Let \(\alpha = (x_1, x_2)\) and \(\beta = (y_1, y_2)\) be vectors in \(\RR^2\) such that \[ x_1 y_1 + x_2 y_2 = 0, \quad x_1^2 + x_2^2 = y_1^2 + y_2^2 = 1. \] Prove that \(\mathfrak{B} = \{\alpha, \beta\}\) is a basis for \(\RR^2\). Find the coordinates of the vector \((a, b)\) in the ordered basis \(\mathfrak{B} = \{\alpha, \beta\}\). (The conditions on \(\alpha\) and \(\beta\) say, geometrically, that \(\alpha\) and \(\beta\) are perpendicular and each has length 1.)
Solution
1. Proof that \(\mathfrak{B} = \{\alpha, \beta\}\) is a basis for \(\mathbb{R}^2\):
1.1. Endow \(\mathbb{R}^2\) with the standard Euclidean inner product \(\langle u, v \rangle = u_1 v_1 + u_2 v_2\).The given conditions state that \(\mathfrak{B} = \{\alpha, \beta\}\) is an orthonormal set:\[ \langle \alpha, \alpha \rangle = 1, \qquad \langle \beta, \beta \rangle = 1, \qquad \langle \alpha, \beta \rangle = 0. \]
1.2. Suppose \(c_1 \alpha + c_2 \beta = (0, 0)\) for scalars \(c_1, c_2 \in \mathbb{R}\).Taking the inner product with \(\alpha\):\[ 0 = \langle (0, 0), \alpha \rangle = \langle c_1 \alpha + c_2 \beta, \alpha \rangle = c_1 \langle \alpha, \alpha \rangle + c_2 \langle \beta, \alpha \rangle = c_1(1) + c_2(0) = c_1. \]Similarly, taking the inner product with \(\beta\):\[ 0 = \langle (0, 0), \beta \rangle = \langle c_1 \alpha + c_2 \beta, \beta \rangle = c_1 \langle \alpha, \beta \rangle + c_2 \langle \beta, \beta \rangle = c_1(0) + c_2(1) = c_2. \]Thus \(c_1 = c_2 = 0\), so \(\mathfrak{B}\) is linearly independent.
1.3. Since \(\dim(\mathbb{R}^2) = 2\) and \(\mathfrak{B}\) consists of 2 linearly independent vectors, \(\mathfrak{B}\) is a basis for \(\mathbb{R}^2\).
2. Coordinates of \((a, b)\) with respect to \(\mathfrak{B}\):
2.1. Let \(v = (a, b) \in \mathbb{R}^2\), and write \(v = c_1 \alpha + c_2 \beta\).Taking the inner product of \(v\) with \(\alpha\):\[ c_1 = \langle v, \alpha \rangle = a x_1 + b x_2. \]
2.2. Taking the inner product of \(v\) with \(\beta\):\[ c_2 = \langle v, \beta \rangle = a y_1 + b y_2. \]
2.3. Thus \((a, b) = (a x_1 + b x_2)\alpha + (a y_1 + b y_2)\beta\), so the coordinate vector is:\[ [(a, b)]_\mathfrak{B} = \begin{pmatrix} a x_1 + b x_2 \\ a y_1 + b y_2 \end{pmatrix}. \]
3. Conclusion:\(\mathfrak{B}\) is an orthonormal basis, and the coordinates of \((a, b)\) are \((a x_1 + b x_2, a y_1 + b y_2)\). Q.E.D.