1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.


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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.