1. ## The rank-nullity theorem

Let $\displaystyle T: R^n \rightarrow R^m$ be a linear transformation. Then there is a theorem that:

RankT + nullity T = n

This theorem is said to be equivalent to:

dim(Ran(T))+ dim(Kernel(T))= n

[n is the dimension of domain]

So here's my question: how can we use the first one to derive/prove the second?

This is my unsuccessful attempt:

And for nullity(T):

nullity(T) = dim(null(T))

Since null(T) = ker(T)

So, nullity(T) = dim(ker(T))

For rank(T):

rank(T) = dim(row(A))

Since row(A) = null(A) and null(A) = ker(T)

dim(row(A)) = dim(ker(T))

???

I will be great if anyone could show me how to prove this. Thanks.

2. Originally Posted by Roam
Let $\displaystyle T: R^n \rightarrow R^m$ be a linear transformation. Then there is a theorem that:

RankT + nullity T = n

This theorem is said to be equivalent to:

dim(Ran(T))+ dim(Kernel(T))= n

[n is the dimension of domain]

So here's my question: how can we use the first one to derive/prove the second?

For rank(T):

----------
I will be great if anyone could show me how to prove this. Thanks.
By definition, for a linear transformation T, dim(Range(T)) = Rank(T). Whats your definition?

You seem to be confusing a linear transformation with its matrix(after choosing a basis).

P.S: What is A?