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The given problem defines a transformation \( T : \mathbb{R}^2 \to \mathbb{R}^2 \) that involves two steps:

1. **Rotation by \( -3\pi/4 \) radians**: This rotation is clockwise (because of the negative angle) by \( 135^\circ \).
2. **Reflection through the horizontal \( x_1 \)-axis**: This reflection flips points over the \( x_1 \)-axis, changing the sign of their second coordinate (the \( x_2 \)-coordinate).

The hint provided gives the transformation of the standard basis vector \( e_1 = (1, 0) \):
\[
T(e_1) = \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)
\]

### Step 1: Rotation matrix
The matrix for rotation by \( \theta = -3\pi/4 \) radians is:
\[
R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}
\]
For \( \theta = -3\pi/4 \):
\[
R\left(-\frac{3\pi}{4}\right) = \begin{pmatrix} \cos\left(-\frac{3\pi}{4}\right) & -\sin\left(-\frac{3\pi}{4}\right) \\ \sin\left(-\frac{3\pi}{4}\right) & \cos\left(-\frac{3\pi}{4}\right) \end{pmatrix}
= \begin{pmatrix} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}
\]

### Step 2: Reflection matrix
Reflection through the horizontal \( x_1 \)-axis changes the sign of the \( x_2 \)-coordinate, so the reflection matrix is:
\[
M = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
\]

### Step 3: Combined transformation
The combined transformation \( T \) is the result of applying the rotation followed by the reflection. Thus, the transformation matrix is:
\[
T = M \cdot R\left(-\frac{3\pi}{4}\right) = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}
\]

This matrix represents the full transformation \( T \), combining both the rotation and reflection.

### Check with the hint:
The hint tells us \( T(e_1) = \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \). Let's apply \( T \) to \( e_1 = (1, 0) \):
\[
T(e_1) = \begin{pmatrix} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}
\]
This confirms the hint is correct.

Would you like further details or have any questions? Here are 5 related questions to expand this topic:

1. How does the rotation matrix change if the angle is positive?
2. What is the effect of reflecting points across the \( x_2 \)-axis instead of the \( x_1 \)-axis?
3. How would the transformation matrix change if the angle was \( \pi/2 \) instead of \( -3\pi/4 \)?
4. Can you express the transformation in terms of complex numbers for an alternative method?
5. What would be the inverse of this transformation matrix?

**Tip**: Understanding transformation matrices and their compositions is crucial in linear algebra, as they can represent various geometric operations like rotation, reflection, and scaling.

T: ℝ² → ℝ² first rotates points through -3π/4 radians (clockwise) and then reflects points through the horizontal x₁-axis.

Understand the transformation in ℝ² that rotates points by -3π/4 radians clockwise, followed by a reflection through the x₁-axis. This example demonstrates the application of rotation and reflection matrices in linear algebra.

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