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[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
    for $3$: $diag(-1,-1,-1)$2012-07-09
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    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 real2012-07-09
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    okay~~~~~~~~~~~2012-07-09
  • 3
    Counter example of $4: \rm{diag}(1, i, -i).$ Eigenvalues are $\{ 1, i, -i \}.$2012-07-09
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    thanx JD...........'2012-07-09
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    @J.D.: but it is specified as having real entries.2012-07-09
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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.2012-07-09
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    If all entries of $A$ are positive, then so are the inner products of the rows, but they were supposed to be orthogonal. IOW this property never holds (even though for the purposes of this question a counter example will suffice).2012-07-10
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    Given in the answer key that (a) is true .We know, $detA$ is the product of eigenvalues. There are three roots of the characteristic equation. Two roots are conjugate to each other and one is real. How do we prove that it is rational?. Please help me.2017-10-19
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    @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.

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    why $d(Ax, Ay) = \| Ax - Ay \|^2$?2012-07-09
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    The distance between two vectors $a, b$ is the *vector length* of their difference $b - a.$ No? I used the length squared.2012-07-09
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    OK~~~~~ why did u take like that?2012-07-09
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    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.