This note presupposes the reader understands about the equality, discussed in Anton’s Chapter 4, Section 1, fork×nmatricesand column vectorsAuinR^{n}andvinR^{k}, given by( Au) •v=u• (A^{T}v) .Assertion: A vectorx(among the column vectorsxinR^{n}) is a solution to the homogeneous equation(this Ax=0_{k}^{ }0_{k}standing for the zero column vector inR^{k}) if and only if it is a solution to the homogeneous equation(the latter A^{T}Ax=0_{n}0_{n}standing for the zero column vector inR^{n}).

Proof: SupposexsatisfiesAx=0_{k}. Multiply both sides of this equation, on the left, byA^{T}— the result isA^{T}Ax=A^{T}0_{k}. But the RHS here is just0_{n}(why?), whencexultimately satisfiesA^{T}Ax=0_{n}, as required.

Conversely, if insteadxsatisfiesA^{T}Ax=0_{n}, we computeAxindirectly by evaluating its norm (length) ||Ax|| , as follows, using the the very first equality displayed:|| Of course, from ||Ax||^{2}=Ax•Ax=x•A^{T}Ax=x•0_{n}= 0 .Ax||^{2}= 0 we quickly infer, first, that ||Ax|| = 0 , and from this, in turn, thatAx=0_{k}, as desired.

Remark. Looking ahead to Chapter 5, section 5 (q.v.), the assertion above may be viewed as informing us thatthe matrices. Looking ahead a tad further in that chapter, to section 6, we may infer then thatandAA^{T}have the sameAnullspaceAandA^{T}Ahave the same, as well, and hence alsonullitythe same.rank

First posted: 2 Apr 2006. Math 221 Home Contents © 2006 by Fred E.J. Linton