$A$ is not similar to a diagonal matrix over the reals Let $A = \begin{bmatrix} 6 & -3 & -2 \\ 4 & -1 & -2 \\ 10 & -5 & -3 \end{bmatrix} $ then $A$ is not similar to a diagonal matrix over the reals and it is not similar to a diagonal matrix over the complexes.
We know that $A$ is similar to a diagonal matrix over the reals(complexes) if there exist $D$ diagonal matriz and $P$ invertible matrix both $n \times n$ with real entries (complex entries) such that $A = PDP^{-1}$
I find that the inverse of $A$ is $A^{-1} = \frac{1}{2}\begin{bmatrix} -7 & 1 & 4 \\ -8 & 2 & 4 \\ -10 & 0 & 6 \end{bmatrix}$ and i diagonalize the matrix $A$ and got $A = PBP^{-1}$ with $ P = \begin{bmatrix} 1 & \frac{3}{5}-\frac{i}{2} & \frac{3}{5}+\frac{i}{2} \\ 0 & \frac{3}{5}-\frac{i}{2} & \frac{3}{5}+\frac{i}{2} \\ 2 & 1 & 1 \end{bmatrix} $ , $ P^{-1} = \begin{bmatrix} 1 & -1 & 0 \\ -1 + 3i & 1-\frac{i}{2} & \frac{1}{2}-\frac{3i}{2} \\ -1 - 3i & 1+\frac{i}{2} & \frac{1}{2}+\frac{3i}{2} \end{bmatrix} $ and finally i got $D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -i & 0 \\ 0 & 0 & i \end{bmatrix}$ then A is diagonalizable but my question is how can i conclude that $A$ is not similar to a diagonalize matrix over the reals? , please some help for this.
 A: *

*$A$ is diagonalisable to a matrix over $\mathbb{C}$.

*The diagonal components contain complex numbers.

*Matrix diagonals are unique (up to a permutation).


Therefore it is not diagonalisable over $\mathbb{R}$.
A: Suppose for contradiction that $A$ were diagonalizable over the reals.
There would be some real $D = \begin{bmatrix} \beta_1 & 0 & 0 \\ 0 & \beta_2 & 0 \\ 0 & 0 & \beta_3\end{bmatrix}$ similar to $A$.
Furthermore, the $\beta_i$ are real roots of $2-x +2 x^2-x^3$.
Thus, $\beta_1=\beta_2=\beta_3=2$
Hence $D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{bmatrix}$
As a result, what can you say about $\operatorname{Trace(A)}$ and $\operatorname{Trace(D)}$ ?
A: The entries on the diagonal matrix are unique, take the characteristic polynomial of $A$ and $D$ which are the same since they are similar matrices (why? $Det(A- y I ) = Det( P D P^{-1} - y I ) = Det ( P ( D - y I ) P^{-1}) = Det (P) Det (D - yI) Det (P) ^{-1}$ ), so the zeros are the same.
Now you notice that the diagonal entries of $D$ are precisely the zeros with multiplicity of the characteristic polynomial, so $p_A$ determine $D$ up to permutation. If there are some $D$ that has complex entries, no $D$ is possible with real entries.
