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I have a big problem with this problem... :

If $V_m(\mathbb{R})$ is a vector space whose dimension is "$m$" then

  • Proving that "$m$" is even number if and only if exist an endomorphism $J$ of $V_m(\mathbb{R})$ where $J^2=J\cdot J=-identity$.
  • Use this endomorphism to proving that $V_m(\mathbb{R})$ is a complex vector space with dimension half of "$m$", namely, $ \frac{m}{2}$

Please Help!

Willyf
  • 375

4 Answers4

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If $m=2n$ is even, then consider the map $$(x_1,x_2,\ldots,x_{2n-1},x_{2n})\mapsto (x_2,-x_1,\ldots,x_{2n-1},-x_{2n}).$$ This will give you the $J$ map.

Now if you are given such a $J$, then $$\det(J^2) =\det(J)^2 = (-1)^m.$$

Since this is a real vector space, we can deduce that $m$ is even since $(-1)^m$ is a square.

The second question is simple once you see the $J$ map above and already answered.

Spencer Leslie
  • 2,662
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Assume $m=2n$. Then $V$ is isomorphic to the real vector space $\mathbb C^n$, where the map "multiply with $i\in\mathbb C$" is our $J$.

On the other hand assume that such $J$ exists. Then $V$ becomes a $\mathbb C$ vector space via $(a+ib)v:=av+bJv$ (check that). Then $\dim_{\mathbb R}V=2\dim_{\mathbb C}V$ must be even.

Hagen von Eitzen
  • 374,180
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If $m$ is even, then we can construct the endomorphism $$ J = \pmatrix{J_0\\&J_0\\ &&\ddots \\&&&J_0} $$ Where $$ J_0 = \pmatrix{0&-1\\1&0} $$ On the other hand, suppose that a $J$ exists such that $J^2 = -I$. Deduce that $J$ has no real eigenvalues. However, if $m$ is odd, then every endomorphism on $V_m$ has a real eigenvalue. We therefore deduce that $m$ is even.

For the last part, we can show that if we define complex multiplication on $V_m$ by $$ iv := Jv $$ Then $V_m$ is a complex vector space.

Ben Grossmann
  • 225,327
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If $m=2$ we can take $J$ to be a rotation by $\pi$. Now assume by induction that for vector spaces $V$ of even dimension $\leq m$ we can find $J:V\to V$ with $J^2 = -I$. Now let $W$ be of dimension $m+2$. Pick a basis $(e_1,\ldots,e_m,e_{m+1},e_{m+2})$. Then the span $S_1$ of $(e_1,\ldots, e_m)$ and $S_2$ of $(e_{m+1},e_{m+2})$ are both of even dimension $\leq m$, so we can choose $J_1:S_1\to S_1, J_2:S_2\to S_2$ with $J_1^2 = I_{S_1}, J_2^2 = I_{S_2}$. Extend $J_1,J_2$ to all of $W$ by making them fix the basis vectors which have not yet been specified in their definitions. Then $J:= J_1J_2$ satisfies $J^2=-I_{W}$ since $J_1,J_2$ commute.

Thus we have shown that if $m$ is even, we can find $J$ with $J^2=-I$. Next suppose that we have a $J^2 = -I$. Taking the determinant of both sides gives $(\det J)^2 = (-1)^m$. Since the left side is positive, it must be that $m$ is even.

You can now finish the rest of the problem showing isomorphism to $\mathbb{C}^{m/2}$ in the obvious way when $m$ is even.

nullUser
  • 27,877