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?

What you said is a general fact and your proof is correct.
If any matix $\alpha$ is similar to a diagonal matrix $\delta=\text{diag}[\delta_{11},\delta_{22},\ldots,\delta_{nn}]$ i.e. $\exists$ an invertible matrix $B$ such that $\delta=B^{-1}\alpha B$, the diagonal entries of $\delta$ must be the eigenvalues of $\alpha$. You can prove this fact easily.
We have $\alpha B=B\delta$. Let the columns of $B$ be $B_i$. Then$$\alpha B=\alpha[B_1~B_2~\ldots B_n]=[\alpha B_1~\alpha B_2\ldots\alpha B_n]$$and$$B\delta=[\delta_{11}B_1~\delta_{22}B_2\ldots\delta_{nn}B_n]$$Equating the columns we get $\alpha B_i=\delta_{ii}B_i$, i.e. $B_i$ are eigenvectors of $\alpha$ and $\delta_{ii}$ their corresponding eigenvalues.
