How to determine if this function is one of the inner product

Question

For p = (p1, p2, p3) and v = (q1, q2, q3) at R3, prove that function h(p, q) = p1q2 + 3p2q3 + 4p3q1 is one of the inner product at R3.

for this question, i have found that this function pass the 2nd axiom <(p,q),w> = <(p,w),(q,w)>

and 3rd axiom < kp,q> = k< p , q>

But i found this function doesn't pass the 1st axiom < p ,q > = < q , p >

since < p , q> = p1q2 + 3p2q3 + 4p3q1 and < q, p> = q1p2 + 3q2p3 + 4q3p1

and 4th axiom < p , p > larger than or equal to 0 with < p, p> = 0 if and only if p = 0 for example if we have p = (x,0,0) with x is any real number, we can get 0.

So, by those explanation am i right or wrong? Please do explain since i'm not really good at linear algebra, Thanks!

Answer 1

You can represent this map by the matrix $$\begin{pmatrix}0 & 1/2 & 2 \\ 1/2 & 0 & 3/2 \\ 2 & 3/2 & 0\end{pmatrix}$$The second principal minor is $0\cdot 0 - (1/2)(1/2) = -1/4 < 0$, so by Sylvester's criterion it is not positive-definite. We don't have an inner product.