Extended Echelon Form and Four Subspaces

Associated with any matrix, there are four fundamental subspaces: the column space, row space, (right) null space, and left null space. We describe a single computation that makes readily apparent bases for all four of these subspaces. Proofs of these results rely only on matrix algebra, not properties of dimension. A corollary is the equality of column rank and row rank. Bringing a matrix to reduced row-echelon form via row operations is one of the most important computational processes taught in an introductory linear algebra course. For example, if we augment a square nonsingular matrix with an identity matrix of the same size and row-reduce the resulting n × 2n matrix, we obtain the inverse in the final n columns: [A | In] RREF −−→ [ In | A −1 ] . What happens if A is singular, or perhaps not even square? We depict this process for an arbitrary m × n matrix A: M = [A | Im] RREF −−→ N = [B | J ] = [ C K 0 L ] . The columns are partitioned into the first n columns, followed by the last m columns. The rows are partitioned so that C has a leading one in each row. We refer to N as the extended echelon form of A. While it is transparent that C contains information about A, it is less obvious that L also contains significant information about A. Given a matrix A, four associated subspaces are of special interest: the column space C (A), the (right) null space N (A) = {x | Ax = 0}, the row space R (A), and the left nullspace N ( At ) . These are Strang’s four subspaces, whose interesting properties and important relationships are described in the “Fundamental Theorem of Linear Algebra” (see [4] and [5]). Bases for all four of these subspaces can be obtained easily from C and L . Additionally, we obtain the result that row rank and column rank are equal. We only know of one other textbook [3] besides our own [1], that describes this procedure. (See also Lay’s paper [2].) So, one purpose of this note is to make this approach better known. Several informal observations about extended echelon form are key. As we perform the row operations necessary to bring M to reduced row-echelon form, the conversion of Im to J records all of these row operations, as it is the product of the elementary matrices associated with the row operations. Indeed, B = J A (see Lemma 1). Second, J is nonsingular, which we can see by its row-equivalence with Im , or recognized as a product of elementary matrices, each of which is nonsingular. Third, the entries of a row of L are the scalars in a relation of linear dependence on the rows of A, so each row of L is an element of N ( At ) . We will see that the rows of L are a basis for N ( At ) . http://dx.doi.org/10.4169/amer.math.monthly.121.07.644 MSC: Primary 15A03, Secondary 97H60; 15A23 644 c © THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121 We begin our last observation by considering that there can be many different sequences of row operations that bring a matrix to reduced row-echelon form. If A does not have rank m, then part of this variety is due to the many ways row operations can produce zero rows. However, because the reduced row-echelon form of a matrix is unique, we conclude that J is unique. Thus, J can be interpreted as a canonical transformation of the rows of A to reduced row-echelon form. Lemma 1. For any matrix A, with the notation as above, B = J A and J is nonsingular. For vectors x and y, Ax = y if and only if Bx = Jy. Proof. Since N is obtained from M via row operations, there is a nonsingular matrix R (a product of elementary matrices) such that N = RM . The last m columns of this matrix equality give J = R, and hence J is nonsingular. If we replace R by J in the matrix equality, then the first n columns give B = J A. The equivalence follows from substitutions and the invertibility of J : Ax = y ⇐⇒ J Ax = Jy ⇐⇒ Bx = Jy. Informally, the equivalence simply says that if we solve the system Ax = y for x by augmenting the coefficient matrix A with the vector y and row-reducing, then we obtain a matrix in reduced row-echelon form representing an equivalent system having B as coefficient matrix and Jy as the vector of constants. Because A and B are row-equivalent, and because B and C differ only in the zero rows of B, it should be apparent that N (A) = N (C) and R (A) = R (C). While the individual rows of L are easily explained as elements of the the left null space of A, together they form a basis for the left null space. Less obvious to a student is that the null space of L is the column space of A! We now establish these two results about L carefully. Our second purpose is to give proofs that establish these set equalities without ever exploiting the properties of the subspaces as vector spaces, in contrast to the arguments on dimension used in [2]. This gives us a fundamental result about the dimensions of these subspaces as a corollary. The equivalence of Lemma 1 is our primary tool in the proofs of the next two lemmas. Lemma 2. If A is a matrix with L as above, then C (A) = N (L). Proof. Suppose that y ∈ C (A). Then there exists a vector x such that Ax = y. Therefore, [ Cx 0 ] = [ Cx 0x ] = Bx = Jy = [ K y Ly ] . If C has r rows, then the last m − r entries of this vector equality imply that y ∈ N (L). Conversely, suppose that y ∈ N (L). Because C is in reduced row-echelon form, there is a vector x such that Cx = K y. Then