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

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?

• Let $W$ be an invariant subspace. What do you know about $T\lvert_W \colon W \to W$? Jan 12, 2017 at 17:12
• I don't get what we know about $T|_W$? Jan 12, 2017 at 17:19

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 $P_W(x)=0$.
• 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 $a_n<0$. Jan 12, 2017 at 17:22