invariant spaces I have a linear transformation $T: \mathbb R^n \to \mathbb R^n $ which is invertible.
There is also an invariant subspace $W$ that implies $\dim W = k $ 
I need to prove that there is an invariant subspace that its dimension is $n-k$
Any suggestions? I think that I need to find basis that $T$ is represented by blocks diagonal matrix but I do not know how to do it. thanks for helpers!
 A: Consider an $n \times n$ matrix $A$ in Jordan canonical form.  Then you get invariant subspaces of any dimension $k$ from $1$ to $n$ by requiring the last $n-k$ entries  to be $0$.
Unfortunately you're working in ${\mathbb R}^n$ instead of ${\mathbb C}^n$, so things are more complicated.  But you might be able to work with the real Jordan Form (see http://en.wikipedia.org/wiki/Jordan_canonical_form#Real_matrices)
A: Let $(w_1,w_2,\ldots,w_k)$ be a basis of $W$.  And let $v_{1}\not\in W$.  We can create an invariant subspace with basis $v_1,T(v_1),T^2(v_1),\ldots,T^{r_1}(v_1)$ for some $r$.  In such a manor we can create a basis of $\mathbb{R}^{n}$ of the form $(w_1,\ldots,w_k,v_1,\ldots,T^{r_1}(v_1),\ldots,v_m,\ldots,T^{r_m}(v_m))$  Then span$(v_1,\ldots,T^{r_1}(v_1),\ldots,v_m,\ldots,T^{r_m}(v_m))$ is an invariant subspace with dimension $n-k$
Below is a hint that I gave earlier that doesn't lead anywhere.
Then$(T(u_1),T(u_2),\ldots,T(u_k))$ is basis of $W$ (because $T$ is invertable and $W$ is invarant) and can be extended to a basis of $\mathbb{R}^n$, say $(T(u_1),T(u_2),\ldots,T(u_k),v_1,v_2,\ldots,v_{n-k})$.  Then what can we say about span$(T^{-1}(v_1),T^{-1}(v_2),\ldots,T^{-1}(v_{n-k}))$?
