Mathematical Methods for Physics and Engineering : A Comprehensive Guide

8.18 SIMULTANEOUS LINEAR EQUATIONS

where theAijandbihave known values. If all thebiare zero then the system of

equations is calledhomogeneous, otherwise it isinhomogeneous. Depending on the

given values, this set of equations for theNunknownsx 1 ,x 2 ,...,xNmay have

either a unique solution, no solution or infinitely many solutions. Matrix analysis

may be used to distinguish between the possibilities. The set of equations may be

expressed as a single matrix equationAx=b, or, written out in full, as











A 11 A 12 ... A 1 N

A 21 A 22 ... A 2 N

..

.

..

.

..

.

..

.

AM 1 AM 2 ... AMN





















x 1

x 2

..

.

xN











=















b 1

b 2

..

.

bM















.

8.18.1 The range and null space of a matrix

As we discussed in section 8.2, we may interpret the matrix equationAx=bas

representing, in some basis, the linear transformationAx=bof a vectorxin an

N-dimensional vector spaceVinto a vectorbin some other (in general different)

M-dimensional vector spaceW.

In general the operatorAwill mapanyvector inV into some particular

subspaceofW, which may be the entire space. This subspace is called therange

ofA(orA) and its dimension is equal to therankofA. Moreover, ifA(and

henceA)issingularthen there exists some subspace ofVthat is mapped onto

the zero vector 0 inW; that is, any vectorythat lies in the subspace satisfies

Ay= 0. This subspace is called thenull spaceofAand the dimension of this

null space is called thenullityofA. We note that the matrixAmustbe singular

ifM=Nandmaybe singular even ifM=N.

The dimensions of the range and the null space of a matrix are related through

the fundamental relationship

rankA+ nullityA=N, (8.119)

whereNis the number of original unknownsx 1 ,x 2 ,...,xN.

Prove the relationship (8.119).

As discussed in section 8.11, if the columns of anM×NmatrixAare interpreted as the
components, in a given basis, ofN(M-component) vectorsv 1 ,v 2 ,...,vNthen rankAis
equal to the number of linearly independent vectors in this set (this number is also equal
to the dimension of the vector space spanned by these vectors). Writing (8.118) in terms
of the vectorsv 1 ,v 2 ,...,vN, we have

x 1 v 1 +x 2 v 2 +···+xNvN=b. (8.120)

From this expression, we immediately deduce that the range ofAis merely the span of
the vectorsv 1 ,v 2 ,...,vNand hence has dimensionr=rankA.