Suppose that $V$ is a complex vector space and $T:V\to V$ is linear.Show that $T$ has an invariant subspace of dimension $j$ for each $j=1,2,\ldots \dim V$.
What happens if $V$ is a real vector space ?
Attempt:
If $V$ is a vector space over $\Bbb C$ then the characteristic polynomial of $T$ has a root which will be an eigen value say $\lambda$ corresponding to eigen value $v_0$. Then the $\text{span}\{v_0\}$ is a $1-$ dimensional invariant subspace of $T$.
The same holds for $V$ to be a vector space over $\Bbb R$ if $\dim V$ is odd.
But I can't proceed further.Any hints will be much appreciated.
Best Answer
Over the complex numbers every linear operator can be brougth in upper triangular form. See here. You can also think about the Jordan-Normal-Form if you're familiar with it. Now in this form the matrix leaves the spaces $\{e_1\}\subset\{e_1,e_2\}\subset\ldots\subset\{e_1,\ldots,e_n\}$ invariant. By $e_i$ I mean the canonical basis.
In the real numbers this is no longer true. For example think about a rotation in 2 dimensions which doesn't leave any subspace (except zero) invariant.