Finding values for which a bilinear form is an inner product

Question

I am trying to find the values (if any) of p and q for which the following satisfies the definition of an inner product:

$$ \left \langle \mathbf{z}, \mathbf{w} \right \rangle = z_1\overline{w_1} + {\color{Green}{p}}z_1\overline{w_2} + {\color{Red}{q}}z_2\overline{w_1} + 9z_2\overline{w_2}$$

To start with I want $\left \langle \mathbf{z}, \mathbf{w} \right \rangle = \overline{\left \langle \mathbf{w}, \mathbf{z} \right \rangle}$ to be satisfied. Would this mean (using ${\color{Green}{p}}$) that ${\color{Red}{q} }=\overline{{\color{Green}{p}}}$?

And therefore $\left \langle \mathbf{z}, \mathbf{w} \right \rangle = z_1\overline{w_1} + {\color{Green}{p}}z_1\overline{w_2} + \overline{{\color{Green}{p}}}z_2\overline{w_1} + 9z_2\overline{w_2}$ ?

Assuming I am correct I want $\left \langle \mathbf{z}, \mathbf{z} \right \rangle \geq 0$

Hence: $$\left \langle \mathbf{z}, \mathbf{z} \right \rangle = z_1\overline{z_1} + {\color{Green}{p}}z_1\overline{z_2} + \overline{{\color{Green}{p}}}z_2\overline{z_1} + 9z_2\overline{z_2}$$

Therefore: $$\left \langle \mathbf{z}, \mathbf{z} \right \rangle = |z_1 + {\color{Green}{p}}z_2|^2 + 9|z_2|^2 - |{\color{Green}{p}}z_2|^2$$

Therefore if $|{\color{Green}{p}}|^2 \leqslant 9$ then I have a valid hermitian inner product on $\mathbb{C^{2}}$, correct?

Answer 1

That's correct. Well, you should have $|p|^2 <9$ (strictly), or we have $\langle z,z \rangle = 0$ for a non-zero $z$.

Another way to see this is as follows: $$ \langle z,w \rangle = w^* \underbrace{\pmatrix{1&p\\q&9}}_A z $$ This bilinear form will be an inner product if and only if the matrix $A$ is positive definite. This occurs if and only if the matrix is Hermitian and all principal minors are positive. That is, it is sufficient to check that $p = \overline q$, that the upper-left entry is positive (it is, since $1>0$) and that $\det(A) > 0$, i.e. $$ 1\cdot 9 - pq = 9 - |p|^2 > 0 \implies |p|^2 < 9 $$ Whenever all these conditions are held, $\langle z,w \rangle$ will indeed define an inner product.

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For an example of things going wrong when $|p| = 3$: check $\langle z,z \rangle$ when $z = (p,-1)^T$.