Equivalent condition for diagonalizable in real matrix I found some equivalent statement of diagonalizable in a complex case from sufficient and necessary conditions for matrix to be diagonalize or triangular answered by @SheldonAxler. 
After seeing this I want to know some equivalence statement of diagonalizable in real case. 
I know 

$A \in M_{n}(\mathbb{R})$ is a diagonalizable iff $A$ has $n$ linearly independent eigenvectors. 

So for $A\in M_n(\mathbb{R})$, If $A$ has $n$ distinct eigenvalues then $A$ is diagonalizable.   But we know that in real case even though $A$ has some multiple roots, it can be diagonalizable. 
For example 
\begin{align}
  A = \begin{pmatrix}
     4 & 2 & 2 \\
     2 & 4 & 2 \\
     2 & 2 & 4 
  \end{pmatrix}, Q =   \begin{pmatrix}
    -\frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\
    \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\
    0 & \frac{2}{\sqrt{6}} & \frac{1}{\sqrt{3}} 
  \end{pmatrix} 
\end{align}
then $D = Q^{-1} A Q$ with $D = \operatorname{diag}(2,2,8)$
Please let me know some equivalent statement of diagonalization of a matrix(or transformation)
 A: "[So, for example, rank()= can be also equivalent statement] " is mistaken. Consider $n = 2$ and the matrix
$$
\pmatrix{0 & -1 \\ 1 & 0}
$$
which has rank $2$, but has no (real) eigenvectors, because its characteristic polynomial is 
$$
p(x) = x^2 + 1
$$
whose roots are $x = \pm i$. 
Similarly, the matrix 
$$
\pmatrix{1 & 0 \\ 0 & 0}
$$
has rank $1$, hence is not full-rank, but is nonetheless diagonalizable (indeed, it's already diagonal!). So "full-rank" is pretty much unrelated to diagonalizability over the reals. 
In fact ... diagonalizability has a name because it characterizes something different from all those other conditions. 
One often-useful case for real matrices is that symmetric matrices are diagonalizable. 
Using Jordan normal form, it's possible to see that if $p$ and $m$ are the characteristic and minimal polynomials, respectively, for a matrix $B$, then $B$ is diagonal if and only if all factors of $p$ are linears, and they appear in $m$ only to the first power. But that's hardly ever useful in practice. 
