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I need help with this question:

Let $V$ be a finite vector space where $ \dim V = n $, over the complex numbers

and let $ T: V\to V $ be a linear transformation. Prove that $ V = \ker(T^n) \oplus Im(T^n) $

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    $\begingroup$ Is $n$ the dimension of $V$? Otherwise it's false: just consider $\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$. $\endgroup$ – Najib Idrissi Feb 9 '14 at 10:43
  • $\begingroup$ yes... thanks!! fixed it... $\endgroup$ – Tomer Amir Feb 9 '14 at 10:43
  • $\begingroup$ related: math.stackexchange.com/q/919231/173147 $\endgroup$ – glS Jun 5 '18 at 18:38
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Note: The scalar field is irrelevant for the result.

Hint: Consider the sequences $\DeclareMathOperator{\im}{im}$

$$\begin{gather}\ker T^1 \subset \ker T^2 \subset \dotsb \subset \ker T^n,\\ \im T^1 \supset \im T^2 \supset \dotsb \supset \im T^n. \end{gather}$$

The rank formula (rank-nullity) says $V = \ker T^k \oplus \im T^k \iff \ker T^k \cap \im T^k = \{0\}$. Show that if that happens, then you also have $\ker T^m \cap \im T^m = \{0\}$ for all $m > k$, and that $n$ is the latest exponent at which that can possibly happen.

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  • $\begingroup$ Could you please elaborate, I mean what if we don't assume that the iff is true? $\endgroup$ – jreing Feb 11 '14 at 10:09
  • $\begingroup$ The task is to prove that for some $1 \leqslant k \leqslant n$ you have $\ker T^k \cap \operatorname{im} T^k = \{0\}$. Translate $\ker T^k \cap \operatorname{im} T^k = \{0\}$ into a relation between the dimensions of say $\operatorname{im} T^k$ and $\operatorname{im} T^{k+1}$. $\endgroup$ – Daniel Fischer Feb 11 '14 at 10:20
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    $\begingroup$ I think the hint, while asking to show things that are true, leads to problems that are harder to answer than the original question. I can show that $\ker T^{k+1}\cap\im T^{k+1}\neq\{0\}\implies \ker T^k\cap\im T^k\neq\{0\}$, but only be showing first that the left condition is equivalent to the apparently stronger $\ker T\cap\im T^{k+1}\neq\{0\}$. And the way this helps proving the crucial claim at the end of the answer is not so clear either. $\endgroup$ – Marc van Leeuwen Feb 11 '14 at 10:57
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Note that in general, although kernel and image of an endomorphism have complementary dimensions, they need not form a direct sum. What you must show here is that for the case of $T^n$ (with $n$ the dimension of the space), kernel and image do always form a direct sum.

This can be shown by showing that the intersection of these subspaces is reduced to $\{0\}$. Suppose for a contradiction there is a nonzero vector $w\in\ker(T^n)\cap\operatorname{im}(T^n)$. Then $w=T^n(v)$ for some $v$, with obviously $v\neq0$. The condition on $w$ gives $T^{2n}(v)=0$, so $m=\min\{\,i\in\Bbb N\mid T^i(v)=0\,\}$ is well defined and $n<m\leq2n$ (the first inequality is the one that will be useful). Because the dimension of the space is only $n$, the $m$ nonzero vectors $v,T(v),T^2(m),\ldots,T^{m-1}(v)$ must be linearly dependent. Then one can write one of those vectors, say $T^i(v)$ as linear combination of the vectors after it in the list. Now apply the linear operator $T^{m-1-i}$ to this relation; this maps $T^i(v)$ to $T^{m-1}(v)\neq0$, but it kills all the vectors occurring in the linear combination giving $T^i(v)$; this is a contradiction.

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    $\begingroup$ The argument can be simplified by using the observation that $ \ker(T^{n}) = \ker(T^{2 n}) $, which is due to the fact that $$ \ker(T^{m}) = \ker(T^{m + 1}) $$ for all $ m \in \Bbb{N}_{\geq n} $. As $ v \in \ker(T^{2 n}) = \ker(T^{n}) $, we obtain $$ w = {T^{n}}(v) = 0_{V}, $$ and as $ w \in \ker(T^{n}) \cap \im(T^{n}) $ is arbitrary, we can conclude that $$ \ker(T^{n}) \cap \im(T^{n}) = \{ 0_{V} \}. $$ $\endgroup$ – Transcendental Sep 15 '15 at 23:14
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This is just an attempt to make explicit what the answer by Daniel Fischer seems to suggest. Consider $$ \{0\}=\ker(T^0)\subseteq\ker(T^1)\subseteq\cdots\subseteq\ker(T^n)\subseteq\ker(T^{n+1}) $$ There are $n+1$ symbols '$\subseteq$', so for dimension reasons they cannot all be strict inclusions. Now $\ker(T^i)=\ker(T^{i+1})$ means that there does not exist $v$ with $T^i(v)\neq 0$ and $T^{i+1}(v)=0$, and this is equivalent to $\def\im{\operatorname{im}}\im(T^i)\cap\ker(T)=\{0\}$. Since the subspace $\im(T^i)$ decreases as $i$ increases, we see that once one instance of '$\subseteq$' is nonstrict (one has equality there), the ones further to the right will be as well. It follows that the last '$\subseteq$' is certainly an equality, in other words $\im(T^n)\cap\ker(T)=\{0\}$. This means that $T$ restricted to $\im(T^n)$ is injective, and so it implies $\im(T^n)\cap\ker(T^n)=\{0\}$.

The desired result now follows using rank-nullity.

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