Little question about eigenvalues

for a diagonal matrix, the eigenvalues are the numbers on the diagonal. so what do you think can happen?

 

another point of view is that eigenvalues are roots of the characteristic polynomial. so if one root is zero can other roots be non zero?

 

Well, you could have something like:
## \lambda^3 + 18\lambda^2 + 81\lambda##
Where you have eigenvalues ##\lambda_1 = 0## and ##\lambda_2 = -9##.
So, I certainly think you can have an eigenvalue of zero and other nonzero eigenvalues. And also, I'm not sure what you mean by the determinant being zero; to get a characteristic polynomial, you have to take the determinant and set it equal to zero to solve for the eigenvalues in the first place.
In other words, you find eigenvalues using:
##\det(A - \lambda I_A) = 0## (Given a matrix A and its identity matrix, I_A).

 

*EDIT*
I misspoke above; I listed two eigenvalues, whereas a ##3 \times 3## matrix (with a cubic characteristic polynomial) would have three eigenvalues. Of course, in the case I listed above, two of the eigenvalues, ##\lambda_2## and ##\lambda_3## would be the same (λ₂ = λ₃ = -9). I was treating them as roots of a cubic, where (with three solutions) there are only two roots, given two of them are the same. In the context of matrix algebra, though, all eigenvalues should be accounted for.

 

Well, if there is an eigenvalue ## \lambda=0 ##, then Det(A-0I_A)=Det(A)=0.

 

Why is that a problem? Having zero determinant just means that it's not invertible.