Why is this diagonal matrix not possible over the reals Let $\alpha = \begin{pmatrix}
7 &3  &-4 \\ 
 -2&-1  &2 \\ 
 6&2  &-3 
\end{pmatrix}$ over the reals.
Show that there does not exist a invertible real matrix $\beta$, so that $\delta = \beta ^{-1} \alpha \beta$ is a diagonal matrix
My "proof"
the characterisitc polynomial is $-\lambda ^3 + 3\lambda ^2 -\lambda + 3=-(\lambda - 3)(\lambda ^2 +1)$. The solutions would then be $3,i,-i$. Now, we have that $\delta = \beta ^{-1} \alpha \beta$ where the diagonal values of $\delta$ would then be eigenvalues of $\alpha$. if both $\alpha , \beta$ are real $3 \times 3$-matrices then $\delta$ must also be a real $3 \times 3$-matrix. But that is not possible since the eigenvalues of $\alpha$ are  $3,i,-i$ and thus $\beta$ must be complex.....
Now I am stuck since I would think I would have to show that the diagonal of $\delta$ must be the eigenvalues of $\alpha$. Is this a generel fact or should this be proven? If so - how?

The characteristic polynomial is satisfied by $\alpha$. $$(\alpha-3I)(\alpha^2+I)=0$$ This is still true if $\alpha$ is replaced by the diagonal matrix $\delta$ $$(\delta-3I)(\delta^2+I)=\beta^{-1}(\alpha-3I)(\alpha^2+I)\beta=0$$ If $\delta$ consists of the diagonal terms $x$, then $$(x-3)(x^2+1)=0$$ Over the reals, this only gives $x=3$ as a solution, so $\delta=3I$ and $\alpha=\beta(3I)\beta^{-1}=3I$, which is not the case.
