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Is the statement true?

There is a $3\times 3 $ real orthogonal matrix with all non zero entries.

for orthogonality, $AA^T=A^TA=I_3$, please give me hint

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Why was this question downvoted? – Git Gud Jan 15 '13 at 17:42
Consider ${1\over 3}\left[\matrix{1&-2&2\cr 2&-1&-2\cr 2&2&1}\right]$. Or, consider appropriate rotations of the standard unit vectors in $\Bbb R^3$. Or, take two orthonormal vectors in a plane that is not perpendicular to any coordinate axis and extend to an orthonormal basis of $\Bbb R^3$. – David Mitra Jan 15 '13 at 17:47
up vote 9 down vote accepted

Gordon Pall found all rational orthogonal matrices, 3 by 3, in 1940, see PALL_PDF

Given an odd number $$ n = a^2 + b^2 + c^2 + d^2,$$ the matrix $$ \frac{1}{n} \begin{pmatrix} a^2 + b^2 - c^2 - d^2 & 2(-ad+bc) & 2(ac+bd) \\ 2(ad+bc) & a^2 - b^2 + c^2 - d^2 & 2(-ab+cd) \\ 2(-ac+bd) & 2(ab+cd) & a^2 - b^2 - c^2 + d^2 \end{pmatrix}$$ is orthogonal, and all rational orthogonal matrices can be written in this manner. This is formula (10) on page 755 of Pall's article, the second page of the pdf.

Right, so you want no zeroes in the matrix. Take, for example, $d=0,$ and nonzero $a,b,c$ which do not make a Pythagorean triple in any order. It would be enough to just add on the condition $c^2 \geq 1 + a^2 + b^2.$

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Of course there is. Choose one vector with nonzero entries, say $(1,1,1)$, choose another one that is orthogonal to it, and that none of the three $2\times2$ minors they define vanishes, say $(1,2,-3)$. Now take their cross product to find a third vector orthogonal to the first two, and which has no zero entries by assumption; in this case it is $(-5,4,1)$. Now normalise everything (divide by square roots of $3,14,42$, respectively) and put the results as the columns of you matrix $A$.

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At first sight i was wondering why do we have to choose second vector such that none of the three minors they define vanishes... Then i realized while computing third column that if some minor is zero then corresponding entry in cross product is zero... This is a wonderful answer.. I would have accepted this answer.. Very natural.. – Praphulla Koushik Feb 10 at 14:44
square root of $42$... – cello Mar 5 at 9:44
@cello: thanks, corrected. – Marc van Leeuwen Mar 5 at 10:51

Almost all unit vectors have a non-zero entry. Almost anything that you try should work. For example, if $A=(a, b, c)$ is a vector of length 1 with $a\neq -1$, you can verify that

$$ \begin{pmatrix} a & b & c \\ b & \frac {b^2-a-1}{a+1} & \frac {bc}{a+1} \\ c & \frac {bc}{a+1} & \frac {c^2-a-1}{a+1} \end{pmatrix}$$

describes an orthonormal set. This is the reflection of all the standard unit vectors across the line $e_1 +A$.

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I do not understand what does it mean to say reflection of standard unit vectors across the line $e_1+A$.. What line are we talking about? – Praphulla Koushik Feb 10 at 14:39
@PraphullaKoushik The line consisting of vectors of the form $k ( e_1 + A) $ for some constant $k$. – Calvin Lin Mar 31 at 16:00

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