0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

$u:=(\frac{1}{\sqrt{n}},\ldots,\frac{1}{\sqrt{n}})$ 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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Can the downvoter explain please? This answer is correct and was also the first to be posted $\endgroup$ – Federico Nov 20 '18 at 18:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.