Section 4.2 Rank and Nullity
Subsection 4.2.1 Some fundamental subspaces
Subsection 4.2.2 The rank-nullity theorem
Given a linear map with finite dimensional, there is a fundamental theorem relating the dimension of to the dimensions of and
Proof.
Let and recall that if is any basis for then
First consider the case that is injective. This means that so that By Proposition 4.1.3, the set is linearly independent, and since this set spans it is a basis for so its cardinality equals the dimension of the image, i.e., Thus and we see that
Now consider the case where Let be a basis for hence Since is a linearly independent set, by
[provisional cross-reference: prop-extend-independent-set-to-basis]
, it can be extended to a basis for To establish the theorem, we need only show that Since is a basis for but we recall that so that Thus we know To obtain an equality, we need only show that the set is linearly independent.
Suppose to the contrary, that the set is linearly independent. Then there exists scalars not all zero, so that
By linearity, this says which means But this in turn says that implying the full set is linearly dependent, contradicting that it is a basis for This completes the proof.
Let’s do a simple example.
Example 4.2.2.
The domain has dimension 4 with standard basis so
One easily checks that At the very least we know that and since we must have Now since is a linearly independent set, we know that which means that by Theorem 4.2.1. It follows that is a basis for the nullspace.
Subsection 4.2.3 Computing rank and nullity
Let be a matrix. Then defines a linear map 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 We know that the image of any linear map is obtained by taking the span of where is any basis for the domain. Indeed if we choose the to be the standard basis vectors (with a 1 in the th coordinate and zeroes elsewhere), then is simply the th column of the matrix Thus is the column space of However to determine the rank of we would need to know which columns form a basis. We’ll get to that in a moment.
The nullspace of is the set of solutions to the homogeneous linear system You may recall that a standard method to deduce the solutions is to put the matrix 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.
Proposition 4.2.3.
- Given the variables
in the system a 1 in the th column of the reduced row-echelon form of called a pivot, means that the variable is a constrained variable while the remaining variables are free variables. Thus if there are pivots, there are free variables, and it follows that - The pivot columns of
(the columns of in which there is a pivot in the reduced row-echelon form of ) can be taken as a basis of the column space of - The row rank of
(number of linearly independent rows) equals the column rank of (number of linearly independent columns).
Subsection 4.2.4 Elementary Row and Column operations
- The matrix
is put in reduced row echelon form by a sequence of elementary row operations. - Each elementary column operation can be achieved by right multiplication of
( ) by an elementary matrix. - Every elementary matrix is invertible and its inverse in again an elementary matrix of the same type.
- The rank of an
matrix is unchanged by elementary row or column operations, that is and for appropriately sized elementary matrices
Every invertible matrix is a product of elementary matrices, and this leads to the
Algorithm 4.2.4.
To determine whether an matrix is invertible and if so find its inverse, reduce to row-echelon form the "augmented" matrix
Exercises Exercises
1.
- Are the row spaces of
and the same? - Are the column spaces of
and the same? - If
is the reduced row-echelon form of are the nonzero rows of a basis for the row space of - If
is the reduced row-echelon form of is the column space of the same as the column space of
Answer.
yes; yes; yes (why?); no; If then and
2.
Given an matrix show that there exist (appropriately sized) elementary matrices so that has the form