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Let $A$ be an $3\times 3$ orthogonal matrices with real entries,Then which are true

  1. $\det A$ is rational number

  2. $d(Ax,Ay)=d(x,y)$ for any two vector $x,y\in \mathbb{R}^3$ where $d$ is ussual eucledean distance.

  3. All entries off $A$ are positive.

  4. All eigen values of $A$ are real.

determinant of orthogonal matrix is $\{1,-1\}$ so 1 is true, modulas of eigen values of an orthogonal matrix is $1$ so 4 may not be true always, I am not getting anything about 2 and 3. Thank you.

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for $3$: $diag(-1,-1,-1)$ –  Belgi Jul 9 '12 at 22:59
    
for 4: I think that since $A,A^t$ are similar and since the eigenvalues of $A^-1$ are $1/$ by eigenvalues of A you get that if $x$ is eigenvalue then $x=\frac{1}{x}$ hence $x=1$ or $x=-1$ in particular $x$ is real –  Belgi Jul 9 '12 at 23:03
    
okay~~~~~~~~~~~ –  Bunuelian Trick Jul 9 '12 at 23:06
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Counter example of $4: \rm{diag}(1, i, -i).$ Eigenvalues are $\{ 1, i, -i \}.$ –  user2468 Jul 9 '12 at 23:24
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Sorry. $\pmatrix{1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0}$ has the eigenvalues $\{ 1, i, -i \}.$ This is a rotation matrix, but you can also think of it as obtained by similarity transformation from the matrix in my comment above. –  user2468 Jul 9 '12 at 23:46
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3 Answers 3

up vote 2 down vote accepted

Hint: For 2 (work out the details as exercise)

First, we show that if $A$ is orthogonal then $\| Ax \|^2 = \|x\|^2.$ Write $\|Ax \|^2 = (Ax)^{T}(Ax).$ Distribute the transpose and recall that $A^{T}A = I.$ Done.

Now, write $$d(Ax, Ay) = \| Ax - Ay \|^2 = (Ax-Ay)^{T}(Ax-Ay).$$ Simplify to get $$(Ax)^{T}(Ax)-2(Ax)^{T}(Ay)+(Ay)^{T}(Ay) = \| Ax \|^2 - 2 (Ax)^{T}(Ay) + \| Ay \|^2$$ Similarly show that $ d(x, y) = \| x \|^2 - 2 x^{T}y + \| y \|^2.$

Now, you can match the terms one to one, except for the middle terms $(2 (Ax)^{T}(Ay)$ and $2 x^{T}y)$. Well, you further simplify $(Ax)^{T}(Ay)$ and recall that $A^{t} A = I.$ Done.

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why $d(Ax, Ay) = \| Ax - Ay \|^2$? –  Bunuelian Trick Jul 9 '12 at 23:18
    
The distance between two vectors $a, b$ is the vector length of their difference $b - a.$ No? I used the length squared. –  user2468 Jul 9 '12 at 23:21
    
OK~~~~~ why did u take like that? –  Bunuelian Trick Jul 9 '12 at 23:28
    
Well, if $A$ preserves vector length, then it preserves length squared, so no harm done. By the way, to show that $A$ preserves vector length (rather than length squared) consider: $$\| Ax \| = \sqrt{(Ax)^{T}(Ax)} = \sqrt(x^{T}A^{T}Ax) = \sqrt{x^{T}x} = \| x \|.$$ –  user2468 Jul 9 '12 at 23:30
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For 4, think about a rotation matrix.

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The orthogonal matrix represents orthogonal transformation, which preservers inner product, so 2 is true.

3 is false, the counter example is the rotation matrix.

4 is also false, the eigenvalues are lying on the unit circle, not necessary real.

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