Show that $(\left (\begin{array}{cc} x_1\\ x_2 \end{array}\right),\left(\begin{array}{cc} y_1\\ y_2\end{array}\right)) = x_1\overline y_1+2x_2\overline y_2$ is a scalar product in $C^2$

I don't really know how to start:

Let $x=(x_1 ,x_2 $) and $y=(y_1 ,y_2)$ then $\langle x,y\rangle$ = $x_1 ∗ \overline y_1 +x_2 ∗ \overline y_2 $ . How do I get $2x_2 ∗ \overline y_2$? what properties of inner product do I have to apply?

Hints are appreciated!


Observe that

$$x_1\overline{y_1}+2x_2\overline{y_2}=(x_1\;x_2)\begin{pmatrix}1&0\\0&2\end{pmatrix}\overline{\binom{y_1}{y_2}}= {x^t}A\,\overline y$$

and thus the above defines a (complex) inner product (=a Hermitian inner product) since $\;A\;$ is Hermitian (in fact it is simply (real) symmetric, which obviously is enough)

  • $\begingroup$ Thanks! So I don't need to show the property of inner product, right? $\endgroup$ – needhelp Oct 27 '14 at 14:29
  • $\begingroup$ As long as you know the above, no: you don't need that. $\endgroup$ – Timbuc Oct 27 '14 at 14:40
  • $\begingroup$ Thanks a lot! I was struggling to prove the positivity of the above. $\endgroup$ – needhelp Oct 27 '14 at 14:41
  • $\begingroup$ How come? That's perhaps the easiest: $$x^tA\overline x= |x_1|^2+2|y_1|^2\ge 0\;$$ $\endgroup$ – Timbuc Oct 27 '14 at 14:48

What you have is only a scalar product on $\mathbb R^2$, it is not a scalar product on $\mathbb C^2$.

You have to prove that your mapping satisfies the conditions of scalar products.

This means the mapping $$(\left (\begin{array}{cc} x_1\\ x_2 \end{array}\right),\left(\begin{array}{cc} y_1\\ y_2\end{array}\right)) \mapsto x_1y_1+2x_2y_2$$

must be linear in both factors and positive definite. In order for the mapping to be a inner product on $\mathbb C^2$, it would also need to satisfy the complex conjugate property, and this particular mapping does not.

  • $\begingroup$ I just edited my question again. I forget to put $\overline y_1$ and $\overline y_2$ in the question. Do i have to show all the property of inner product such as homogeneity, additivity and symmetry? Thanks! $\endgroup$ – needhelp Oct 27 '14 at 11:55
  • $\begingroup$ @needhelp That is exactly what you must prove. Note that this is an inner product which is different from the standard inner product defined by $x_1\bar{y_1} + x_2\bar{y_2}$. It is still an inner product, though. $\endgroup$ – 5xum Oct 27 '14 at 11:58
  • $\begingroup$ Yeah, my problem is I kind of know what to prove but I don't know know how to start off. Thanks for your hints! $\endgroup$ – needhelp Oct 27 '14 at 12:01
  • 1
    $\begingroup$ @needhelp Just go through the axioms (homogenicity, additivity, symmetry...). None should be hard to prove. $\endgroup$ – 5xum Oct 27 '14 at 12:02

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