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

Proof

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

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

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

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

Writing

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

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

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

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

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