Question

The standard ordered basis of ℝ² is {e₁, e₂}. Let T : ℝ² → ℝ2 be the linear transformation such that T reflects the points through the line x₁ = -x₂. The standard matrix of T is:
1. \(\left( {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right)\)
2. \(\left( {\begin{array}{*{20}{c}} 0&-1\\ -1&0 \end{array}} \right)\)
3. \(\left( {\begin{array}{*{20}{c}} -1&0\\ 0&-1 \end{array}} \right)\)
4. \(\left( {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right)\)

Answer 1

Correct Answer - Option 2 : \(\left( {\begin{array}{*{20}{c}} 0&-1\\ -1&0 \end{array}} \right)\)

Concept:

If x₁ and x₂ are the components of a vector X with respect to a standard basis, this means

X = [x₁, x₂]^T = x₁ e₁ + x₂ e₂ where {e1, e2} is standard ordered basis.

Reflection theorem:

Let T : R² → R² be a linear transformation given by reflecting vectors over the line x₂ = m x₁. Then the matrix of T is given by 

\(\frac {1}{1+{m^2}} \left[ {\begin{array}{*{20}{c}} {1-m^2}&2m\\ 2m&m^2-1 \end{array}} \right]\)

Calculation:

Given T is the linear transformation that reflects the points through the line x1 = -x2

⇒ m = -1

The line x1 = -x2 is the bisector of the second and fourth quadrant and the reflection through this line is represented by a matrix 

\(T = \left( {\begin{array}{*{20}{c}} 0&-1\\ -1&0 \end{array}} \right)\)