Show that $\langle x,x\rangle > 0$ if $x \neq0$ Let $x=(x_1,x_2)$ and $ y=(y_1,y_2)$ be vectors in the vector space $C^2$ over $C$ and define $\langle \cdot,\cdot\rangle : C^2 \times C^2 \rightarrow C$ by
$ \langle x,y\rangle =3x_1 \overline{y_1}+(1+i)x_1 \overline{y_2}+(1-i)x_2 \overline{y_1}+x_2 \overline{y_2} $
Show that $\langle x,x\rangle > 0$ if $x \neq0$.
So what I have so far is:
$\begin{aligned} \langle x,x\rangle &= 3x_1 \overline{x_1}+(1+i)x_1 \overline{x_2}+(1-i)x_2 \overline{x_1}+x_2 \overline{x_2} \\
&= 3|x_1|^2 + |x_2|^2 +x_1 \overline x_2 + ix_1\overline x_2 - ix_2 \overline x_1 +x_2 \overline x_1 \\
&=3|x_1|^2 + |x_2|^2 + (1+i)(x_1 \overline x_2 - i\overline x_1 x_2)
\end{aligned}$
Now $3|x_1|^2 + |x_2|^2 >0$ if $x \neq0$, but I don't know how to show that $(1+i)(x_1 \overline x_2 - i\overline x_1 x_2) >0$ as well.
Any guidance would be appreciated.
Thank you.
 A: You have
$$
\langle x,x\rangle = 3|x_1|^2 + |x_2|^2 + 2 \operatorname{Re}((1+i)x_1 
\overline{x_2}) \\
\ge 3|x_1|^2 + |x_2|^2 - 2 |(1+i)x_1 
\overline{x_2})| \\
 = 3|x_1|^2  - 2 \sqrt 2 |x_1| |x_2|+ |x_2|^2
$$
and that is zero only for $x_1=x_2 = 0$ because the quadratic form
$$
 (x, y) \mapsto 3 x^2  - 2 \sqrt 2 xy+ y^2
$$
is positive definite, and that is because the matrix
$$
 \begin{pmatrix} 3 & -\sqrt 2 \\ -\sqrt 2 & 1 \end{pmatrix}
$$
is positive definite, and that is because its entries on the main diagonal and its determinant are strictly positive.
A: Let $x_1=a+bi$,  $x_2=c+di$.  Then
$$(1+i)(a+bi)(c-di)+(1-i)(a-bi)(c+di)=$$
$$(1+i)((ac+bd)+(bc-ad)i)+(1-i)((ac+bd)+(ad-bc)i)=$$
$$((ac+bd)-(bc-ad))+((ac+bd)+(bc-ad))i+((ac+bd)-(ad-bc)+((ad-bc)-(ac+bd))i=$$
$$(ac+bd-bc+ad+ac+bd-ad+bc)+(ac+bd+bc-ad+ad-bc-ac-bd)i=$$
$$2ac+2bd$$
So it isn't always greater than 0 by itself,  you need to show that
$$3(a^2+b^2)+(c^2+d^2)>-(2ac+2bd)$$
as long as at least one of $a,b,c,d\neq 0$
Can you finish?
A: \begin{align*}
\langle x,x\rangle &= 3|x_1|^2 + |x_2|^2 + 2 \operatorname{Re}((1+i)x_1 
\overline{x_2}) \\
&\ge 3|x_1|^2 + |x_2|^2 - 2 |(1+i)x_1 
\overline{x_2})| \\
&= 3|x_1|^2  - 2 \sqrt 2 |x_1| |x_2|+ |x_2|^2
\end{align*}
So,
$$\langle x,x\rangle = 0 \implies 3|x_1|^2  - 2 \sqrt 2 |x_1| |x_2|+ |x_2|^2=0$$
Now, calculate the determinant of
$$3|x_1|^2  - 2 \sqrt 2 |x_1| |x_2|+ |x_2|^2$$
to see that this can be $0$ if and only if $x_1=x_2=0$.
A: $$
\begin{align}
\langle x,x\rangle
&=3x_1\bar{x}_1+(1+i)x_1\bar{x}_2+(1-i)x_2\bar{x}_1+x_2 \bar{x}_2\tag1\\[6pt]
&=3|x_1|^2+|x_2|^2+2\mathrm{Re}((1+i)x_1\bar{x}_2)\tag2\\[6pt]
&\ge3|x_1|^2+|x_2|^2-2\sqrt2|x_1||x_2|\tag3\\[3pt]
&=\frac52|x_1|^2+\frac45|x_2|^2-2\sqrt2|x_1||x_2|+\frac12|x_1|^2+\frac15|x_2|^2\tag4\\
&=\left(\sqrt{\tfrac52}\,|x_1|-\sqrt{\tfrac45}\,|x_2|\right)^2+\frac12|x_1|^2+\frac15|x_2|^2\tag5\\
&\ge\frac12|x_1|^2+\frac15|x_2|^2\tag6
\end{align}
$$
Explanation:
$(1)$: definition of the inner product
$(2)$: $x+\bar{x}=2\mathrm{Re}(x)$
$(3)$: triangle inequality
$(4)$: $3=\frac52+\frac12$ and $1=\frac45+\frac15$
$(5)$: recognize the square of a difference
$(6)$: the square of a real number is greater than or equal to $0$
A: Here is an alternative approach that requires less brute-force computation.  Note that we can rewrite the definition of $\langle x, y \rangle$ as
$$\begin{bmatrix} \bar y_1 & \bar y_2 \end{bmatrix} \begin{bmatrix} 3 & 1+i \\ 1-i & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$$
Let's let $M$ denote that $2\times 2$ matrix in the middle.  Calculate the characteristic polynomial of $M$ to find that $M$ has two distinct real eigenvalues, namely $\lambda_1 = 2 + \sqrt 3$ and $\lambda_2 = 2 - \sqrt 3$.  It follows that there are  $v, w \in \mathbb C^2$ such that $M v = (2 + \sqrt 3) v$ and $M w = (2 - \sqrt 3) w$.  These eigenvectors are necessarily linearly independent, so comprise a basis $\mathcal B$.  (Note that we don't need to actually find the eigenvectors, although we could if we want to.)
Now everything can be re-expressed in terms of this basis $\mathcal B$.  That is, let $x', y'$ be the result of applying the change-of-basis matrix $S = \begin{bmatrix}v & w \end{bmatrix}^{-1}$ to $x, y$ respectively.   Then we have
$$\langle x, y \rangle = \begin{bmatrix} \bar y'_1 & \bar y'_2 \end{bmatrix} \begin{bmatrix}2 + \sqrt 3 & 0 \\ 0 & 2-\sqrt 3 \end{bmatrix} \begin{bmatrix}x'_1 \\ x'_2 \end{bmatrix}$$
In this form, it is straightforward to verify that $\langle x, x \rangle > 0$ for $x \ne 0$.
The moral(s) of the story:

*

*Everything is easier if you choose an appropriate basis.

*At the end of the day, the property that we want to prove comes down to the fact that the two eigenvalues of $M$ are both positive real numbers.

