# Show that $T$ has an invariant subspace of dimension $j$ for each $j=1,2,\ldots \dim V$.

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.

• Have you tried induction? – Mnifldz Jan 13 '17 at 3:45
• Tried@Mnifldz; But if I assume existence of a n-1 dimensional subspace how does it guarantee existence of $n$ dimensional subspace(invariant) – Learnmore Jan 13 '17 at 4:15

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.