# The existence of an orthonormal matrix

Let $$U$$ be an orthonormal matrix. The dimension of $$U$$ is equal to $$n \times n$$. The first row of $$U$$ has the following form: $$\bigg( \frac{1}{\sqrt{n}}, \ldots , \frac{1}{\sqrt{n}} \bigg). \tag{1}$$ How can I prove that $$U$$ with property (1) exists?

Take your row vector (call it $$v_1$$) and extend this set to a basis of $$\mathbb{R}^n$$. Then orthonormalize this basis, (using Gram-Schmidt), obtaining {$$v_1, \dots, v_n$$}. Then, the matrix U such that its jth row is $$v_j$$ is an orthonormal matrix.
$$u:=(\frac{1}{\sqrt{n}},\ldots,\frac{1}{\sqrt{n}})$$ is a unit vector in your vector space. You can extend it to a basis $$\{u,v_{2},\ldots,v_{n}\}$$ of $$V$$ and then apply the Gram-Schmidt algorithm on this basis to obtain an orthonormal basis: $$\{e_{1},e_{2},\ldots,e_{n}\}$$ of $$V$$, where $$e_{1}=u$$. The matrix with rows $$e_{1},\ldots e_{n}$$ will then be a unitary matrix.
Consider the vectors $$v_1=e_1+\dots+e_n$$, $$v_i=e_i$$ for $$i=2,\dots,n$$ and apply Gram-Schmidt to them. Then you obtain an orthonormal basis whose first vector is $$(e_1+\dots+e_n)/\sqrt n$$, which is precisely the vector you want.