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?

Question

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 $0,1,1$.

Answer 1

If $A$ is your matrix above, then

$$ \det(A- \lambda E)=- \lambda(\lambda-1)^2.$$

Answer 2

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

$$\det(A - \lambda I)=\begin{vmatrix} 1- \lambda & 0 & 0\\ 0 & 1- \lambda & 0\\ 0 & 0 & -\lambda \end{vmatrix} = -\lambda(\lambda-1)^2$$

Answer 3

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 $\mathbb Q$).