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Null Space

Module by: Doug Daniels, Steven J. Cox. E-mail the authors

Summary: This module defines the null space, shows an example of what one is, and describes how to find one given an arbitrary matrix.

Null Space

Definition 1: Null Space
The null space of an m-by-n matrix A A is the collection of those vectors in n n that A A maps to the zero vector in m m . More precisely, 𝒩A= x n Ax=0 𝒩 A A x 0 x n

Null Space Example

As an example, we examine the matrix A A:

A=( 0100 -1010 0001 ) A 0100 -1010 0001
(1)

It is fairly easy to see that the null space of this matrix is:

𝒩A= t( 1 0 1 0 ) tR 𝒩 A t t 1 0 1 0
(2)

This is a line in 4 4 .

The null space answers the question of uniqueness of solutions to Sx=f S x f . For, if Sx=f S x f and Sy=f S y f then S(xy)=SxSy=ff=0 S x y S x S y f f 0 and so (xy)𝒩S x y 𝒩 S . Hence, a solution to Sx=f S x f will be unique if, and only if, 𝒩S=0 𝒩 S 0 .

Method for Finding the Basis

Let us now exhibit a basis for the null space of an arbitrary matrix A A. We note that to solve Ax=0 A x 0 is to solve A red x=0 A red x 0 . With respect to the latter, we suppose that

c j j=1r j 1 r c j
(3)
are the indices of the pivot columns and that
c j j=r+1n j r 1 n c j
(4)
are the indices of the nonpivot columns. We accordingly define the rr pivot variables
x c j j=1r j 1 r x c j
(5)
and the nr n r free variables
x c j j=r+1n j r 1 n x c j
(6)

One solves A red x=0 A red x 0 by expressing each of the pivot variables in terms of the nonpivot, or free, variables. In the example above, x 1 x 1 , x 2 x 2 , and x 4 x 4 are pivot while x 3 x 3 is free. Solving for the pivot in terms of the free, we find x 4 =0 x 4 0 , x 3 = x 1 x 3 x 1 , x 2 =0 x 2 0 , or, written as a vector,

x= x 3 1010 x x 3 1010
(7)
where x 3 x 3 is free. As x 3 x 3 ranges over all real numbers the x above traces out a line in 4 4 . This line is precisely the null space of AA. Abstracting these calculations we arrive at:

Definition 2: A Basis for the Null Space
Suppose that A A is m-by-n with pivot indices c j j=1r j 1 r c j and free indices c j j=r+1n j r 1 n c j . A basis for 𝒩A 𝒩 A may be constructed of nr n r vectors z 1 z 2 z n - r z 1 z 2 z n - r where z k z k , and only z k z k , possesses a nonzero in its c r + k c r + k component.

A MATLAB Observation

As usual, MATLAB has a way to make our lives simpler. If you have defined a matrix A and want to find a basis for its null space, simply call the function null(A). One small note about this function: if one adds an extra flag, 'r', as in null(A, 'r'), then the basis is displayed "rationally" as opposed to purely mathematically. The MATLAB help pages define the difference between the two modes as the rational mode being useful pedagogically and the mathematical mode of more value (gasp!) mathematically.

Final thoughts on null spaces

There is a great deal more to finding null spaces; enough, in fact, to warrant another module. One important aspect and use of null spaces is their ability to inform us about the uniqueness of solutions. If we use the column space to determine the existence of a solution xx to the equation Ax=b A x b . Once we know that a solution exists it is a perfectly reasonable question to want to know whether or not this solution is the only solution to this problem. The hard and fast rule is that a solution xx is unique if and only if the null space of AA is empty. One way to think about this is to consider that if Ax=0 A x 0 does not have a unique solution then, by linearity, neither does Ax=b A x b . Conversely, if (Az=0) and (z0) and (Ay=b) A z 0 z 0 A y b then A(z+y)=b A z y b as well.

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A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

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