The rank theorem ↩

Metadata

shorthands: {}
aliases: [Rank–nullity theorem]
created: 2021-11-14 02:50:26
modified: 2022-01-11 16:06:43

Suppose that

and

and

is a linear map.
We can see that the kernel of

is a subspace of and the image of

is a subspace of .
For these, the following relation holds for the dimensions:

Where is the nullity and is the rank of the linear map.

Proof

Let be a basis for and be a basis for . Choose such that . We claim that is a basis for . If this is true, the theorem is proven since , and .

First we need to show that is linearly independent. Suppose . Since are in , this implies:

Since are linearly independent in ,

Implies are all zero.
But then

so are all zero.
This proves that

is linearly independent.

Now we need to prove that these vectors span . Let . Then so

Set . Since , we see that . This is in and is a unique linear combination of