Show that the result given in Exercise 12 (that for bilinear forms \(f, g\) on a finite-dimensional vector space, there exist linear operators \(T_1, T_2\) such that \(f(\alpha, \beta) = g(T_1\alpha, \beta) = g(\alpha, T_2\beta)\)) need not be true if \(g\) is singular.
Solution
Goal: Provide an explicit counterexample of bilinear forms \(f\) and \(g\) on a finite-dimensional vector space \(V\) with \(g\) singular such that no linear operators \(T_1, T_2\) satisfy \(f(\alpha, \beta) = g(T_1\alpha, \beta) = g(\alpha, T_2\beta)\).
1. Definition of the space and bilinear forms:Proof:
1.1. Let \(V = \mathbb{R}^2\) with standard basis \(\{e_1, e_2\}\).
1.2. Define the singular bilinear form \(g: \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}\) with matrix \([g] = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\):\[g(\alpha, \beta) = x_1 y_1 \quad \text{for } \alpha = (x_1, x_2), \, \beta = (y_1, y_2).\]
1.3. Since \(\det[g] = 0\) (and \(g(e_2, \beta) = 0\) for all \(\beta\)), \(g\) is singular (degenerate).
1.4. Define the bilinear form \(f: \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}\) with matrix \([f] = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\):\[f(\alpha, \beta) = x_2 y_2 \quad \text{for } \alpha = (x_1, x_2), \, \beta = (y_1, y_2).\]
2. Non-existence of the operator \(T_1\):Proof:
2.1. Suppose for contradiction that there exists a linear operator \(T_1: \mathbb{R}^2 \to \mathbb{R}^2\) such that \(f(\alpha, \beta) = g(T_1\alpha, \beta)\) for all \(\alpha, \beta \in \mathbb{R}^2\).
2.2. For any vector \(\alpha \in \mathbb{R}^2\), let \(T_1\alpha = (u_1, u_2) \in \mathbb{R}^2\).
2.3. Then for \(\beta = e_2 = (0, 1)\), the value of \(g(T_1\alpha, e_2)\) is\[g(T_1\alpha, e_2) = u_1 \cdot 0 = 0.\]
2.4. Evaluating \(f\) on \(\alpha = e_2 = (0, 1)\) and \(\beta = e_2 = (0, 1)\):\[f(e_2, e_2) = 0 \cdot 0 + 1 \cdot 1 = 1.\]
2.5. Thus \(1 = f(e_2, e_2) = g(T_1 e_2, e_2) = 0\), which is a contradiction.
2.6. Therefore no such linear operator \(T_1\) exists.
3. Non-existence of the operator \(T_2\):Proof:
3.1. Similarly, suppose for contradiction that there exists a linear operator \(T_2\) such that \(f(\alpha, \beta) = g(\alpha, T_2\beta)\) for all \(\alpha, \beta\).
3.2. For any \(\beta\), let \(T_2\beta = (v_1, v_2)\).
3.3. For \(\alpha = e_2 = (0, 1)\), \(g(e_2, T_2\beta) = 0 \cdot v_1 = 0\).
3.4. But \(f(e_2, e_2) = 1 \neq 0\), giving \(1 = 0\), a contradiction.
3.5. Thus no such linear operator \(T_2\) exists.
4. Conclusion:Proof:The factorization \(f(\alpha, \beta) = g(T_1\alpha, \beta) = g(\alpha, T_2\beta)\) fails when \(g\) is singular.