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Suppose that $V$ is a real vector space and $T:V\to V$ is a linear operator having no eigen values.

Prove that every subspace of $V$ invariant under $T$ has even dimension.

Attempt: If $V$ has dimension odd so the characteristic polynomial of $T$ also has odd degree and hence has a real root which is false

Hence $\dim V$= even.

But that's not helping me here,I should deal with invariant subspace here.But how should I do it?

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    Let $W$ be an invariant subspace. What do you know about $T\lvert_W \colon W \to W$? – 2017-01-12
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    I don't get what we know about $T|_W$? – 2017-01-12

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If $W$ is invariant by $T$, $T_{\mid W}$ , the restriction of $T$ to $W$ does not have real eigenvalues too. Suppose that its dimension is odd, the degree of the characteristic polynomial $P_W$ of $T_{\mid W}$ is odd, so it has a real root, contradiction, since a root of $P_W$ is an eigenvalue.

Write $P_W=a_0+a_1x+..+a_nx^n$, if $a_n>0$ $lim_{x\rightarrow-\infty}P_W(x)=-\infty$ and $lim_{x\rightarrow+\infty}P_W(x)=+\infty$ or the converse if $a_n<0$. So IVT theorem implies there exists $x$ such that $P_W(x)=0$.

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    Why it does not have real eigen values? – 2017-01-12
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    if $n$ is the degree of $P_W$ and $a_n>0$ $lim_{x\rightarrow-\infty}P_W(x)=-\infty$ and $lim_{x\rightarrow+\infty}P_W(x)=+\infty$ or the converse if $a_n<0$. – 2017-01-12