Corresponding eigenvalues complex/real endomorphism

Question

Let $A:\mathbb{C}^n\rightarrow \mathbb{C}^n$ be a complex linear map that we write $A=A_1+iA_2$. Under the identification $\mathbb{C}^n\simeq\mathbb{R}^{2n}$ we can see $A$ as a real endomorphism $$B=\begin{pmatrix} A_1 & -A_2\\ A_2 & A_1 \end{pmatrix}.$$ How can I relate the (complex) spectrum of $A$ with the one of $B$?

Answer 1

If $V$ is a complex vector space, then $\mathbb{C}\otimes_\mathbb{R} V$ (considered as a complex vector space via the first coordinate) is naturally isomorphic to $V\oplus \overline{V}$, where $\overline{V}$ is $V$ with its scalar multiplication conjugated. Explicitly, this isomorphism is the map $\varphi:\mathbb{C}\otimes_\mathbb{R} V\to V\oplus \overline{V}$ given by $\varphi(z,v)=(zv,\overline{z}v)$ (where $\overline{z}v$ is using the original scalar multiplication of $V$, not the conjugated one of $\overline{V}$); this is easily verified to be an isomorphism when $V$ is 1-dimensional and then the general case follows since every vector space is a direct sum of 1-dimensional vector spaces.

Applying this to $V=\mathbb{C}^n$, it follows that the complexification of $B$ is just the direct sum of $A$ and the conjugate $\overline{A}$. It follows that the spectrum of $B$ is the union of the spectrum of $A$ and its conjugate.

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Here's another way to think about it. Let $J:\mathbb{R}^{2n}\to\mathbb{R}^{2n}$ be the multiplication by $i$ map, when we identify $\mathbb{R}^{2n}$ with $\mathbb{C}^n$. Note that $BJ=JB$ since $A$ is $\mathbb{C}$-linear. Let $\lambda=a+bi\in\mathbb{C}$ and write $p(x)=(x-\lambda)(x-\overline{\lambda})$. Since $B$ has real entries, $\lambda$ is an eigenvalue of $B$ iff $p(B)$ is not invertible. Now observe that $$(B-(a+bJ))(B-(a-bJ))=p(B)$$ since $p$ has real coefficients (so all the $J$s on the left-hand side will end up cancelling and all that matters is that $J^2=-1$ and $J$ commutes with $B$). Thus $\lambda$ is an eigenvalue of $B$ iff either $B-(a+bJ)$ or $B-(a-bJ)$ is not invertible. But when we think of our vector space as $\mathbb{C}^n$ instead, $B-(a+bJ)$ is just $A-\lambda$ and $B-(a-bJ)$ is just $A-\overline{\lambda}$. Thus $\lambda$ is an eigenvalue of $B$ iff either $\lambda$ or $\overline{\lambda}$ is an eigenvalue of $A$.