Let $A$ is a real square matrix. If $A$ has an eigenvalue equal to $1$ and has algebraic multiplicity $1$ then $A$ is diagonalizable. 
Let $A$ be a square matrix of order $n$ over $\mathbb R$. If $A$ has an eigenvalue equal to $1$ and has algebraic multiplicity $1$ then $A$ is diagonalizable (True\False).

My try -
It is false because we can use the counter example
$A= \left(\begin{matrix}
1 & 0 & 0 \\
0 & 0 & -1 \\
0 & 1 & 0 \\
\end{matrix} \right)
$
The characteristic polynomial is $\det(\lambda I-A)= \left|\begin{matrix}
\lambda -1 & 0 & 0 \\
0 & \lambda & 1 \\
0 & -1 & \lambda \\
\end{matrix} \right|
$
$p(\lambda)=(\lambda -1 ) (\lambda^2 +1)$ so the eigenvalue is equal to $1$ and we have only one algebraic multiplicity but we cannot factorize the characteristic polynomial so it is not diagonalizable
My questions are:

*

*is my way correct?


*are there any other ways?
It is important for me to know as much ways as possible because for example here it took me a long time to figure this matrix out and it was mostly guessing randomly.
Thank you!
 A: If the claim was true then for any $\lambda$ different from $0$ , and for any square matrix $A$, if $\lambda$ has multiplicity $1$ then $A$ would be diagonalizable.
This is so, because one would look at the matrix $\frac{1}{\lambda}A$ and use the claim. Of course if $\frac{1}{\lambda}A$ is diagonalizable then so is $A$. In particular, for any $A$, if its characteristic polynomial has a single root different from $0$ it would be diagonalizable. Now, if $A$ has $0$ as it characteristic polynomial root, then $A$ can be approximated by another matrices without $0$ as characteristic polynomial root and single roots (this is so because roots are continuously depended on matrix entries) and if each approximation is diagonalizable then, since everything is in finite dimensional space and bounded , hence compact, we can find sub-sequence of diagnoalizable matrices which is convergent and it must be convergent to diagonalization of $A$. So we would have following claim - if $A$ has a single root it is diagonalizable. But then arguing as above, i.e approximating ANY matrix by matrices with simple roots we would get to conclusion that $A$ must be diagonalizable. In words, any matrix $A$ is diagonalizable. Which is of course not the case.
