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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.
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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.

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