When finding eigenvalues of a matrix $A$ , what if $\det(A-\lambda I)$ is $0$ regardless of the value of $\lambda$. Then, how can I find eigenvalues? Just like the title says, I don't know how to write an example matrix here to look like matrix. If this makes sense, $A$ can be $[1 0 0;0 1 0;0 1 0]$ (like in MATLAB syntax) then if we find determinants of $A-\lambda Ι$ we get $0$ for every $\lambda$. In MATLAB it says the eigenvalues are $0,1,1$.
 A: Two answers
First one based on a bit of theory
Having $\det(A - \lambda I)=0$ for all $\lambda$ is impossible (at least if you work in the field of the reals $\mathbb R$). $\det(A - \lambda I)$ is a non-zero polynomial function that can't be always vanishing.
Second one based on computation
$$\det(A - \lambda I)=\begin{vmatrix}
1- \lambda & 0 & 0\\
0 & 1- \lambda & 0\\
0 & 0 & -\lambda
\end{vmatrix} = -\lambda(\lambda-1)^2$$
A: If $A$ is your matrix above, then
$$ \det(A- \lambda E)=- \lambda(\lambda-1)^2.$$
A: Let $A = (a_{ij}),\ i,j=1,\ldots,n$. From the definition of determinant, it's clear that $\det(A-\lambda I)$ is a polynomial in $\lambda$. Moreover, it is of degree $n$:
\begin{align}\det(A-\lambda I) &= (a_{11} - \lambda)\ldots(a_{nn}-\lambda) + \text{lower degree terms} \\ &= (-1)^n\lambda^n + \text{lower degree terms}
\end{align}
In particular, it is not a zero polynomial and it has at most $n$ roots in the field you are interested in.
As a side note, in the case of a finite field of characteristic $p$, it is possible that the determinant vanishes for all values of the field, without it being a zero polynomial, but it's not possible for the fields of characteristic $0$ (fields containing $\mathbb Q$).
