Show that $\langle x,x\rangle > 0$ if $x \neq0$

Question

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.

Answer 1

$$ \begin{align} \langle x,x\rangle &=3x_1\bar{x}_1+(1+i)x_1\bar{x}_2+(1-i)x_2\bar{x}_1+x_2 \bar{x}_2\tag1\\[6pt] &=3|x_1|^2+|x_2|^2+2\mathrm{Re}((1+i)x_1\bar{x}_2)\tag2\\[6pt] &\ge3|x_1|^2+|x_2|^2-2\sqrt2|x_1||x_2|\tag3\\[3pt] &=\frac52|x_1|^2+\frac45|x_2|^2-2\sqrt2|x_1||x_2|+\frac12|x_1|^2+\frac15|x_2|^2\tag4\\ &=\left(\sqrt{\tfrac52}\,|x_1|-\sqrt{\tfrac45}\,|x_2|\right)^2+\frac12|x_1|^2+\frac15|x_2|^2\tag5\\ &\ge\frac12|x_1|^2+\frac15|x_2|^2\tag6 \end{align} $$ 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 $0$

Answer 2

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?

Answer 3

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.

Answer 4

\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 $x_1=x_2=0$.

Answer 5

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:

1. Everything is easier if you choose an appropriate basis.
2. 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.