# Inner product over the $C^2$

Let a, b, c, d ∈ C and consider the vector space $C^2$

Suppose inner product is defined as:

$⟨x, y⟩ = ax_1\bar y_1 + bx_2\bar y_1 + cx_1\bar y_2 + dx_2\bar y_2$

I am trying to find all a, b, c, d such that the definition above is false.

I believe I checked all the points in definition of inner product over complex field and found only

a = b = c = d = 0

as it contradicts $⟨x, x⟩ = 0 \implies x=0$

Can you find anything else?

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What happens when $b=c=d=0$ and $a\neq 0$? –  Brad Apr 7 '14 at 17:17
in this case ⟨x,x⟩ = 0 iff x = 0 –  tmac_balla Apr 7 '14 at 18:08
Can you prove that? –  Brad Apr 7 '14 at 21:13

Let $$M=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
You have$$\langle x,y\rangle=xMy^*$$
You want $\langle x,x\rangle=0\implies x =0$. Suppose that $M$ isn't invertible. Take any $v\in\operatorname{Ker}(M)\setminus\{0\}$. $v\not=0$ and $v^*Mv=0$ so $M$ does not represent a scalar product. You therefore need $M$ to be invertible for it to represent a scalar product.
So taking any matrix $M$ with $\det M = 0$ would give you a counterexample.