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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.

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

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