Throughout this part, $n$ denotes an integer greater than or equal to 3. We say that a matrix $M = (m_{i,j})_{1 \leqslant i,j \leqslant n}$ in $\mathcal{M}_n(\mathbb{R})$ is a Hankel matrix if there exists $a = (a_0, \ldots, a_{2n-2}) \in \mathbb{R}^{2n-1}$ such that for all $i$ and $j$ in $\{1, \ldots, n\}$, $m_{i,j} = a_{i+j-2}$. Such a matrix is denoted $M = H(a)$. Show that if $\lambda \in \mathbb{R}^*$ then the $n$-tuple $(\lambda, \ldots, \lambda)$ is not the ordered $n$-tuple of eigenvalues of a Hankel matrix of size $n$.
Throughout this part, $n$ denotes an integer greater than or equal to 3. We say that a matrix $M = (m_{i,j})_{1 \leqslant i,j \leqslant n}$ in $\mathcal{M}_n(\mathbb{R})$ is a Hankel matrix if there exists $a = (a_0, \ldots, a_{2n-2}) \in \mathbb{R}^{2n-1}$ such that for all $i$ and $j$ in $\{1, \ldots, n\}$, $m_{i,j} = a_{i+j-2}$. Such a matrix is denoted $M = H(a)$.
Show that if $\lambda \in \mathbb{R}^*$ then the $n$-tuple $(\lambda, \ldots, \lambda)$ is not the ordered $n$-tuple of eigenvalues of a Hankel matrix of size $n$.