In the last post we defined the column and row space of a matrix as the span of the columns (in the case of the column space) or rows (for the row space) of the matrix. There’s a third important subspace of a matrix, called the *null space*, which is the set of all vectors which maps to zero. That is, if we think of an matrix as a function from to , then the null space of , , is simply the kernel of the map.

The dimension of the null space is sometimes called the *nullity* of the matrix. There’s an important relationship between the column space, row space, and null space which we’ll now state and prove: if is an matrix, then .

We begin by assuming is a basis for . Since is a subspace of , we may extend to a basis for all of by adding properly chosen vectors. Call these vectors so that is a basis for . Since this is a basis, let and suppose

Now we multiply on the left by . Since we have

where the last step follows from the fact each of are mapped to zero by .

Since our choice of was arbitrary, spans . If we can now show that are linearly independent we’ll have that and have proven our theorem.

So now we want to find the such that

Now since is a basis for , we have that , so is a linearly independent set which spans , so where . Since , we have our result.

Notice that if , clearly , so that . If is non-singular, however, it has a trivial null space and so . Combining this with the above theorem we have that if is non-singular,

Likewise, if is non-singular. (Apply the previous argument with .)

Picking up from last time, we have if and are the non-singular matrices such that , then

So the dimension of the column space equals the dimension of the row space. This common value is called the *rank* of , and is denoted . For this reason the theorem we proved above is known as the *rank-nullity theorem*.

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My query relates to the last part of your blog where you prove that the dim of the col space = the dim of the row space.

I cannot see from what you have done before why the following are immediate:

dim(CS(A))=dim(CS(AQ))

and

dim(RS(PA))=dim(RS(A))

Comment by Eugene Kernan — August 4, 2009 @ 2:44 pm |