,1)?
– Amerov
Nov 25 at 22:24
How did you know that the columns of the m. are f(1,0)and f(0,1)?
– Amerov
Nov 25 at 22:24
This is always true.
– Chris Custer
Nov 25 at 22:27
This is always true.
– Chris Custer
Nov 25 at 22:27
can you give me a simple example to know how the result should look like ?
– Amerov
Nov 25 at 22:33
can you give me a simple example to know how the result should look like ?
– Amerov
Nov 25 at 22:33
For the first column, $f(1,0)=(costheta,sintheta)$. So we have $begin{pmatrix}costheta &*\sintheta &*end{pmatrix}$, so far. You try the second column.
– Chris Custer
Nov 25 at 22:39
For the first column, $f(1,0)=(costheta,sintheta)$. So we have $begin{pmatrix}costheta &*\sintheta &*end{pmatrix}$, so far. You try the second column.
– Chris Custer
Nov 25 at 22:39
it should be (x1,x2) if it's wrong please don't blame me XD XD XD
– Amerov
Nov 25 at 22:46
it should be (x1,x2) if it's wrong please don't blame me XD XD XD
– Amerov
Nov 25 at 22:46
|
show 23 more comments
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Your question is very difficult to read! You really have to learn MathJax. See this link for a good tutorial.
– TonyK
Nov 25 at 22:08
You have a description there of the result of applying $f$ to an arbitrary column vector $(x_1,x_2)^T$. Can you write that result as a matrix times $(x_1,x_2)^T$ in any way?
– Arthur
Nov 25 at 22:08
@TonyK thanks..
– Amerov
Nov 25 at 22:25