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 R^k , given by (Au) • v = u • (A^Tv) . Assertion: A vector x (among the column vectors x in Rⁿ ) is a solution to the homogeneous equation Ax = 0_k  (this 0_k standing for the zero column vector in R^k ) if and only if it is a solution to the homogeneous equation A^TAx = 0_n (the latter 0_n standing for the zero column vector in Rⁿ ).

Proof: Suppose x satisfies Ax = 0_k . Multiply both sides of this equation, on the left, by A^T — the result is A^TAx = A^T0_k . But the RHS here is just 0_n (why?), whence x ultimately satisfies A^TAx = 0_n , as required.
Conversely, if instead x satisfies A^TAx = 0_n , we compute Ax indirectly by evaluating its norm (length) ||Ax|| , as follows, using the the very first equality displayed: ||Ax||² = Ax • Ax = x • A^TAx = x • 0_n = 0 . Of course, from ||Ax||² = 0 we quickly infer, first, that ||Ax|| = 0 , and from this, in turn, that Ax = 0_k , 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 A^TA have the same nullspace. Looking ahead a tad further in that chapter, to section 6, we may infer then that A and A^TA have the same nullity, as well, and hence also the same rank.

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First posted: 2 Apr 2006.

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Contents © 2006 by Fred E.J. Linton