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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 $K$?

hardmath
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Isidor Konrad Maier
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2 Answers2

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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.

Matt Samuel
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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.

Trevor Gunn
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    @hardmath I just picked random polynomials that looked like they'd give a small determinant. The fact that the determinant is degree $0$ was an accident. I just made an error copying the matrix. – Trevor Gunn Nov 28 '19 at 15:46