Prove that for an NxN matrix with all elements along the leading diagonal equal to 2?

Question

and all other elements 1, one of the Eigen values is N+1 and all other
Eigen values are 1.

Answer 1

The one part is easy:

If you take A-I you get a matrix with all 1s everywhere, so when you row reduce this to find the e-vectors, you see that it has rank 1 with (N-1) e-vectors e1-ek for k =2..N. Basis for e-space = {(1,-1,000000), (1,0,-1,000000), etc.}.

The other part is easy also. A*(1,1,1,1,1...1) = (N+1,N+1,...,N+1) = (N+1)*(1,1,1,1...1) so N+1 is an eigenvalue.

What about any others? You found them all! 1 w/ multiplicity (N-1) and (N+1) w/ multiplicity 1, so total of N....all e-values!