The Rank-Nullity Theorem

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.