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I know is quite easy to find a matrix $A\in\mathbb{R}^{2,2}$ that is diagonalisable if the base field is $\mathbb{C}$, but not diagonalisable if the base field is $\mathbb{R}$. The easiest example can be: $$\begin{pmatrix} 0&-1 \\ 1&0 \end{pmatrix}$$ because then we have the eigenvalues equation in the form of $\lambda^2+1=0$.

But what if we would like to find a matrix $B\in\mathbb{C}^{2,2}$ that is diagonalisable if the base field is $\mathbb{R}$, but not diagonalisable if the base field is $\mathbb{C}$? Is this even possible? I came across this question in a math question bank and I have huge concerns about it.

I would like to ask you one more question. What if we would like to find a matrix $B\in\mathbb{Q}^{2,2}$ that is diagonalisable if the base field is $\mathbb{R}$, but not diagonalisable if the base field is $\mathbb{Q}$? Will this matrix do? $$\begin{pmatrix} \pi&0 \\ 0&\pi \end{pmatrix}$$ Thank you very much.

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That is impossible: if a matrix is diagonisable in a field K, it is also diagonisable in any field $L$ that contains $K$. The change of basis matrix that will lead to the diagonal form will not change. It is just a matter of extending the scalars from $K$ to $L$.

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  • $\begingroup$ Thank you very much. How about the second part? $\endgroup$ – marco11 May 10 '15 at 23:47
  • $\begingroup$ The general answer I gave is all so about that case. Isn't it clear? $\endgroup$ – Bernard May 10 '15 at 23:52
  • $\begingroup$ It is clear, but is my example to the second case valid? (I meant the last part) $\endgroup$ – marco11 May 10 '15 at 23:55
  • $\begingroup$ Well, it's not a matrix in $M_2(\mathbf Q)$, so it's meaningless to try to diagonalise it over $\mathbf Q$. The field which you'll compute must contain the matrix coefficients. $\endgroup$ – Bernard May 10 '15 at 23:58
  • $\begingroup$ Ah, you are right. What would you say about matrix $\begin{pmatrix} 0&2\\1&0\end{pmatrix}$? Then the eigenvalue equation is $\lambda^2-2=0$ and so it is not diagonalisable in $\mathbb{Q}$ but diagonalisable in $\mathbb{R}$. Am I right? $\endgroup$ – marco11 May 11 '15 at 0:01
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This answers only the second part as the first is answered in another answer.

Your example with $\pi$ does not work as the matrix is not even defined over the rationals.

What you would need is a rational matrix whose eigen-values are not rational (but real). This exists.

Take for example \begin{pmatrix} 0 & 2 \\ 1 & 0 \end{pmatrix} The characteristic polynomial is $X^2 - 2$. The matrix is diagonalizable over the reals (the polynomial decomposes into linear factors, and the multiplicity of each eigenvector is $1$) but it cannot be diagonalizable over the rationals as the eigenvectors are not rational, and for a matrix to be diagonalizable over a given field you need that the characteristic polynomial factors completely over that field.

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  • $\begingroup$ I see, meanwhile this came up in the comments. Oh well. As it is written I might as well leave it. $\endgroup$ – quid May 11 '15 at 0:05
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    $\begingroup$ Thank you very much for your answer. Now everything is absolutely clear. I just suggested that matrix, but because of your answer I actually understand why it works. $\endgroup$ – marco11 May 11 '15 at 0:06

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