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Let $L$ be a lattice in $\mathbb R^d$ and $v_1$ be a shortest nonzero vector in $L$. Is it always possible to find a basis of $L$ containing $v_1$?

It is easy to complete it into a basis of $\mathbb R^d$ but why is it always possible to complete it into a basis of the lattice?

If possible, I would like to see a counterexample or a direct proof (rather than a special case of a very complicated theorem). But please let me know if this is highly nontrivial. Any reference is welcome!

It seems that I have found a proof. Here is a sketch: project $L$ along $v_1$ to get a new lattice $L_1$. Find a basis of $L_1$ and lift them to $L$. Those $d-1$ vectors together with $v_1$ will form a basis of $L$.

taylor
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  • A shortest nonzero vector in ${\bf Z}^5$ is $(2,0,0,0,0)$. Can you extend that to a basis for ${\bf Z}^5$? – Gerry Myerson Sep 10 '22 at 02:09
  • @GerryMyerson: perhaps I am misunderstanding notation or forgetting something obvious, but aren't the shortest nonzero vectors in $\mathbb{Z}^5$ $(1, 0, 0, 0, 0), (0, 1, 0, 0, 0)$, etc? – user1090793 Sep 10 '22 at 04:08
  • Sorry, I was thinking of a different situation. Consider the sublattice of ${\bf Z}^5$ consisting of all $5$-tuples where all the entries have the same parity. The shortest vectors in it are $\pm(2,0,0,0,0),\dots,\pm(0,0,0,0,2)$, but they don't span it since $(1,1,1,1,1)$ is in the sublattice but not in their span. But that's not the same as extending a single vector to a basis, which is what you are interested in. – Gerry Myerson Sep 10 '22 at 07:13

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