Matrix diagonalisable in R, but not in C.

Question

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.

Answer 1

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 $L$.

Answer 2

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.