Proving an image and annihilator of a kernel are equal

[Adgorn]

Homework Statement

Suppose T:V→U is linear and V has finite dimension. Prove that I am T^t = (Ker T)⁰

Homework Equations

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

The Attempt at a Solution

I first proved I am T^t ⊆ (Ker T)⁰. Let u be an arbitrary element of Ker T and T^t(Φ )∈ I am T^t, then T^t(Φ)(u)=Φ(T(u))=Φ(0)=0. Thus, since all elements of I am T^t map all elements of Ker T into 0, I am T^t ⊆ (Ker T)⁰.

I now only need to proof dim((Ker T)⁰) = dim I am T^t. 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)?

 

[zwierz]

V has finite dimension

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

 

[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