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?
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?
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.