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I've been trying to complete the following problem, but I just can't think of any ideas.

If $\mathcal{L}$ is a linear operator on vector space $V$, where $dimV=n$ and $dim(rang(L))=k$, prove that characteristic polynomial of $\mathcal{L}$, $f_{\mathcal{L}}(\lambda)$ is divisible by $\lambda^{n-k}$.

The only thing I know is that $dim(Ker(\mathcal{L}))=n-k$, but couldn't come up with anything to do with that fact. Could we say, that $dim(Ker(\mathcal{L}-\lambda\mathcal{E}))=dim(Ker(\mathcal{L}))=n-k$, which would then lead us to the proof? Any hints will be appreciated.

On the other hand let's say this can be shown (I'm guessing it can). Would you mind giving me an example of such linear operator, preferably just a matrix of one? I really haven't encountered a lot of examples where there is $\lambda=0$ as eigenvalue, with algebraic multiplicity of $n-k$.

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You already have your answer, just use a base fitting with a decomposition $E = \ker L \oplus F$, then try to look what the matrix will look like in this base, this gives you an immediate answer.
Another possibility is that the geometric multiplicity of $0$ (here $\dim \ker L = n-k$) is always smaller than the algebraic one which means the multiplicity of $0$ in your characteristic polynomial.

If you want a definition, let $\lambda \in \text{Sp}(u)$, we define $m_g(\lambda) = \dim E_\lambda$ the geometric multiplicity and $m_a (\lambda) $ the algebraic multiplicity as the multiplicity of $\lambda$ as a root of $\chi_u$.