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Download Hp3577MotherboardDrivers APK For PC,Laptop And Laptop.Q:

$L=\{x \in \mathbb{R}^n: \exists y \in \mathbb{R}^n (\langle x,y \rangle=0) \text{ and } \Vert x \Vert > 0\}$

I am working on this simple problem and don’t know where to start.
$$L=\{x \in \mathbb{R}^n: \exists y \in \mathbb{R}^n (\langle x,y \rangle=0) \text{ and } \Vert x \Vert > 0\}$$
Thank you in advance

A:

We assume that $x
eq 0$ and $\|x\|=1$. Take $z=x/\|x\|$. Then we have
$$
\left\langle x,y\right\rangle=\left\langle z,y\right\rangle=0
$$
and
$$
\|x\|=\|z\|.
$$
Can you finish?

A:

Assume that $x$ and $y$ are linearly independent and $\|x\|
eq\|y\|$, then there exists a $\lambda\in \mathbb{R}$ such that $\langle x,y \rangle=\lambda(\langle x,y \rangle)=0$. This contradicts that $\langle x,y \rangle=0$ only if $\|x\|=\|y\|$.
If $x$ and $y$ are linearly dependent, they are $\lambda x+\mu y$ for some $\lambda,\mu\in \mathbb{R}$. Suppose $x$ is unit-normed, then $\mu
eq0$, $\mu\langle x,y \rangle=0$, and $\langle x,y \rangle=0$, from where we get that $\|x\|=\|y\|$. If $y$ is unit-normed, the reasoning is the same.
Then $L=\{x\in \mathbb{R}^n \mid x$ and $
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