a linear map $T ∈ L(F^{n} )$ that has exactly m eigenvalues.

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So I run into the question:

For each m ∈ {1, ..., n}, give an example of a linear map

$T ∈ L(F^{n} )$

that has exactly m eigenvalues.

And I came up with the matrix A which looks like this:

$$ A= \left(\begin{matrix}\lambda_{1}&0&\cdot&\cdot&\cdot&\cdot&0\\0&\lambda_{2}&\cdot&\cdot&\cdot&\cdot&\cdot\\\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\\\cdot&\cdot&\cdot&\lambda_{m-1}&\cdot&\cdot&\cdot\\\cdot&\cdot&\cdot&\cdot&1&\cdot&\cdot\\\cdot&\cdot&\cdot&\cdot&\cdot&1&0\\0&\cdot&\cdot&\cdot&\cdot&0&1\end{matrix}\right) $$

Would it have m eigenvalues? Am I on the right track?