# Thread: Vector space / linear transformation question

1. ## Vector space / linear transformation question

I've been going over this question for a while, and can't seem to make head or tail of it. Any help would be appreciated!

Let V be a vector space s.t. dim(V) = n. T: V -> V is a linear transformation.

Define $\displaystyle {K_i} = \ker {T^i},{\rm{ }}i \ge 0$, where $\displaystyle {T^0}$ is the identity mapping. I need to show that $\displaystyle {K_i} \subseteq {K_{i + 1}}$ (I've done that bit); and then deduce that $\displaystyle \exists m \ge 0:{K_m} = {K_{m + 1}}$, and so $\displaystyle {K_m} = {K_{m + j}}$ for all $\displaystyle j \ge 1$, and so finally $\displaystyle V = {K_m} \oplus {T^m}\left( V \right)$.

In particular, I don't understand why the kernal can't just keep growing - can you not just keep including one more element in successive linear transformations each time? If your vector space has an infinite number of elements in it (e.g. the real numbers), then why does there have to come a point at which the kernel doesn't grow any bigger?

Define $\displaystyle {K_i} = \ker {T^i},{\rm{ }}i \ge 0$, where $\displaystyle {T^0}$ is the identity mapping. I need to show that $\displaystyle {K_i} \subseteq {K_{i + 1}}$ (I've done that bit); and then deduce that $\displaystyle \exists m \ge 0:{K_m} = {K_{m + 1}}$, and so $\displaystyle {K_m} = {K_{m + j}}$ for all $\displaystyle j \ge 1$, and so finally $\displaystyle V = {K_m} \oplus {T^m}\left( V \right)$.