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?
Best Answer
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 $P_W(x)=0$.