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Let $A$ be an orthogonal $3 \times 3$ matrix with real entries. Pick out the true statements:

a. The determinant of $A$ is a rational number.

b. $d(Ax,Ay) = d(x, y)$ for any two vectors $x, y \in \mathbb{R}^3$, where $d(u, v)$ denotes the usual Euclidean distance between vectors $u, v \in \mathbb{R}^3$.

c. All the entries of $A$ are positive.

d. All the eigenvalues of $A$ are real.

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Yup, it's a homework question. You'll need to describe how you've tried to figure it out before anyone will help you, probably. –  Paul Z Aug 7 '12 at 1:10
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Well, @poton, my solution awaits your reply to Paul Z. –  ncmathsadist Aug 7 '12 at 1:20
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@ncmathsadist - by this and his previous question, it looks like the OP isn't particularly interested in putting in even minimal effort, but I'll be glad to be proved wrong. –  nbubis Aug 7 '12 at 1:28
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Tres annoying, nbubis. –  ncmathsadist Aug 7 '12 at 1:31
    
i think a is true.but no idea about the other options –  poton Aug 7 '12 at 2:24
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1 Answer

up vote 0 down vote accepted

(a) True. $A$ being orthogonal means $A^tA = I$. Hence $\mathrm{det} (A^tA) = \mathrm{det}(I)$; i.e., $\mathrm{det}(A)^2 = 1$. So $\mathrm{det}(A) = \pm 1$. That is, its determinant is a rational number.

(b) True: $d(Ax,Ay) = \vert\vert Ax - Ay \vert\vert = \vert\vert A(x-y) \vert\vert$, on the one hand. On the other hand, $d(x,y) = \vert\vert x-y \vert\vert$. So it's enough to prove that $\vert\vert Au \vert\vert = \vert\vert u \vert\vert$ for every vector $u$, right? And, since both guys in this last equation are positive numbers, it's enough to see that $\vert\vert Au \vert\vert^2 = \vert\vert u \vert\vert^2$.

This fact can be deduced as follows:

$$ \vert\vert Au \vert\vert^2 = \langle Au, Au \rangle = (Au)^tAu = u^tA^tAu = \langle u, A^tAu \rangle = \langle u,u \rangle = \vert\vert u \vert\vert^2 \ . $$

(c) False. For instance,

$$ \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$

is an orthogonal $3\times 3$ real matrix and not all of its entries are positive.

(d) Shame on me. See Paul Z.'s comment.

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As tempting as it is to leave this up as misdirection for someone who's using the site to do his math homework, (d) is certainly not true. Consider $\begin{pmatrix}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}$; its eigenvalues are $1$ and $\pm {\hat i}$. –  Paul Z Aug 7 '12 at 5:17
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