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Let a $\mathbb{R}-$vector space $V$, with $\dim V<\infty$

If exist $J:V\to V$ linear such that $J^2=-I_V=$ operator identity, then $\dim V=2n$, $n\in\mathbb{N^*}$.

How I proof this?

Kempa
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    The statement is ill-defined; what is $n$? Besides, you're expected to post your own thoughts, workings, research, etc.; as stated, this comes off as a "do my homework for me" post. – PrincessEev Jun 30 '22 at 03:02
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    What is $I$, and what have you tried? Try reformulating your question, including your work so far. – Bertrand Wittgenstein's Ghost Jun 30 '22 at 03:02
  • I search at this site something to help me, and i fail. I tried did this question (is a lot questions of like that that i did lately), and this one I cant did it, so I'm here no to ask for "do my homework", but for a help. – Kempa Jun 30 '22 at 03:09
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    Take determinants of both sides. Can a real number square to $-1$? I'm certain this is a duplicate, though too lazy to find a dupe. Close-voters might want to look for one. – anon Jun 30 '22 at 03:15
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    The minimal polynomial has degree $2$? So the characteristic polynomial is even degree. – suckling pig Jun 30 '22 at 03:21
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    See https://math.stackexchange.com/questions/1572449/prove-that-the-dimension-of-v-is-even-and-that-l-dim-v-2-0 also https://math.stackexchange.com/questions/2025678/there-is-no-real-3-times3-matrix-a-for-which-a2-i – Gerry Myerson Jun 30 '22 at 03:32
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    ...and https://math.stackexchange.com/questions/75963/question-about-negative-matrix-involution and https://math.stackexchange.com/questions/2007643/let-a-be-an-n-times-n-matrix-with-real-entries-such-that-a2-i-0-then and probably several more. – Gerry Myerson Jun 30 '22 at 03:39

1 Answers1

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The hypothesis

$J^2 = -I \tag 1$

implies the eigenvalues of $J$ lie in the set $\{i, -i\}$, for if

$Jv = \lambda v, \tag 2$

then

$-v = -Iv = J^2v = J(Jv) = J(\lambda v) = \lambda Jv = \lambda(\lambda v) = \lambda^2 v, \tag 3$

from which we infer that

$\lambda^2 = -1, \tag 4$

and thus that

$\lambda \in \{i, -i\}. \tag 5$

Now if $\dim V$ is odd, then the characteristic polynomial of $J$ has at least one real root, since it is of odd degree. But this contradicts our work above; hence $\dim V$ must be even.

Robert Lewis
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