## A versus ATA

This note presupposes the reader understands about the equality, discussed in Anton’s Chapter 4, Section 1, for k × n matrices A and column vectors u in Rn and v in Rk , given by
(Au) • v = u • (ATv) .
Assertion: A vector x (among the column vectors x in Rn ) is a solution to the homogeneous equation
Ax = 0k
(this 0k standing for the zero column vector in Rk ) if and only if it is a solution to the homogeneous equation
ATAx = 0n
(the latter 0n standing for the zero column vector in Rn ).

Proof: Suppose x satisfies Ax = 0k . Multiply both sides of this equation, on the left, by AT — the result is ATAx = AT0k . But the RHS here is just 0n (why?), whence x ultimately satisfies ATAx = 0n , as required.
Conversely, if instead x satisfies ATAx = 0n , we compute Ax indirectly by evaluating its norm (length) ||Ax|| , as follows, using the the very first equality displayed:
||Ax||2 = AxAx = xATAx = x0n = 0 .
Of course, from ||Ax||2 = 0 we quickly infer, first, that ||Ax|| = 0 , and from this, in turn, that Ax = 0k , as desired.

Remark. Looking ahead to Chapter 5, section 5 (q.v.), the assertion above may be viewed as informing us that the matrices A and ATA have the same nullspace. Looking ahead a tad further in that chapter, to section 6, we may infer then that A and ATA have the same nullity, as well, and hence also the same rank.
 First posted: 2 Apr 2006. Math 221 Home  Contents © 2006 by Fred E.J. Linton