raytyler 20181101 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/  
raytyler 20181101 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 login  
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blabla0001 20170325 09:10:25  Notice that the char. equation for this matrix is
blabla0001 20170325 09:11:29 
Sorry about this. everything is fine

 
neo55378008 20120818 13:26:23  Some simple matrix properties should help.
The trace of a matrix is invariant, so
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 20181027 02:12:16 
But 1 + i  i is not equal to 0. The eigenvalues are actually 1, .

 
fizix 20101109 17:49:37  By spectral theory only a symetric matrix have a full real eigenbasis. This matrix is not symetric
shka 20180703 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 nonsymmetric diagonalizable real matrix. For example,\r\n\r\n[ 5 4 ] \r\n[ 4 5 ] \r\n\r\nis ANTIsymmetric 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.

 
neutrinosrule 20081005 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.  
blah22 20080411 18:59:32  Out of curiosity is it OK to say since it is not hermitian it's eigenvalues aren't real?
chrisfizzix 20081014 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 20081016 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
has real eigenvalues 0 & 3, but it's not Hermitian (real, but not diagonal).

 
petr1243 20080315 16:46:02  Using det(AI) = 0, we will simply get:
,n=1,2,3, Solving this we get:
, n=1,2,3
So:
These 3 eigenvalues satisfy all choices, except for choice B.
 
dumbguy 20071010 18:11:05  Could some one maybe post the classic why the other answers are wrong?
dumbguy 20071010 18:12:48 
I mean why the other ones are right

alemsalem 20100926 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

 
jax 20051207 11:14:12  Where do you get ?
nitin 20061121 00:54:14 
Trivial:
The characteristic equation is
. Therefore
.

Zephyr3.14 20070407 23:25:30 
does this work? Then wouldn't
so
?
Why isnt it just ?

Richard 20070913 14:45:46 
Putting Euler to work,
.
But because,
and ,
you really don't arrive at anything that interesting.

jburkart 20071102 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: . Where's the error?

Jeremy 20071103 16:13:17 
Link

 