Is semi-simplicity for real matrices equivalent with diagonalizable over $\mathbb{C}$?

Question

I know that for the complex matrices $\mathbb{C}^{n \times n}$, being semi-simple is equivalent with being diagonalizable since $\mathbb{C}$ is algebraically closed. I was wondering if something of this nature can be said about semi simple elements in $\mathbb{R}^{n \times n}$. For example that the semi-simple elements in $\mathbb{R}^{n \times n}$ are exactly the matrices that are diagonalizable over $\mathbb{C}$. If this is not true, a counter example would be nice.

Answer 1

It's important to say what exactly we mean when we say that an element is semisimple. I'll use the definition given here: a matrix $A$ is semisimple if every $A$-invariant subspace $U \subset \Bbb R^n$ has an $A$-invariant complement $U'$. Under this defintion, it is indeed the case that if $A$ is semisimple over $\Bbb R$, then it must be diagonalizable over $\Bbb C$ and vice versa. That is, $A$ is semisimple over $\Bbb R$ iff it is also semisimple over $\Bbb C$.

The proof that if $A$ is diagonalizable over $\Bbb C$ then it must also be semisimple over $\Bbb R$ is relatively simple and is "left as an exercise to the reader".

The other implication can be proved as follows. Suppose that $A$ is semisimple over $\Bbb R$. Consider a minimal $A$-invariant subspace $U$. Note that every minimal $A$-invariant subspace $U$ must satisfy one of the following conditions:

1. $U$ is of of dimension $1$ and satisfies $U \subset \ker(A - \lambda I)$ for some $\lambda \in \Bbb R$,
2. $U$ is of dimension $2$ and satisfies $U \subset \ker p(A)$ for some irreducible quadratic $p(x)$.

In case $1$, let $u_1$ span $U$, and let $U'$ be an $A$-invariant complement of $U$. Consider a basis $\mathcal B = \{u_1,\dots,u_n\}$ of $\Bbb R^n$ for which $U' = \operatorname{span}\{u_2,\dots,u_n\}$. The matrix of $A$ relative to this basis has the form $$ [A]_{\mathcal B} = \pmatrix{\lambda &0\\0&A'}. $$

In case $2$, let $u$ be a non-zero element of $U$. Suppose that $U \subset \ker p(A)$ with $p(x) = x^2 - bx - c$. We find that the vectors $u_1 = u$ and $u_2 = A u$ span $U$. Let $U'$ be an $A$-invariant complement of $U$. Consider a basis $\mathcal B = \{u_1,\dots,u_n\}$ of $\Bbb R^n$ for which $U' = \operatorname{span}\{u_3,\dots,u_n\}$. The matrix of $A$ relative to this basis has the form $$ [A]_{\mathcal B} = \pmatrix{C &0\\0&A'}, $$ where $C$ is the matrix $$ C = \pmatrix{0&c\\1&b}. $$ Because $p$ is an irreducible quadratic, we find that $C$ must be diagonalizable over $\Bbb C$.

In either case, we can proceed inductively to show that $A$ must be diagonalizable over $\Bbb C$.

---

The matrix $C$ of case 2 is maybe a bit "more familiar" if we choose our basis differently. Suppose that $p(x) = (x - h)^2 + k^2$. If we select $u_1 = u$ and $u_2 = hu + kAu$, then we instead end up with the matrix $$ C = \pmatrix{h&-k\\k&h} $$ which is the matrix corresponding to multiplication in $\Bbb C \cong \Bbb R^2$ by $h + ik$.