In understanding these subspaces, one of the first questions that might come to mind is: “How big are they?” Well, the size of a subspace is measured by its dimension, and the following theorem shows that if you know the dimension of either or , then you immediately also know the dimension of the other one.
Let be a linear transformation between vector spaces over , where is finite dimensional. Then
[Aside: This is sometimes also called the “Rank-Nullity theorem” because is the rank of (see below), and is often referred to as the nullity of .]
Let , and choose a basis of the kernel, where . By Corollary 2.37, we can extend this linearly independent set to a basis of , by adding vectors . Now I claim that is a basis for .
Since the image of is spanned by the images of the basis vectors, and for any , this shows spans .
To show is linearly independent, assume we have scalars such that
, where last equality follows from the linearity of . In particular, this means So by the linear independence of the , we have for all . This proves is linearly independent, and hence is a basis for . Therefore, as required. ∎
We will define the rank of a matrix differently to that used in MATH105. The “rank” of is defined as
The next theorem shows that this definition is equivalent to the one used in MATH105.
Let be a matrix. Then
The left equality is true by Theorem 4.16. The proof of the right hand equality is omitted from this module, but we include it below for the interested reader.
Next, by Theorem 2.49, the number equals the number of non-zero rows of , and therefore the image of is contained in the -dimensional subspace . So . In other words:
Since this argument applies to any matrix, it applies to the transpose , which tells us the reverse inequality is true (since the transpose operation exchanges the rows and columns of a matrix). Hence we must have equality. ∎
An immediate consequence of this Theorem is that .
Let be the linear transformation defined by differentiation of the single variable. For example, . Let ; this is the standard basis for .
Compute ,
Find a basis for the kernel of ,
Find a basis for the image of ,
Verify the dimension theorem for .
[End of Exercise]
Let be a square matrix. Then the following three conditions are equivalent to each other:
The rows of are linearly independent
The columns of are linearly independent
is invertible.
The above corollary to Theorem 4.25 is used regularly in statistics during the process of multiple linear regression (see MATH235 and MATH452).