Does $\mathbb{C}^n$ "contain" $\mathbb{R}^n$?

Question

I have a question of the form "Let $T : V → V$ be a normal operator, where $V$ is a finite dimensional inner product space over $\mathbb{R}$. Show that if $T^3(v) = 0$ for some $v \in V$, then $T(v) = 0$." from an exam. Now , i know that if $T$ is normal , on a complex vector space , matrix of $T$ for some basis can be diagonalized by Spectral Theorem. But the question requires the space is real. Now , my question is that if $C^n$ can be seen as containing $\mathbb{R}^n$ , then by extending diagonal matrix has zero eigenvalues as $T^3$ having a zero output requires , i think , all diagonal entries are zero , hence $T$ is zero linear map. Thanks

Answer 1

To answer your title question, $\Bbb{C}^n$ contains $\Bbb{R}^n$ and in some sense this containment is unique: there is a unique field homomorphism $\Bbb{R}\to\Bbb{C}$, and this may be used for each coordinate.

For your application, there are subtleties. What you can certainly do is consider the matrix of $T$ and let that matrix act on $\Bbb{C}^n$, and diagonalize it using the spectral theorem. In general, this cannot be converted back to a diagonalization over $V$, since some eigenvalues may be complex. You can still make this work along the lines you are suggesting, but you are right to be careful!

(Edit: Actually, it is not true that if $T^3$ has a single zero output, that it is necessarily the zero map (much less $T$). But this shouldn't be needed.)