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The original question is: Column Vectors orthogonal implies Row Vectors also orthogonal?

A counterexample with zero entries is given in one post. However, my question is whether pairwise orthogonal columns with no zero entries imply pairwise orthogonal rows?

Community
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Bohan Lu
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  • @WillJagy if the matrix is square and has zero determinant, then one of the columns has all 0s because any set of nonzero orthogonal vectors is linearly independent. – Matt Samuel Dec 09 '14 at 20:24
  • The only case when it is generally true is when the columns are orthonormal and the matrix is square. – Algebraic Pavel Dec 10 '14 at 10:37

1 Answers1

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The answer is no. Here's a counterexample: $$ \pmatrix{1&2\\-1&2} $$

Ben Grossmann
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  • Is there a systematic way of constructing these counterexamples with any n*n square matrices with nonzero pairwise orthogonal columns? – Bohan Lu Dec 09 '14 at 20:49
  • Yes. Start with an orthonormal basis of $n$ column-vectors, and multiply one of them by $2$. At least, I think that should work. – Ben Grossmann Dec 09 '14 at 21:17