GREPhysics.NET: GR9277 #98

GR9277 #98

Problem

GREPhysics.NET Official Solution

\prob{98}

The matrix A, shown above, has three eigenvalues $\lambda_i$ defined by $A\nu_i=\lambda_i\nu_i$ . Which of the following statements is NOT true?

$\lambda_1+\lambda_2+\lambda_3=0$
$\lambda_1,\lambda_2$ , and $\lambda_3$ are all real numbers.
$\lambda_2\lambda_3=+1$ for some pair of roots
$\lambda_1\lambda2+\lambda_2\lambda_3+\lambda_3\lambda_1=0$
$\lambda_i^3=1,i=1,2,3$

Quantum Mechanics $\Rightarrow$ }Characteristic Equation

The characteristic equation of the matrix solves for the eigenvalues. It is $-\lambda(\lambda^2)+1=0$ . Not all solutions are real, since $\lambda = e^{2i\pi/3} = \cos(2\pi/3)+i\sin(2\pi/3)$ , where Euler's relation is used.

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Comments

neutrinosrule
2008-10-05 20:21:41

once you have the characteristic equation, you can note that the only possible real root is 1... therefore the other two roots must be imaginary (since a third order polynomial must always have 3 roots). No need to waste time finding the roots.

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blah22
2008-04-11 18:59:32

Out of curiosity is it OK to say since it is not hermitian it's eigenvalues aren't real?

chrisfizzix
2008-10-14 13:24:31

I agree; this test is about physics, and the physical result that corresponds to this question is from quantum mechanics. Only Hermitian matrices give you observables - that is, all real eigenvalues. This matrix is not Hermitian, easily seen as it is real but not equal to its transpose. Thus, the statement that all its eigenvalues are real is clearly false.

student2008
2008-10-16 15:18:47

You're not quite right. Although Hermitian matrices definitely have only real eigenvalues (and the orthonormal basises composed of eigenvectors), the opposite statement is erroneous. For example, the matrix
$\begin{bmatrix}1 & 1\\2 & 2\end{bmatrix}$
has real eigenvalues 0 & 3, but it's not Hermitian (real, but not diagonal).

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petr1243
2008-03-15 16:46:02

Using det(A-I $\lambda$ ) = 0, we will simply get:

$\lambda_n = 1^(\frac{1}{3})$ ,n=1,2,3, Solving this we get:

$\lambda_n = e^(\frac{2i\pi n}{3})$ , n=1,2,3

So: $\lambda_1 = -\frac{1}{2} + i\frac{sqrt{3}}{2}$

$\lambda_2 = -\frac{1}{2} - i\frac{sqrt{3}}{2}$

$\lambda_3 = 1$

These 3 eigenvalues satisfy all choices, except for choice B.

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dumbguy
2007-10-10 18:11:05

Could some one maybe post the classic why the other answers are wrong?

dumbguy
2007-10-10 18:12:48

I mean why the other ones are right

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jax
2005-12-07 11:14:12

Where do you get $\lambda = e^{\frac{2i \p}{3}}$ ?

nitin
2006-11-21 00:54:14

Trivial:

The characteristic equation is

$\begin{eqnarray*}\lambda^3&=&1\\&=&e^{2\pi i}\end{eqnarray*}$ . Therefore

$\lambda=e^{\frac{2\pi i}{3}}$ .

Zephyr3.14
2007-04-07 23:25:30

does this work? Then wouldn't
$1^{n} = 1 = e^{2 \pi i}$
$1 = e^{2 \pi i / n}$
$1 = \cos\left(\frac{2 \pi}{n}\right) + i \sin\left(\frac{2 \pi}{n}\right)$
$Re[1] = 1 = \cos\left(\frac{2 \pi}{n}\right)$
so
$\cos(\theta) = 1 \ \ \forall \theta$ ?

Why isnt it just $\lambda_{1} = \lambda_{2} = \lambda_{3} = 1$ ?

Richard
2007-09-13 14:45:46

$1=e^{2\pi{i}}$
$(1)^n=(e^{2\pi{i}})^n=e^{2\pi{in}} \hspace{1cm}\forall{n}\in{N}$
Putting Euler to work,
$1=\cos(2\pi{n})+i\sin(2\pi{n})$ .
But because, $\sin(2\pi{in})=0 \forall{n}\in{N}$
and $\cos(2\pi{in})=1 \forall{n}\in{N}$ ,
you really don't arrive at anything that interesting.

jburkart
2007-11-02 00:45:49

In a complex analysis course you will learn about branch cuts and how you have to be careful with roots. I can also prove the following: $i=\sqrt{-1}=\sqrt{1/-1}=\sqrt{1}/\sqrt{-1}=1/i=-i$ . Where's the error?

Jeremy
2007-11-03 16:10:38

But wait! If $i=-i$ , are you really getting even with "an eye for an eye?" This reminds me of the wonderful wild-eyed stranger problems I had in calculus. Get great laughs and see how well you know calculus: Find " target="_blank">http://www.dougshaw.com/findtheerror/">Find the error!. Absolutely priceless!

Jeremy
2007-11-03 16:13:17

Link

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Could some one maybe post the classic why the other answers are wrong?

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