Suppose T is a transformation from ℝ2 to ℝ2. Find the matrix A that induces T if T is:
a) rotation by 5/4π
b) reflection over the line y=−4/5x
a) Ah, rotation! Bringing a whole new meaning to the term "twist and shout." To find the matrix A for a rotation by 5/4π, we need to channel our inner dancers. The good news is that rotations are pretty straightforward.
First, let's break down the rotation angle, 5/4π. If we divide it by π, we get 5/4. Now, let's convert it to degrees to make it more relatable. Multiplying 5/4 by 180 gives us 225 degrees. A 225-degree rotation is like facing southwest instead of north.
To get the matrix A for our transformation, we can use the following formula:
A = |cosθ -sinθ|
|sinθ cosθ|
Plugging in our rotation angle of 225 degrees, we get:
A = |cos(225) -sin(225)|
|sin(225) cos(225)|
Now, let me whip out my calculator. Ah, here we go:
A = |-√2/2 √2/2|
|-√2/2 -√2/2|
Ta-da! That's the matrix A that induces a rotation by 5/4π in ℝ2.
b) Reflection! It's like looking into a mirror, but without any "bad hair days" regrets. If we want to reflect over the line y = -4/5x, we need the perfect mirror matrix A to make it happen.
To create the matrix A for a reflection, we can follow a simple formula:
A = |cos(2θ) sin(2θ)|
|sin(2θ) -cos(2θ)|
But wait! We don't know θ yet. No worries, my friend! We can determine it by examining the line of reflection, which is y = -4/5x. If we express it in terms of θ, that is arctan(-4/5), we can find θ.
Let me reach for my trusty calculator once again... *rustling noises* ... Ah-ha!
θ ≈ -38.66 degrees (approximately)
Now, we can plug in θ into the reflection matrix formula:
A = |cos(-77.32) sin(-77.32)|
|sin(-77.32) -cos(-77.32)|
Pause for effect... drum rolls, please!
A = | 0.287 -0.957 |
|-0.957 -0.287 |
And there you have it! The matrix A that induces a reflection over the line y = -4/5x in ℝ2. Now, go and reflect on that!
To find the matrix A that induces the given transformations, we can consider the effect of the transformation on the standard basis vectors of ℝ^2.
a) Rotation by 5/4π:
Let's start with the standard basis vectors, i and j.
To find the effect of the transformation T on the vector i = (1, 0), we rotate it counterclockwise by an angle of 5/4π. So, we get:
T(i) = (cos(5/4π), sin(5/4π))
Using the trigonometric identities cos(5/4π) = -sqrt(2)/2 and sin(5/4π) = -sqrt(2)/2, we have:
T(i) = (-sqrt(2)/2, -sqrt(2)/2)
Similarly, for the vector j = (0, 1), we have:
T(j) = (cos(5/4π + π/2), sin(5/4π + π/2))
= (-sin(5/4π), cos(5/4π))
Using the trigonometric identities sin(5/4π) = -sqrt(2)/2 and cos(5/4π) = sqrt(2)/2, we have:
T(j) = (-sqrt(2)/2, sqrt(2)/2)
The matrix A that induces the rotation by 5/4π can be obtained by writing the transformed basis vectors T(i) and T(j) as columns:
A = [(-sqrt(2)/2, -sqrt(2)/2), (-sqrt(2)/2, sqrt(2)/2)]
b) Reflection over the line y = -4/5x:
For a reflection, the transformation T is linear, which means the matrix A will be a 2x2 matrix.
Let's start with the standard basis vectors, i and j.
To find the effect of the transformation T on the vector i = (1, 0), we reflect it over the line y = -4/5x. The resulting vector will have the same x-coordinate but a negated y-coordinate. So, we have:
T(i) = (1, 0)
For the vector j = (0, 1), we reflect it over the line y = -4/5x. The resulting vector will have a negated x-coordinate but the same y-coordinate. So, we have:
T(j) = (-1, 1)
The matrix A that induces the reflection can be obtained by writing the transformed basis vectors T(i) and T(j) as columns:
A = [(1, -1), (0, 1)]
a) To find the matrix A that induces a rotation by 5/4π, we can use the following steps:
Step 1: Let's consider a vector v in ℝ2. We want to find the new coordinates of v after it has been rotated by 5/4π.
Step 2: The rotation can be thought of as a linear transformation from ℝ2 to ℝ2, which can be represented by a matrix. We need to find the matrix A that represents this rotation.
Step 3: The key observation is that the coordinates of the rotated vector can be obtained by multiplying the original vector by the matrix A.
Step 4: Let's consider a vector v = [x, y] in ℝ2. After rotating the vector by 5/4π, let's call the new vector w = [a, b]. We need to find the values of a and b.
Step 5: To find the matrix A, we can start by considering the unit vectors i = [1, 0] and j = [0, 1], which represent the x and y axes, respectively.
Step 6: After rotating i by 5/4π, the resulting vector should point in the direction of a. Similarly, after rotating j by 5/4π, the resulting vector should point in the direction of b.
Step 7: We can find the values of a and b by applying the rotation formula:
a = cos(5/4π) * 1 - sin(5/4π) * 0
b = cos(5/4π) * 0 + sin(5/4π) * 1
Step 8: Simplifying these equations gives us:
a = cos(5/4π) = cos(π + 1/4π) = -cos(1/4π) = -1/sqrt(2)
b = sin(5/4π) = sin(π + 1/4π) = -sin(1/4π) = -1/sqrt(2)
Step 9: Therefore, the matrix A that induces the rotation by 5/4π is:
A = [a, b] = [-1/sqrt(2), -1/sqrt(2)]
b) To find the matrix A that induces a reflection over the line y = -4/5x, we can follow these steps:
Step 1: Let's consider a vector v in ℝ2. We want to find the new coordinates of v after it has been reflected over the line y = -4/5x.
Step 2: The reflection can be thought of as a linear transformation from ℝ2 to ℝ2, which can be represented by a matrix. We need to find the matrix A that represents this reflection.
Step 3: The key observation is that the coordinates of the reflected vector can be obtained by multiplying the original vector by the matrix A.
Step 4: Let's consider a vector v = [x, y] in ℝ2. After reflecting the vector over the line y = -4/5x, let's call the new vector w = [a, b]. We need to find the values of a and b.
Step 5: To find the matrix A, we can start by considering the unit vectors i = [1, 0] and j = [0, 1], which represent the x and y axes, respectively.
Step 6: After reflecting i over the line y = -4/5x, the resulting vector should point in the direction of a. Similarly, after reflecting j over the line y = -4/5x, the resulting vector should point in the direction of b.
Step 7: We can find the values of a and b by considering how the unit vectors i and j are reflected:
a = reflection of i over y = -4/5x
b = reflection of j over y = -4/5x
Step 8: To reflect a vector over the line y = -4/5x, we can use the formula:
a = x - 2 * (y - (-4/5)x)
b = (-4/5)x - 2 * (y - (-4/5)x)
Step 9: Simplifying these equations gives us:
a = x + 8/5y
b = -4/5x + 3y
Step 10: Therefore, the matrix A that induces the reflection over the line y = -4/5x is:
A = [a, b] = [[1, 8/5]
[-4/5, 3]]