Matrix diagonalisable in R, but not in C. 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.
 A: 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 $L$.
A: 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.     
