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\n$$L=\\{x \\in \\mathbb{R}^n: \\exists y \\in \\mathbb{R}^n (\\langle x,y \\rangle=0) \\text{ and } \\Vert x \\Vert > 0\\}$$
\nThank you in advance<\/p>\n
\neq 0$ and $\\|x\\|=1$. Take $z=x\/\\|x\\|$. Then we have
\n$$
\n\\left\\langle x,y\\right\\rangle=\\left\\langle z,y\\right\\rangle=0
\n$$
\nand
\n$$
\n\\|x\\|=\\|z\\|.
\n$$
\nCan you finish?<\/p>\n
\neq\\|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 $\\|x\\|=\\|y\\|$.
\nIf $x$ and $y$ are linearly dependent, they are $\\lambda x+\\mu y$ for some $\\lambda,\\mu\\in \\mathbb{R}$. Suppose $x$ is unit-normed, then $\\mu
\neq0$, $\\mu\\langle x,y \\rangle=0$, and $\\langle x,y \\rangle=0$, from where we get that $\\|x\\|=\\|y\\|$. If $y$ is unit-normed, the reasoning is the same.
\nThen $L=\\{x\\in \\mathbb{R}^n \\mid x$ and $
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