Let \(( \mid )\) be the standard inner product on \(C^2\). Prove that there is no non-zero linear operator on \(C^2\) such that \((\alpha \mid T\alpha) = 0\) for every \(\alpha\) in \(C^2\). Generalize.
Solution
1. The Complex Polarization Identity:
1.1. Let \(V\) be an inner product space over the complex field \(\mathbb{C}\) with inner product \((\cdot \mid \cdot)\), and let \(T: V \to V\) be a linear operator.
setup.
1.2. For any \(\alpha, \beta \in V\) and each \(k \in \{0, 1, 2, 3\}\), expand the quadratic form at \(\alpha + i^k \beta\):\[ \big(\alpha + i^k \beta \mid T(\alpha + i^k \beta)\big) = (\alpha \mid T\alpha) + i^k (\alpha \mid T\beta) + \overline{i^k} (\beta \mid T\alpha) + (\beta \mid T\beta). \]
conjugate-linearity in the first argument and linearity in the second argument.
1.3. Multiply by \(i^k\) and sum over \(k \in \{0, 1, 2, 3\}\):- \(\sum_{k=0}^3 i^k = 1 + i - 1 - i = 0\),- \(\sum_{k=0}^3 i^{2k} = 1 - 1 + 1 - 1 = 0\),- \(\sum_{k=0}^3 i^k \overline{i^k} = \sum_{k=0}^3 |i^k|^2 = 4\).Thus:\[ \sum_{k=0}^3 i^k \big(\alpha + i^k \beta \mid T(\alpha + i^k \beta)\big) = 4 (\alpha \mid T\beta). \]
sum of powers of \(i\).
2. Proof that \((v \mid Tv) = 0\) for all \(v\) forces \(T = 0\):
2.1. By hypothesis, \((v \mid Tv) = 0\) for all \(v \in V\).Applying this to the four vectors \(v_k = \alpha + i^k \beta\) gives:\[ (\alpha \mid T\beta) = \frac{1}{4} \sum_{k=0}^3 i^k (v_k \mid T v_k) = \frac{1}{4} \sum_{k=0}^3 i^k \cdot 0 = 0 \quad \text{for all } \alpha, \beta \in V. \]
polarization identity from 1.
2.2. For each \(\beta \in V\), choose \(\alpha = T\beta\):\[ \|T\beta\|^2 = (T\beta \mid T\beta) = 0 \implies T\beta = 0. \]
positive definiteness of inner products.
2.3. Since \(T\beta = 0\) for every \(\beta \in V\), \(T\) is the zero operator (\(T = 0\)).
definition of the zero operator.
3. Generalization and Comparison with Real Spaces:
3.1. Generalization: The result holds for any complex inner product space \(V\) (of arbitrary dimension):If \((v \mid Tv) = 0\) for all \(v \in V\), then \(T = 0\).
polarization holds over any complex inner product space.
3.2. Real vs Complex Distinction: Over a real inner product space, this property fails:For example, on \(\mathbb{R}^2\) with standard inner product, the \(90^\circ\) rotation \(T(x, y) = (-y, x)\) satisfies \((v \mid Tv) = -xy + yx = 0\) for all \(v \in \mathbb{R}^2\), yet \(T \neq 0\).
skew-symmetric operators over \(\mathbb{R}\).
4. Conclusion:There is no non-zero linear operator on \(\mathbb{C}^2\) (or any complex inner product space) such that \((\alpha \mid T\alpha) = 0\) for all \(\alpha\). Q.E.D.
1 through 3.