+2353 It's me again 
Have a look at the following matrix; $$\begin{pmatrix}
11&0&0&0&0\\
0&11&1&1&1\\
0&1&11&1&1\\
0&1&1&11&1\\
0&1&1&1&11\\
\end{pmatrix}$$
Find the eigenvalues of this matrix.
Obviously $$\lambda_1 = 11 \mbox{ and } \lambda_2 = 10$$
What's the easiest approach to find the others?
(I think it's only one eigenvalue with am(3) given the set-up of the matrix.)
Reinout 
+26393 ( 11.000000, 0.000000, 0.000000, 0.000000, 0.000000,
0.000000, 11.000000, 1.000000, 1.000000, 1.000000,
0.000000, 1.000000, 11.000000, 1.000000, 1.000000,
0.000000, 1.000000, 1.000000, 11.000000, 1.000000,
0.000000, 1.000000, 1.000000, 1.000000, 11.000000 )
Matrix D: eigenvalue
( 10.000000, 0.000000, 0.000000, 0.000000, 0.000000,
0.000000, 10.000000, 0.000000, 0.000000, 0.000000,
0.000000, 0.000000, 10.000000, 0.000000, 0.000000,
0.000000, 0.000000, 0.000000, 11.000000, 0.000000,
0.000000, 0.000000, 0.000000, 0.000000, 14.000000 )
Matrix V: eigenvector
( 0.000000, 0.000000, 0.000000, 1.000000, 0.000000,
0.007685, -0.707107, 0.499941, 0.000000, 0.500000,
0.007685, 0.707107, 0.499941, 0.000000, 0.500000,
0.699338, 0.000000, -0.510809, 0.000000, 0.500000,
-0.714708, 0.000000, -0.489073, 0.000000, 0.500000 )
Vector D: eigenvalue - real
( 10.000000,
10.000000,
10.000000,
11.000000,
14.000000 )
Vector e: eigenvalue - imag
( 0.000000,
0.000000,
0.000000,
0.000000,
0.000000 )
Matix: eigenvalue - real,imag
( 10.000000, 0.000000,
10.000000, 0.000000,
10.000000, 0.000000,
11.000000, 0.000000,
14.000000, 0.000000 )
+893 Easiest is to put the matrix into a calculator, like my TI-89, and ask it for the eigenvalues, 14,10,10,10,11, but I suppose that's not what you're asking.
On paper they drop out easily by fiddling around with determinants, (rather than deriving and solving the characteristic equation).
I'm going to describe my calculation if that's OK. I don't have the patience to mess around with the LaTex.
Subtract L (read lambda) from each diagonal element and expand along the top row, that gets you (11-L) times a 4 by 4.
Add columns 2,3 and 4 to the first column, that get you 14-L down the first column which can then be removed as a common factor. You now have (11-L)(14-L) outside a 4 by 4 which has 1's down its first column.
Subtract the top row from rows 2,3 and 4, and you are left with a determinant which has 1's across the top row, 10-L down the remainder of the lead diagonal and zero's elsewhere. Expand that down the first column and then expand the resulting 3 by 3 and you finish up with (11-L)(14-L)(10-L)(10-L)(10-L), which you can now equate to zero.
BTW, I've just started to look at your earlier asymptotic stability problem. It's not something I've encountered in the past, but I'll see if I can get anywhere with it.
+26393 ( 11.000000, 0.000000, 0.000000, 0.000000, 0.000000,
0.000000, 11.000000, 1.000000, 1.000000, 1.000000,
0.000000, 1.000000, 11.000000, 1.000000, 1.000000,
0.000000, 1.000000, 1.000000, 11.000000, 1.000000,
0.000000, 1.000000, 1.000000, 1.000000, 11.000000 )
Matrix D: eigenvalue
( 10.000000, 0.000000, 0.000000, 0.000000, 0.000000,
0.000000, 10.000000, 0.000000, 0.000000, 0.000000,
0.000000, 0.000000, 10.000000, 0.000000, 0.000000,
0.000000, 0.000000, 0.000000, 11.000000, 0.000000,
0.000000, 0.000000, 0.000000, 0.000000, 14.000000 )
Matrix V: eigenvector
( 0.000000, 0.000000, 0.000000, 1.000000, 0.000000,
0.007685, -0.707107, 0.499941, 0.000000, 0.500000,
0.007685, 0.707107, 0.499941, 0.000000, 0.500000,
0.699338, 0.000000, -0.510809, 0.000000, 0.500000,
-0.714708, 0.000000, -0.489073, 0.000000, 0.500000 )
Vector D: eigenvalue - real
( 10.000000,
10.000000,
10.000000,
11.000000,
14.000000 )
Vector e: eigenvalue - imag
( 0.000000,
0.000000,
0.000000,
0.000000,
0.000000 )
Matix: eigenvalue - real,imag
( 10.000000, 0.000000,
10.000000, 0.000000,
10.000000, 0.000000,
11.000000, 0.000000,
14.000000, 0.000000 )
+2353 I'm sorry Bertie, but do you mind writing it out?
I know it is a lot to ask, but I'm completely lost at the part where you start taking (14-L) out and adding rows and such. I think I don't remember enough rules about computing the determinant to be able to follow you by your explanation.
(And yes, I am not allowed to use a calculator on my exam).
