Can a transcendent matrix have an algebraic spectrum? Let $K$ be an algebraically closed field (e.g $\mathbb{A}$) and $K'/K$ a transcendent field extension (e.g. $\mathbb{C}/\mathbb{A}$). Let $A\in K'^{n\times n}$ b...--prophetes.ai

Can a transcendent matrix have an algebraic spectrum? Let $K$ be an algebraically closed field (e.g $\mathbb{A}$) and $K'/K$ a transcendent field extension (e.g. $\mathbb{C}/\mathbb{A}$). Let $A\in K'^{n\times n}$ be a matrix over K', which has at least one entry from $K'\setminus K$. Is it still possible that all eigenvalues of $A$ lie in $K$?

Sure. The matrix could be nilpotent, hence all the eigenvalues would be $0$. It could also be upper or lower triangular with only elements of $K$ on the diagonal.