Suppose T is a transformation from ℝ2 to ℝ2. Find the matrix A that induces T if T is reflection over the line y=−3/2x.
Please give me the solution
To find the matrix A that induces the reflection over the line y = -3/2x, we can start with the standard basis vectors in ℝ², which are [1, 0] and [0, 1].
Step 1: Reflection of [1, 0]
To reflect [1, 0] over the line y = -3/2x, we can consider the line passing through the origin and perpendicular to y = -3/2x, which has a slope of -2/3. Let's call this line L₁.
The reflection of [1, 0] over L₁ will result in a vector that is perpendicular to L₁, passing through the origin.
Since L₁ has a slope of -2/3, we can find the slope of the vector perpendicular to L₁ by taking the negative reciprocal: 3/2.
So, the reflection of [1, 0] over y = -3/2x is a vector that has a slope of 3/2 and passes through the origin.
Let's call this vector v₁.
To find v₁, we can find the intersection point of L₁ with the x-axis, which is obtained when y = 0. Solving for x, we get:
-3/2x = 0 -> x = 0
Hence, the point of intersection is (0, 0), which is the origin.
Therefore, v₁ = [0, 0].
Step 2: Reflection of [0, 1]
To reflect [0, 1] over the line y = -3/2x, we need to find the line perpendicular to y = -3/2x, which passes through the origin.
Again, this line will have a slope of -2/3. Let's call this line L₂.
The reflection of [0, 1] over L₂ will result in a vector that is perpendicular to L₂, passing through the origin.
Therefore, the reflection of [0, 1] over y = -3/2x is a vector that has a slope of 3/2 and passes through the origin.
Let's call this vector v₂.
To find v₂, we can find the intersection point of L₂ with the y-axis, which is obtained when x = 0. Solving for y, we get:
y = -3/2 * 0 -> y = 0
Hence, the point of intersection is (0, 0), which is the origin.
Therefore, v₂ = [0, 0].
Step 3: Constructing the matrix A
Now that we have v₁ = [0, 0] and v₂ = [0, 0], we can construct the matrix A using these column vectors.
A = [v₁, v₂] = [0, 0; 0, 0]
So, the matrix A that induces the transformation T, which is the reflection over the line y = -3/2x, is:
A = [0, 0; 0, 0].
To find the matrix A that induces the transformation T, we need to understand how T behaves on the standard basis vectors of ℝ2.
Let's start by considering the standard basis vectors u1 = (1, 0) and u2 = (0, 1) in ℝ2.
1. Finding the image of u1:
When u1 is reflected over the line y = -3/2x, it ends up on the opposite side of the line with the same perpendicular distance from the line. In this case, the line y = -3/2x acts as the mirror.
The point on the line y = -3/2x that is closest to the origin is the origin itself (0, 0). The distance from the origin to the line is the shortest distance between them, which can be found using the formula: distance = |y - (-3/2x)| / sqrt(1 + (-3/2)^2) = |-3/2x| / sqrt(1 + 9/4) = |-3/2x| / sqrt(13/4) = (3/2) |x| / (sqrt(13)/2) = (3/sqrt(13)) |x|.
So when we reflect the point u1 = (1, 0) over the line y = -3/2x, it ends up at (-1, 0), which has the same x-coordinate but a negative y-coordinate.
2. Finding the image of u2:
Applying the same reflection over the line y = -3/2x to u2 = (0, 1) gives us (0, -1).
Now, we have the images of the basis vectors u1 and u2: (-1, 0) and (0, -1) respectively.
The matrix A that induces the transformation T can be constructed using these image points as columns.
A = [(-1, 0), (0, -1)] = [ -1 0 ]
[ 0 -1 ]
Therefore, the matrix A that induces the transformation T of reflection over the line y = -3/2x is:
A = [ -1 0 ]
[ 0 -1 ]
google provides many examples.
A better transformation would be R2:D2 :-)