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Proof or counterexample. Here \(v, w, z\) are vectors in a real inner product space \(H\).
a) Let \(v, w, z\) be vectors in a real inner product space. If \(\langle v, w\rangle=0\) and \(\langle v, z\rangle=0\), then \(\langle w, z\rangle=0\).
b) If \(\langle v, z\rangle=\langle w, z\rangle\) for all \(z \in H\), then \(v=w .\)
c) If \(A\) is an \(n \times n\) symmetric matrix then \(A\) is invertible.
in Mathematics by Platinum (97.7k points) | 389 views

1 Answer

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a. Counterexample: Let \(H=\mathbb{R}^2\), \(v=(1, 0)\), \(w=(0, 1)\) and \(z=(1, 1)\). Then \(\langle v, w\rangle=0\) and \(\langle v, z\rangle=0\), but \(\langle w, z\rangle=1 \neq 0\).

b. Counterexample: Let \(H=\mathbb{R}^2\), \(v=(1, 0)\) and \(w=(1, 1)\). Then \(\langle v, z\rangle=\langle w, z\rangle\) for all \(z \in H\), but \(v \neq w\).

c. True: Since \(A\) is symmetric, it is a self-adjoint operator and thus has an orthonormal basis of eigenvectors. Since the eigenvalues are non-zero, the matrix is invertible.

by Bronze Status (8.8k points)

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