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Section 4.2 Rank and Nullity

Subsection 4.2.1 Some fundamental subspaces

Let T:VW be a linear map between vector spaces over a field F. We have defined the kernel of T, ker(T)=Null(T), (also called the nullspace) and noted that it is a subspace of the domain V. The image of T, Im(T), is a subspace of the codomain W.

Subsection 4.2.2 The rank-nullity theorem

Given a linear map T:VW, with V finite dimensional, there is a fundamental theorem relating the dimension of V to the dimensions of ker(T) and Im(T).

Proof.

Let n=dimV, and recall that if {v1,,vn} is any basis for V, then Im(T)=Span({T(v1),,T(vn)}).
First consider the case that T is injective. This means that ker(T)={0}, so that nullity(T)=0. By Proposition 4.1.3, the set {T(v1),,T(vn)} is linearly independent, and since this set spans Im(T), it is a basis for Im(T), so its cardinality equals the dimension of the image, i.e., rank(T). Thus rank(T)=n, and we see that
n=dimV=n+0=rank(T)+nullity(T).
Now consider the case where ker(T){0}. Let {u1,,uk} be a basis for ker(T), hence nullity(T)=k. Since {u1,,uk} is a linearly independent set, by [provisional cross-reference: prop-extend-independent-set-to-basis], it can be extended to a basis for V:
{u1,,uk,uk+1,,un}
To establish the theorem, we need only show that rank(T)=nk. Since {u1,,uk,uk+1,,un} is a basis for V, Im(T)=Span({T(u1),,T(un)}), but we recall that u1,,ukker(T), so that Im(T)=Span({T(uk+1),,T(un)}). Thus we know rankTnk. To obtain an equality, we need only show that the set {T(uk+1),,T(un)} is linearly independent.
Suppose to the contrary, that the set is linearly independent. Then there exists scalars aiF, not all zero, so that
i=k+1naiT(ui)=0.
By linearity, this says T(i=k+1raiui)=0, which means i=k+1raiuiker(T). But this in turn says that i=k+1raiuiSpan({u1,,uk}) implying the full set {u1,,un} is linearly dependent, contradicting that it is a basis for V. This completes the proof.
Let’s do a simple example.

Example 4.2.2.

Consider the linear map T:P3(R)P3(R) given by T(f)=f+f, where f and f are the first and second derivatives of f.
The domain has dimension 4 with standard basis {1,x,x2,x3}, so
Im(T)=Span{T(1),T(x),T(x2),T(x3)}
One easily checks that Im(T)=Span{0,1,2+2x,6x+3x2}=Span{1,x,x2}. At the very least we know that rank(T)3, and since T(1)=0, we must have nullity(T)1. Now since {1,x,x2} is a linearly independent set, we know that rank(T)=3 which means that nullity(T)=1 by Theorem 4.2.1. It follows that {1} is a basis for the nullspace.

Subsection 4.2.3 Computing rank and nullity

Let AMm×n(F) be a matrix. Then T(x)=Ax defines a linear map T:FnFm. Indeed in Subsection 4.3.2, we shall see how to translate the action of an arbitrary linear map between finite-dimensional vectors spaces into an action of a matrix on column vectors.
Let’s recall how to extract the image and kernel of the linear map xAx. We know that the image of any linear map is obtained by taking the span of T(e1),,T(en) where {e1,,en} is any basis for Fn, the domain. Indeed if we choose the ei to be the standard basis vectors (with a 1 in the ith coordinate and zeroes elsewhere), then T(ej) is simply the jth column of the matrix A. Thus Im(T) is the column space of A. However to determine the rank of A, we would need to know which columns form a basis. We’ll get to that in a moment.
The nullspace of T, is the set of solutions to the homogeneous linear system Ax=0. You may recall that a standard method to deduce the solutions is to put the matrix A in reduced row-echelon form. That means that all rows of zeros are at the bottom, the leading nonzero entry of each row is a one, and in every column containing a leading 1, all other entries are zero. These leading ones play several roles.

Subsection 4.2.4 Elementary Row and Column operations

The following are a series of facts about elementary row and column operations on an m×n matrix A.
  • The matrix A is put in reduced row echelon form by a sequence of elementary row operations.
  • Each elementary row operation can be achieved by left multiplication of A (AEA) by an m×m elementary matrix.
  • Each elementary column operation can be achieved by right multiplication of A (AAE) by an n×n elementary matrix.
  • Every elementary matrix is invertible and its inverse in again an elementary matrix of the same type.
  • The rank of an m×n matrix is unchanged by elementary row or column operations, that is rank(EA)=rank(A) and rank(AE)=rank(A) for appropriately sized elementary matrices E.
Every invertible matrix is a product of elementary matrices, and this leads to the

Exercises Exercises

1.
Let A be an m×n matrix and E an elementary matrix of the appropriate size.
  • Are the row spaces of A and EA the same?
  • Are the column spaces of A and AE the same?
  • If R is the reduced row-echelon form of A, are the nonzero rows of R a basis for the row space of A?
  • If R is the reduced row-echelon form of A, is the column space of R the same as the column space of A?
Answer.
yes; yes; yes (why?); no; If A=[1111], then R=[1100], and
Span([11])Span([10])
2.
Given an m×n matrix A, show that there exist (appropriately sized) elementary matrices U,V so that UAV has the form
UAV=[Ir000].
where Ir is an r×r identity matrix with r=rank(A), and the other entries are all zeros.