\begin{array}{l}{\text { 1. Let } n \geq 3 \text { be an integer and consider the } n \times n \text { matrix } A \text { with entries } a_{i j}=i-j .} \\ {\text { (a) Show that } A \text { has three complex eigenvalues and that } A \text { is diagonalisable. }}\end{array}
Matrix is skew symmetric and looks like:
$$\begin{pmatrix}0&-1&-2&\cdots&1-n\\1&0&-1&\cdots&2-n\\2&1&0&\cdots&3-n\\\vdots&\vdots&\vdots&\ddots&\vdots\\n-1&n-2&n-3&\cdots&0\end{pmatrix}$$
$$ $$
For the first part: I know how to show that a real skew symmetric matrix has only complex eigenvalues, but how would one show that there are three distinct eigenvalues?
For the second part: I believe that this has something to do with the minimal polynomial based on the lecture notes which state a relation between diagonalizability and minimal polynomials. How would one compute the minimal polynomial of such a matrix though?
For the second part, as $A$ is real skew-symmetric, it is normal. Therefore it is unitarily diagonalisable over $\mathbb C$.
For the first part, as $A$ is diagonalisable, its rank is equal to its number of nonzero eigenvalues (over $\mathbb C$). Note that $A=\eta e^T-e\eta^T$, where $\eta=(1,2,\ldots,n)^T$ and $e=(1,1,\ldots,1)^T$. Therefore it has rank two and it has two nonzero eigenvalues.
As $A$ is real and skew-symmetric, these two nonzero eigenvalues is a conjugate pair of purely imaginary numbers. Therefore the eigenvalues of $A$ are two simple eigenvalues $iy, -iy$ for some $y\in\mathbb R\setminus0$ and a zero eigenvalue of multiplicity $n-2$.