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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
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What have you tried? Do you know Gram-Schmidt? –  Calvin Lin Jan 15 '13 at 17:44
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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

3 Answers 3

up vote 5 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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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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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 b square roots of $3,14,40$ respectively) and put the results as the columns of you matrix $A$.

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