# 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)

• "[For complex case converse is also true from fundamental theorem of algebra]" is wrong: consider your real example as $\mathbb{C}$ valued matrices... Mar 16, 2020 at 15:32
• @Frobin, Thanks for the comment. From answer from link, I notice that $A$ is diagonalizable iff its minimal polynomial has no repeated roots. Does that mean if there are $n$ distinct eigenvectors then $A$ is diagonalizable? Mar 16, 2020 at 15:37
• You have to be clear on what distinct means. for instance if you have one eigenvector $v$, you can construct an infinity of eigen vectors $\lambda v$ for $\lambda\in\mathbb{R}$ (for instance). What you realy want (and what you said in your question) is a basis of eigenvetors. (Or equivalently $n$ linear independant eigenvectors). Mar 16, 2020 at 15:42

"[So, for example, rank(𝐴)=𝑛 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.
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.