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Let $V$ be a vector space over $\mathbb C$ with dimension $n$.

Let $T:V\to V$ be a linear transformation. I tried to show that $V=\text{KerT}^n \oplus \text{Im}T^n$ .

This is my solution, would like you to verify it:

Since $T$ is over $\mathbb C$, we know that there exists a jordan form $J$ for which $T=M^{-1}JM$.

Therefore $T^n = M^{-1}J^nM$. So I want to show that $V=\text{KerJ}^n \oplus \text{Im}J^n$.

$J^n$ is just each jordan block $n$'th power. Therefore, if the eigenvalue of the block is $\lambda \ne 0$ then the block stays inversible, having a rank of the size of the block. If the eigenvalue of the block is $\lambda = 0$ then since it is the $n$'th power, and because the block is nilpotent, the block becomes $0$.

Now, since each block operates on a different subspace in $V$, we get that for all blocks with eigenvalues $\ne 0$ the relevant subspace is in $\text{Im}T$, and for all blocks with eigenvalues $=0$ the relevant subpace is in $\text{Ker}T$. This is why they sum $\oplus$, and by using the dimensions equations it is easy to show that $\text{dim}V=\text{dimKer}T + \text{dimIm}T$.

Is the direction right? I know it is not very formal. Would also love to see other ideas on this, but mainly comments on my solution, Thanks!

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Your solution has the right idea, but the last two paragraphs need a little work.

Here is one idea:

The Jordan form decomposes $V$ into a product of $J$ invariant subspaces $V= V_1 \oplus \cdots \oplus V_m $ and each subspace $V_k$ is invariant under the corresponding Jordan block $J_k$.

Pick some $x \in V$ and write $x = x_1+\cdots x_m$ with $x_k \in V_k$.

If $J_k$ is invertible, let $v_k = (J_k^n)^{-1} x_k$ and $n_k = 0$.

If $J_k$ is not invertible (and hence $J_k^n = 0$) let $v_k = 0$ and $n_k = x_k$.

Let $v = v_1+\cdots+v_k, n = n_1+\cdots+n_k$. Note that $n \in \ker J^n$.

Then $x = J^n v + n$ is a suitable decomposition.

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  • $\begingroup$ thanks! so Since I have a decomposition for every $x\in V$, for which $v$ is in the image, and $n$ is in the kernel, and for each vector it is either $v=0$ or $n=0$ therefore the sum is trivial? $\endgroup$ – Mickey Jul 5 '17 at 7:14
  • $\begingroup$ I'm not sure what you mean, for a given $x$ you may have $v \neq 0$ and $n \neq 0$. Take $J = \operatorname{diag}(1,0)$. Then if $x=(1,1)^T$ we have $v = (1,0)^T, n=(0,1)^T$. $\endgroup$ – copper.hat Jul 5 '17 at 7:16
  • $\begingroup$ hmm you are right. so the decomposition gives us a way to "split" the vector space in to two subspaces for which each corresponds to the image or the kernel (that's because jordan blocks are invariant subspaces of $V$) and therefore the sum is trivial? $\endgroup$ – Mickey Jul 5 '17 at 7:19
  • $\begingroup$ I don't understand what you mean by 'therefore the sum is trivial'. It just shows that $V$ can be written as ${\cal R}J^n \oplus \ker J^n$. $\endgroup$ – copper.hat Jul 5 '17 at 7:21
  • $\begingroup$ My choice of $n$ for the vector above was a bad one, not to be confused with the dimension of the space. $\endgroup$ – copper.hat Jul 5 '17 at 7:22
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I propose the following solution.

a) Let $P(x)$ the characteristic polynomial of $\phi$, and suppose that $P(x)=x^lQ(x)$ with $Q(0)\not =0$ (we have $l\leq n$). Let $k\geq n$. There exists polynomials $U,V$ such that $U(x)P(x)+V(x)x^k =x^l$. Hence $ U(\phi)P(\phi)+V(\phi)\phi^k =\phi^l=V(\phi)\phi^k$, using Cayley-Hamilton's theorem. Hence, if $x$ is such that $\phi^k(x)=0$, we get $\phi^l(x)=0$, hence $\phi^n(x)=0$.

b) Let $x\in E={Ker}(\phi^n)\cap {Im}(\phi^n)$. Then $\phi^n{x}=0$, and there exist $y$ such that $\phi^n(y)=x$. We have $\phi^{2n}(y)=0$, hence by a) $\phi^n(y)=x=0$.

c) as ${Dim}({Ker}(\phi^n))+{Dim} ({Im}(\phi^n))=n$, we are done.

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