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I am given a unit vector $e=1/\sqrt{n}(1,1,\ldots,1)'$ and the problem is to construct an $n \times n$ (real) unitary matrix $U$ which will contain $e$ as the last column. I understand that there are infinitely many such $U$ ($n>2$). I wonder if there is a very simple closed form for $U$, i.e., very simple $n-1$ vectors orthogonal to each other and to $e$.

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    also, you can always extend the given vector to a basis for $R^n$ and then use the gram schmidt process to get what you want.2011-12-07

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The columns of $\pmatrix{1&1&1&1&1\cr1&-1&1&1&1\cr1&0&-2&1&1\cr1&0&0&-3&1\cr1&0&0&0&-4\cr}$ are pairwise orthogonal. If you divide the 1st column by $\sqrt5$, the second by $\sqrt2$, the third by $\sqrt 6$, the fourth by $\sqrt12$, and the fifth by $\sqrt20$, you should get an orthogonal matrix. Then you just have to move the first column to the far right.

This is the case $n=5$, but the pattern should be clear.

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    @J.M. The reference is great (I know the author is a great matrician:)). Thank you.2011-12-07