How to prove this operator polynomial always havs even dimensional eigenspace 
Suppose $V$ is a finite-dimensional real vector space, and T doesn't have any eigenvector on $V$. Then let $P(x)$ be a nonconstant,real,monic polynomial on $V$. Prove that all the eigenspaces of $P(T)$ are even-dimensional.

My idea is since $T$ doesn't have any eigenvector, this means that the vector space $V$ must be even, and so the complexification of $V$ which is $V_C$ also has an even dimension.
Then since every finite-dimensional complex vector space can be decomposed to the direct sum of several generalized eigenspaces. Then
$$ V_C= G(\lambda_1,P(T_C))\oplus\cdots\oplus G(\lambda_m,P(T_C))$$
where $P(T_C)$ is the complexification of operator polynomial $P(T)$
It's clear that if $P(T)$ doesn't have any eigenvector, it has zero-dimensional eigenspace, so the dimension is even
Suppose $P(T)$ at least has one eigenvector. Then since $P(T), $ is on a real vector space $V$, its eigenvalue should be real, and assume it is $\mu$, so $P(T_C)$ should also have this real eigenvalue. Thus:
$$ V_C= G(\lambda_1,P(T_C))\oplus\cdots\oplus G(\lambda_m,P(T_C))\oplus G(\mu,P(T_C))$$
It's also clear that $G(\lambda_1,P(T_C))\oplus\cdots\oplus G(\lambda_m,P(T_C))$ will have even dimension since for complex eigenvalue comes in pair. Thus $\dim G(\mu,P(T_C))$ is even-dimensional due to $\dim V_C$ is even
Then I totally don't know how to continue, since I don't know how this generalized eigenspace is even-dimensional can deduce that the eigenspace of $P(T)$ is even-dimensional
I also tried to use induction. Suppose $\dim E(\mu, P(T)=1$,it's clear that $T$ preserves $\dim E(\mu, P(T)$. Thus $$ \forall v\in E(\mu, P(T)), Tv=\text{span}(v) $$
This implies that if $\dim E(\mu, P(T)=1$, $T$ must have an eigenvector which is a contradiction. Then suppose the dimension of $\dim E(\mu, P(T)$ is $2n-1$ and $T$ has eigenvectors. Then supposed when $\dim E(\mu, P(T)=2n+1$ I need to prove that at this time $T$ still has (at least one) eigenvector, but I failed to do this.
Thus any helps on this? Thanks!
 A: By the direct sum decomposition on complex vector space, one can get $$ V_C=G(\lambda_1,T_C)\oplus G(\lambda_2,T_C)\oplus \cdots \oplus G(\lambda_m,T_C)$$
and since $T$ doesn’t have any real eigenvalue, the $\lambda_k$ here are all nonreal
Consider the operator $P(T_C)$
Then the decomposition can be rewritten as $$ V_C=G(P(\lambda_1),P(T_C))\oplus G(P(\lambda_2),P(T_C))\oplus \cdots \oplus G(P(\lambda_m),P(T_C))$$
If for all $P(\lambda_k)$ are all nonreal, then $P(T)$  has zero(even) dimensional eigenspace
Then suppose there exists $P(\lambda)$ is real . Then $P(\overline \lambda)=P(\lambda)$ is also real.
For eigenvalue $\lambda$ , suppose $T_C$ has eigenvectors $u_1+iv_1,...,u_j+iv_j\in V_C$ are linearly independent eigenvectors relating to $\lambda$ , and hence $u_1+iv_1,...,u_j+iv_j$ are linearly independent eigenvectors of $P(T_C)$ with respect to real eigenvalue $P(\lambda)$
Similarly, for eigenvalue $\overline \lambda$ ,  $T_C$ will have linearly independent eigenvectors
$u_1-iv_1,...,u_j-iv_j$ since nonreal eigenvalues and eigenvectors comes in pairs.
Thus $u_1-iv_1,...,u_j-iv_j$ are linearly independent eigenvectors of  $P(T_C)$ with respect to eigenvalue $P(\overline \lambda)$
Thus the vector list $(u_1+iv_1,...,u_j+iv_j,u_1-iv_1,...,u_j-iv_j)$ is linearly independent with even length and each of them is a eigenvector of $P(T_C)$ with respect to a real eigenvalue $P(\lambda)$ , so $\dim E(\lambda,P(T_C))$ is even. $\dim E(P(\lambda) ,P(T))=\dim E(P(\lambda),P(T_C))$. The $\dim E(P(\lambda),P(T))$ is even. This result can be generalized to any real eigenvalue of $P(T)$
