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I'm trying to show that $\mu_1 = \mu_2 = 0$ in the below system:

$$\begin{cases} x \mu_1 &= 0\\ y(\mu_1 + \mu_2) &= 0 \\ z \mu_2 &= 0 \\ x^2 + y^2 &= 1\\ y^2+z^2 &= 4 \end{cases}$$

$\textbf{My attempt:}$

If $\mu_1 \neq 0$ then $x=0$. It follows that $y^2 =1$ and $z^2 =3$. Hence $\mu_2=0$ and thus $y(\mu_1 + \mu_2) \neq 0$, which is impossible. If $\mu_2 \neq 0$ then $z=0$. It follows that $y^2 =4 >1$, which is impossible. As such, $\mu_1= \mu_2 =0$ and thus LICQ is satisfied.

This argument can be written alternatively as

$\mu_1 \neq 0 \implies x=0 \implies y^2=1 \implies z^2 = 3$ $\implies \mu_2=0 \implies y(\mu_1 + \mu_2) \neq 0$, which is impossible. Similarly, $\mu_2 \neq 0 \implies z=0 \implies y^2 = 4>1$, which is impossible.

To me, both of them are not ease for reading because the first one uses too many sentences and the second one is terse.

Could you please help me to write this argument more friendly to read? Thank you so much!

Akira
  • 17,367

2 Answers2

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I'd start with the second argument about $\mu_2$:

If $\mu_2 \neq 0$ we'd have $z=0$ by the third equation. The last one then says $y^2=4$ which is impossible as then $1 = x^2 + y^2 \ge y^2=4$. So $\mu_2=0$.

We can then use this right away and say

Hence $x\mu_1= y\mu_1=0$, and if $\mu_1 \neq 0$ we'd have $x=y=0$ contradicting $x^2+y^2=1$. So $\mu_1 = 0$ as well.

Henno Brandsma
  • 242,131
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From $x^2+y^2=1$ it follows that $z^2\geq3$, hence $\mu_2=0$, and then that $\mu_1=0$.

Christian Blatter
  • 226,825