Let V be a finite-dimensional complex vector space, and $T\in L(V)$. Let $W=Res_{\mathbb C/\mathbb R}V$ be the Restriction of $V$. Let S=$Res_{\mathbb C/\mathbb R}T\in L(W)$ be the induced operator on W. Suppose that $T$ has $\lambda$ as a non-real eigenvalue. Prove that $S_C$ has both $\lambda$ and $\overline \lambda$ as eigenvalues
The above is the problem I was trying to solve, but I don't understand the meaning of "restriction of V".
I know that complex vector space can be regarded as the "complexification" of a real vector space.
For example, suppose $V$ is a real vector space over the real number field $\mathbb R$, and $ V \times V=V_{ C}$ can be regarded as the "complexification" of $V$, so $V_C$ is regarded as a complex vector space on a field $\mathbb C$.
Then as so far, my understanding of this "restriction" of a complex vector space $V_C$ over $\mathbb C$ means that $V_C$ is "degrade" back to $V$ and make $V$ to be a real vector space over $\mathbb R$. And then in $V_C$, all vectors are in the form of $u+iv$ and can be scaled by complex numbers such as $a+bi$, but after the restriction, the vectors in $V$ are just $u,v$ and so on and can only be scaled by real numbers. Is this right?
Best Answer
I understand your quandry... but, first, if $V'$ is a complex vector space obtained by complexifying a real vector space $V$, it is not the case that the "restriction of scalars" of $V'$ is $V$ again. It is isomorphic to a direct sum of two copies. The operations of "complexification" and "restriction of scalars from complex to real" are not inverse, but "adjoint", in a categorical sense.
It is structurally better (though not more explanatory at a first pass) to say that the complexification of a real vector space $V$ is $V\otimes_{\mathbb R}\mathbb C$, rather than more ad-hoc, coordinate-dependent descriptions.
In particular, to be clear, if we repeatedly complexify-restrict-complexify-restrict we'll double the real dimension in each cycle of complexify-restrict.