GREPhysics.NET
GR | # Login | Register
   
  GR9277 #98
Problem
GREPhysics.NET Official Solution    Alternate Solutions
\prob{98}
9277_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?

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

Quantum Mechanics}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.

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
There are no Alternate Solutions for this problem. Be the first to post one!
Comments
raytyler
2018-11-01 03:14:57
\\lambda_1 = -\\frac{1}{2} + i\\frac{sqrt{3}}{2}\r\n\r\n\\lambda_2 = -\\frac{1}{2} - i\\frac{sqrt{3}}{2}\r\n\r\n\\lambda_3 = 1\r\n192.168.0.1 http://19216801help.com/NEC
raytyler
2018-11-01 03:13:02
1 + i - i is not equal to 0. The eigenvalues are actually 1, -\\\\\\\\\\\\\\\\frac{1}{2} \\\\\\\\\\\\\\\\pm \\\\\\\\\\\\\\\\frac{\\\\\\\\\\\\\\\\sqrt{3}}{2}i.\r\n192.168.0.1 loginNEC
Benjamin
2018-08-08 07:15:11
Hi there tim\r\n\r\n[url=http://www.pumpkin.com]harry[/url] " target="_blank">http://www.tim.com\">tim\r\n\r\n[url=http://www.pumpkin.com]harry[/url] \r\n\r\nhttp://www.essaywritinglab.co.uk/NEC
blabla0001
2017-03-25 09:10:25
Notice that the char. equation for this matrix is \\lambda^3-1
blabla0001
2017-03-25 09:11:29
Sorry about this. everything is fine
Benjamin
2018-08-08 07:12:18
brillassignment review
Typo Alert!
neo55378008
2012-08-18 13:26:23
Some simple matrix properties should help.
The trace of a matrix is invariant, so \lambda_1 + \lambda_2 + \lambda_3 = 0
The determinant is also equal to the product of the eigenvalues (1 in this case)
All real eigenvalues of one would give the right determinant, but the wrong trace!
Using 1, i, and -i both the trace and determinant are correct.
TheBridge
2018-10-27 02:12:16
But 1 + i - i is not equal to 0. The eigenvalues are actually 1, -\\\\\\\\frac{1}{2} \\\\\\\\pm \\\\\\\\frac{\\\\\\\\sqrt{3}}{2}i.
NEC
fizix
2010-11-09 17:49:37
By spectral theory only a symetric matrix have a full real eigenbasis. This matrix is not symetric
shka
2018-07-03 15:22:19
fizix has asserted the converse of the spectral theorem, which is not true in general. In fact, it\'s easy to find a non-symmetric diagonalizable real matrix. For example,\r\n\r\n[ 5 -4 ] \r\n[ 4 -5 ] \r\n\r\nis ANTI-symmetric but has eigenvalues 3 and -3. The spectral theorem guarantees the diagonalizability of real symmetric matrices, but it does not force all diagonalizable real matrices to be symmetric.
NEC
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.NEC
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).
NEC
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.
NEC
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
alemsalem
2010-09-26 09:04:30
(A) the trace is always the sum of the eigenvalues (sum of the terms on the diognal of the matrix which is 0+0+0), or you can say the trace is the same in all representations.
(E) this is just the eigenvalue equation.
for the rest substitute the roots:
1, exp(i 2 pi/3),exp(-i 2 pi/3).
hint 2*cosine(2 pi/3) = -1
NEC
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
NEC

Post A Comment!
Username:
Password:
Click here to register.
This comment is best classified as a: (mouseover)
 
Mouseover the respective type above for an explanation of each type.

Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .
type this... to get...
$\int_0^\infty$
$\partial$
$\Rightarrow$
$\ddot{x},\dot{x}$
$\sqrt{z}$
$\langle my \rangle$
$\left( abacadabra \right)_{me}$
$\vec{E}$
$\frac{a}{b}$
 
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...