TITLE: Can a transcendent matrix have an algebraic spectrum?
QUESTION [5 upvotes]: 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 $K$?

REPLY [5 votes]: You can construct such a matrix easily enough. Let $K = \mathbf{C}$ and $K' = \mathbf{C}(t)$, then consider
$$ 
\begin{pmatrix}
 t & t+1 \\
 1-t & -t \\
\end{pmatrix}
\begin{pmatrix}
 1 & 0 \\
 0 & 2 \\
\end{pmatrix}
\begin{pmatrix}
 t & t+1 \\
 1-t & -t \\
\end{pmatrix}^{-1} =
\begin{pmatrix}
 2-t^2 & - t - t^2 \\
 t^2 - t & t^2+1 \\
\end{pmatrix}
 $$
This matrix has eigenvalues $1$ and $2$ and eigenvectors $(t, 1-t)$ and $(t + 1, -t)$. There's nothing particularly special about these values, if you were to just pick any eigenvalues and eigenvectors, the resulting matrix would probably give you what you want.