# Proving an image and annihilator of a kernel are equal

Tags:
1. May 19, 2017

1. The problem statement, all variables and given/known data
Suppose T:V→U is linear and V has finite dimension. Prove that Im Tt = (Ker T)0

2. Relevant equations
dim(W)+dim(W0)=dim(V) where W is a subspace of V and V has finite dimension.

3. The attempt at a solution
I first proved Im Tt ⊆ (Ker T)0. Let u be an arbitrary element of Ker T and Tt(Φ )∈ Im Tt, then Tt(Φ)(u)=Φ(T(u))=Φ(0)=0. Thus, since all elements of Im Tt map all elements of Ker T into 0, Im Tt ⊆ (Ker T)0.

I now only need to proof dim((Ker T)0) = dim Im Tt. Since V has finite dimension, dim((Ker T)0) = dim(V)-dim(Ker T), but dim(Im T) = dim(V)-dim(Ker T). Thus, dim((Ker T)0)=dim(Im T). This is where I got stuck, how do I prove dim (Im T) = dim ((Im T)0)?

2. May 20, 2017

### zwierz

this assumption is redundant. Redundant assumptions just obscure a core of the theorem and make proof more complicated than it really is.

Last edited: May 20, 2017
3. May 20, 2017

### zwierz

Hint.
First prove a
Theorem. Let $A:V\to U,\quad w:V\to\mathbb{R}$ be lenear mappings such that $\mathrm{ker}\,A\subseteq \mathrm{ker}\,w$. Then there exist a linear mapping $\Lambda:U\to\mathbb{R}$ such that $w=\Lambda A$.

It is actually a special case of some another general theorem