Show Diagonal Matrix is not possible Consider the real matrix $\alpha = \begin{pmatrix}
7 &3  &-4 \\ 
 -2&-1  &2 \\ 
 6&2  &-3 
\end{pmatrix}$. Show/explain that there does not exist a invertible real matrix $\beta$, so that $\beta ^{-1} \alpha \beta$ is a diagonal matrix
My effort so far
I know that to diagonalize a matirx $A$
Find the eigenvalues of $A$ using the characteristic polynomial.
For each eigenvalue $λ$ of $A$ , compute a basis $B λ$ for the $λ$-eigenspace.
If there are fewer than $n$ total vectors in all of the eigenspace bases $B λ$ , then the matrix is not diagonalizable. I am losing it when it comes to the specific case "Show/explain that there does not exist a invertible real matrix $\beta$, so that $\beta ^{-1} \alpha \beta$ is a diagonal matrix". How would I go about this one in question?
 A: The characteristic polynomial of $\alpha$ is $(X^2+1)(X-3)$ therefore $i$ and $-i$ are eigenvalues of $\alpha$ and thus $\alpha$ is not diagonalizable in $\mathbb{R}$ (but is in $\mathbb{C}$).
A: To address your specific question at the end of your post:
You may go about it by first assuming that such a $\beta$ exists and lead this assumption to a contradiction.
But first of all, it seems your question is about whether the given matrix can be diagonalized over $\mathbb{R}$. Because, over $\mathbb{C}$ you will find $3$ linearly independent (complex) eigenvectors.
Now, back to your question:
Assume there is an invertible $\beta = (b_1\: b_2\: b_3)$ with columns $b_1,b_2,b_3 \in \mathbb{R}^3$.  s.t.
$$\beta^{-1}\alpha\beta=\operatorname{diag}(\lambda_1,\lambda_2,\lambda_3)\Leftrightarrow
\alpha\beta=\beta\operatorname{diag}(\lambda_1,\lambda_2,\lambda_3)$$
But since $$\beta\operatorname{diag}(\lambda_1,\lambda_2,\lambda_3) = (\lambda_1b_1\: \lambda_2b_2\: \lambda_3b_3)$$ this means
$$\alpha(b_1\: b_2\: b_3)=(\lambda_1b_1\: \lambda_2b_2\: \lambda_3b_3)$$
Hence, the columns of $\beta$ form a basis of real eigenvectors of $\alpha$. But since you find only one linearly independent real eigenvector of $\alpha$, such a $\beta$ does not exist.
