Endomorphism/Real/Two-dimensional invariant/Fact/Proof

We may assume ${\displaystyle {}V=\mathbb {R} ^{n}}$, and that ${\displaystyle {}\varphi }$ is described by the matrix ${\displaystyle {}M}$ with respect to the standard basis. If ${\displaystyle {}\varphi }$ has an eigenvalue, then we are done. Otherwise, we consider the corresponding complex mapping, that is,

${\displaystyle \varphi \colon \mathbb {C} ^{n}\longrightarrow \mathbb {C} ^{n},}$

which is given by the same matrix ${\displaystyle {}M}$. This matrix has a complex eigenvalue ${\displaystyle {}a+b{\mathrm {i} }}$, and a complex eigenvector ${\displaystyle {}v\in \mathbb {C} ^{n}}$. In particular, we have

${\displaystyle {}Mv=(a+b{\mathrm {i} })v\,.}$

Writing

${\displaystyle {}v=v_{1}+{\mathrm {i} }v_{2}\,}$

with ${\displaystyle {}v_{1},v_{2}\in \mathbb {R} ^{n}}$, this means

${\displaystyle {}Mv_{1}+{\mathrm {i} }Mv_{2}=Mv=(a+b{\mathrm {i} }){\left(v_{1}+{\mathrm {i} }v_{2}\right)}=av_{1}-bv_{2}+{\mathrm {i} }(av_{2}+bv_{1})\,.}$

Comparing the real part and the imaginary part we can deduce that ${\displaystyle {}Mv_{1},Mv_{2}\in \langle v_{1},v_{2}\rangle }$. Therefore, the real linear subspace ${\displaystyle {}\langle v_{1},v_{2}\rangle }$ is invariant.