Little question about eigenvalues

[Kubilay Yazoglu]

Hey there, I'm thinking about if one of the eigenvalues is zero (means determinant is 0. right?) So, is there any possibility to non-zero eigenvalue also exists?

 

[mathwonk]

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

 

[Kubilay Yazoglu]

Well, now that I'm more inside of these things, I realized that I've asked a stupid question :D Of course there is.

 

[mathwonk]

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?

 

[Kubilay Yazoglu]

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?

Yes

 

[AMenendez]

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).

 

[AMenendez]

*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.

 

[WWGD]

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

 

[Maybe_Memorie]

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