Characteristic Polynomial of Matrix

[iamalexalright]

Homework Statement

Let J be the nxn matrix all of whose entries are equal to 1. Find the minimal polynomial and characteristic polynomial of J and the eigenvalues.


Well, I figure the way I'm trying to do it is more involved then other methods but this is the easiest method for me to start with.

We can find the characteristic equation of J (and thus the eigenvalues) by computing:
det(J - \lambda I)

We also know that the determinant of a nxn matrix can be found by computing:
det(A) = \sum_{\sigma \in S_{n}} sgn(\sigma) \prod_{i = 1}^{n}A_{i,\sigma_{i}}
where S_n is the set of all permutations of (1,2,3,...n). sgn is the signature of the permutation.

Now we want to compute det(J - \lambda I)
I know there will be n! terms after applying Leibniz's formula, half of which will have a positive signature and half that will have an odd signature.

Now, I don't really know how to prove this but with some investigation we can see that there will only be one term with degree n, ie:
det(J - \lambda I) = (1 - \lambda)^{n} + ...

There shouldn't be any terms with degree n - 1...

I'm not sure where to go after this... I realize that this probably isn't the best method to go about this (it's late and I've been working on homework for awhile now...) so any other suggestions would be great (along with suggestions on how to continue with this path if possible).

 

[tiny-tim]

hi iamalexalright!

(have a lambda: λ )

hint: subtract the first row from the other rows

 

[iamalexalright]

Hrmph, could have sworn I thought of that...

so then we are left with a new J with all 1's in the first row and 0's everywhere else

the determinant of this new matrix would then be (since only diagonal terms contribute):
(-1)^{n + 1}(1 - \lambda)\lambda^{n - 1}

Here is my issue... that is the charecteristic equation for the new matrix but there should be a coefficient of n on the smaller degree:

example: n = 3
the formula above gives:
-x^{3} + x^{2}
but I should have:
-x^{3} + 3x^{2}

Don't know how to resolve that.

Also, if n = 3 would the minimal polynomial be:
x(3 - x) = 3x - x²
?

 

[tiny-tim]

so then we are left with a new J with all 1's in the first row and 0's everywhere else

no, you need to do it for J - λI

 

[iamalexalright]

Okay, so that will leave me with a matrix that has:
(1 - λ) in the first diagonal position
-λ for all other diagonals
1's in the rest of the first row
λ's in the rest of the first column
zero's elsewhere

Not sure where to go from here... I can see with small examples that it works out but proving for all n seems tricky for me.

 

[tiny-tim]

have you tried induction?

 

[spamiam]

Check out Avodyne's reply in this (https://www.physicsforums.com/showthread.php?t=314070&page=3) thread for another way to pursue this. You can check to see that J² = nJ, which imposes a condition on the possible values for J's eigenvalues. Then look at the trace tr(J) to find the multiplicity of each of these values.