# Proving $V=\text{KerT}^n \oplus \text{Im}T^n$ for $V$ in $\mathbb C$

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!

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.

• 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? – Mickey Jul 5 '17 at 7:14
• 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 $v = (1,0)^T, n=(0,1)^T$. – copper.hat Jul 5 '17 at 7:16
• 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? – Mickey Jul 5 '17 at 7:19
• I don't understand what you mean by 'therefore the sum is trivial'. It just shows that $V$ can be written as ${\cal R}J^n \oplus \ker J^n$. – copper.hat Jul 5 '17 at 7:21
• My choice of $n$ for the vector above was a bad one, not to be confused with the dimension of the space. – copper.hat Jul 5 '17 at 7:22

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.