Let a be a nonzero vector in \( \mathbb{R}^{2} \). Define \( \tau_{\mathbf{a}}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2} \) by \[ \tau_{\mathbf{a}}(\mathbf{x})=-\mathbf{x}+2 \frac{\mathbf{a} \cdot \mathbf{x}}{\mathbf{a} \cdot \mathbf{a}} \mathbf{a} \text {. } \] 2. Let a be a nonzero vector of \( \mathbb{R}^{2} \). Answer the following questions. (ii) Write \( \mathbf{a}=\left(a_{1}, a_{2}\right) \) and let \( \mathbf{b}=\left(-a_{2}, a_{1}\right) \in \mathbb{R}^{2} \). Show that \( \mathcal{B}=\{\mathbf{a}, \mathbf{b}\} \) is a basis for \( \mathbb{R}^{2} \).

To show that \( \mathcal{B} = \{\mathbf{a}, \mathbf{b}\} \) is a basis for \( \mathbb{R}^2 \), we need to demonstrate that the vectors \(\mathbf{a}\) and \(\mathbf{b}\) are linearly independent. This means that the only solution to the equation \( c_1\mathbf{a} + c_2\mathbf{b} = \mathbf{0} \) (where \( c_1 \) and \( c_2 \) are scalars) must be \( c_1 = 0 \) and \( c_2 = 0 \). We can express this equation as follows: \[ c_1(a_1, a_2) + c_2(-a_2, a_1) = (0, 0). \] This expands to the system of equations: 1. \( c_1 a_1 - c_2 a_2 = 0 \) 2. \( c_1 a_2 + c_2 a_1 = 0 \) We will analyze these equations. If we assume \( c_1 \) and \( c_2 \) are not both zero, we can derive relationships between them. From the first equation, we can solve for \( c_2 \): \[ c_2 = \frac{c_1 a_1}{a_2} \quad (\text{provided } a_2 \neq 0). \] Substitute \( c_2 \) into the second equation: \[ c_1 a_2 + \frac{c_1 a_1}{a_2} a_1 = 0 \implies c_1 \left(a_2 + \frac{a_1^2}{a_2}\right) = 0. \] Since \( \mathbf{a} \) is nonzero, \( a_1 \) and \( a_2 \) cannot both be zero, thus \( a_2 + \frac{a_1^2}{a_2} \neq 0 \). Therefore, it must be that \( c_1 = 0 \), and from the first equation, \( c_2 = 0 \) follows. This confirms that \( \mathbf{a} \) and \( \mathbf{b} \) are linearly independent. As a result, since we have two linearly independent vectors in \( \mathbb{R}^2 \), \( \mathcal{B} = \{\mathbf{a}, \mathbf{b}\} \) is indeed a basis for \( \mathbb{R}^2 \).