2

I came across this question in an exercise in my linear algebra course and am at a loss as to how to solve it.

Let $V$ be a complex vector space. Let $\mathcal{J}:V \to V$ be a linear map such that $$\mathcal{J}^2=-I_V$$ Show that $V=V_{+i} \oplus V_{-i}$, where $$V_{\pm i}=\{v \in V \mid \mathcal{J} v=\pm i v\}$$ Here $i=\sqrt{-1}$.

I initially began by proving that $V_{+i} \cap V_{-i} = \{0\}$ and subsequently following it up with a proof that for all $v \in V$, $v = v_1 +v_2$ where $v_1 \in V_{+i}$ and $v_2 \in V_{-i}$. However, I was unable to fully prove the latter statement that $V = V_{+i} + V_{-i}$.

A natural step would be to define a map $\mathcal{K} = \frac{1}{2}(\mathcal{J} + iI_V)$, proving that $\ker(\mathcal{K}) = V_{-i}$ and that $\mathrm{im}(\mathcal{K}) = V_{+i}$ and then considering $v - i\mathcal{K}v + i\mathcal{K}v$. However, that line of reasoning hit a dead end since I wasn't able to reach the conclusion.

Can someone guide me to the correct proof?

Lt. Commander. Data
  • 1,176
  • 8
  • 27
  • There's nothing to continue. Once you've shown $V_i\cap V_{-i}=0$ and $V=V_i+V_{-i}$ that means $V=V_i\oplus V_{-i}$. Did you mean to imply you haven't yet shown $V_i\cap V_{-i}=0$ or $V=V_i+V_{-i}$? – anon Dec 27 '20 at 04:55
  • @runway44 I meant that I haven't been able to show that $V = V_{-i} + V_{+i}$. – Lt. Commander. Data Dec 27 '20 at 04:57
  • See an answer at https://math.stackexchange.com/questions/3945897/why-it-is-important-to-write-a-function-as-sum-of-even-and-odd-functions/3945918 for some inspiration. – KCd Dec 27 '20 at 05:29

1 Answers1

1

Suppose $v\in V_i\cap V_{-i}$. Then $Jv=iv=-iv$ and dividing $2iv=0$ gives $v=0$.

Note that $J$ is invertible. Thus, given any $v\in V$ there is a $w\in V$ for which $v=Jw$, and thus

$$ v=Jw=\frac{1}{2}\big(J+i\big)w+\frac{1}{2}\big(J-i\big)w. $$

Show that the image of $\frac{1}{2}\big( J\pm i \big)$ is contained in $V_{\pm i}$.

anon
  • 151,657