Bertie

It's not that obvious. For example,

$$a^{7}+b^{7}=(a+b)(a^{6}-a^{5}b+a^{4}b^{2}-a^{3}b^{3}+a^{2}b^{4}-ab^{5}+b^{6}).$$

Third line, 2√(x) + √(x) = 60, so 3√(x) = 60 etc. ?

That's my attempt at the early theory for this question

You have to pretend that the two posts are a single post !

I don't agree with the analysis for the last part of the question, my analysis to follow.

Hi Reinout

Are you still interested in this question or have you got it sorted ?

You need to define exactly what you mean by 'larger'.

You might take it to mean magnitude, (in which case -0.5 is as large as 0.5) or you might take CPhill's dollars argument or Melody's line that all negative numbers are smaller than all positive numbers.

As an alternative, in electrostatics there are things called point charges, and they can be positive or negative. So, if I have a positive point charge of +9 and a negative point charge of -40, which is the larger ? I would say that it was the -40, (but it comes down to a matter of definition).

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.