Kernel/Image of a repeated linear transformation

**robeuler** · March 16th 2009, 07:03 PM

Let V be a vector space of finite dimension (say n). A:V->V is a linear transformation. prove there exists an integer m such that the intersection of im(A^m) and ker(A^m) is {0}.

since the image and kernel are subspace, {0} is always in both.
I know the general strategy is to use dim(im(A^m))+dim(ker(A^m))=n and that after successive applications of a linear transformation the kernel gets smaller. I just don't know how to prove that the kernel gets smaller. I tried to show that dim(ker(A^m))<dim(ker(A^(m-1))) without success.

**ThePerfectHacker** · March 16th 2009, 08:03 PM

Originally Posted by robeuler

Let V be a vector space of finite dimension (say n). A:V->V is a linear transformation. prove there exists an integer m such that the intersection of im(A^m) and ker(A^m) is {0}.

since the image and kernel are subspace, {0} is always in both.
I know the general strategy is to use dim(im(A^m))+dim(ker(A^m))=n and that after successive applications of a linear transformation the kernel gets smaller. I just don't know how to prove that the kernel gets smaller. I tried to show that dim(ker(A^m))<dim(ker(A^(m-1))) without success.

We have the following chain of "inequalities":
$... \subseteq \text{im}(A^3)\subseteq \text{im}(A^2) \subseteq \text{im}(A)$ i.e. $\text{im}(A^{r+1})\subseteq \text{im}(A^r)$ .
$\ker (A)\subseteq \ker (A^2)\subseteq \ker (A^3) \subseteq ...$ i.e. $\ker (A^r)\subseteq \ker (A^{r+1})$ .

Notice that these chains cannot be strictly ascending, meaning that they cannot have $\subset$ instead of $\subseteq$ . The reason is because if $U\subset V\implies \dim (U) < \dim (V)$ for subspaces and these subspace dimensions in our chain consist of natural numbers so there is an upper bound for the higest possible dimension. Therefore, along the upper and lower chain there got to be $=$ somewhere. Say that in the upper chain this occurs at $\text{im}(A^i) = \text{im}(A^{i+1})$ and in the lower chain this occurs at $\ker (A^j) = \ker (A^{j+1})$ . This tells us that $\text{im}(A^i) = \text{im}(A^{i+1}) = \text{im}(A^{i+2}) = ...$ and $\ker (A^j) = \ker (A^{j+1}) = \ker (A^{j+2}) = ...$ . Thus, if we let $k = \max (i,j)$ then $\text{im}(A^k) = \text{im}(A^{k+1})$ and $\ker (A^k) = \ker (A^{k+1})$ . Let $y \in \text{im}(A^k) \cap \ker (A^k)$ , this means $y = A^k(x)$ for some $x\in V$ and $A(y) = 0$ . But then $A(y) = A^{k+1}(x) \implies 0 = A^{k+1}(x)$ . Thus, by definition, $x\in \ker (A^{k+1})$ , however $\ker (A^{k}) = \ker (A^{k+1})$ so $x\in \ker (A^k)$ and this forces $A^k(x) = 0$ . Thus, $y = A^k(x) = 0$ which means $\text{im}(A^k) \cap \ker (A^k) = \{ 0 \}$ .

**ComoUnCocodrilo** · February 5th 2011, 02:33 PM

It is wrong to assume A(y)=0. If y is in that intersection, we only know A^k(y)=0. No?

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