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If $\vec u = (u_1, u_2, u_3)$ and $\vec v = (v_1, v_2, v_3)$ is a vector of $\mathbb{R}^3$, then $f(u, v) =$ $2u_1v_1 + 3u_2v_2 – 2u_2v_2$ is one of the inner product in $\mathbb{R}^3$.

My answer is False, because if we simplify the $f(\vec u,\vec v)$, we can find that

$f(\vec u,\vec v) = 2u_1v_1 + u_2v_2$, as we can see we can't found the $u_3v_3$ operation in the function, so my opinion is since there is no function concerned about the $u_3$ and $v_3$, then the function is not an inner product in $\mathbb{R}^3$,

Please correct me if I'm wrong, it would be very helpful if anyone can give the right answer

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  • $\begingroup$ To clean up your argument. Consider $v = (0,0,1), f(v,v) = 0$ and the inner product of a non-zero vector with itself must be greater than $0.$ $\endgroup$ – Doug M Nov 21 '17 at 18:10
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To be more clear, note that $$f((0,0,1),(0,0,1))=0$$

but $(0,0,1) \neq (0,0,0)$ violating the positive definiteness.

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