2
$\begingroup$

[NBHM_2006_PhD Screening Test 2006_Algebra]

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.

  • 0
    @JyrkiLahtonen, Please help me, sir.2017-10-19

3 Answers 3

3

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.

  • 0
    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 \|.$2012-07-09
3

For 4, think about a rotation matrix.

1

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.