STATISTICS 512 TECHNIQUES OF MATHEMATICS FOR ...

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4) By 3), then 1), (A) =(A 0 A) ≤ (A 0 );
replacing A by A 0 gives (A 0 ) ≤ (A) and
so
(A 0 )=(A);
i.e. row rank = column rank= # of independent
rows or columns. Thus, from now on,
‘rank’ can mean either row rank or column
rank.
5) (AB) ≤ min((A)(B)).
Proof: That (AB) ≤ (A) has been shown.
Using 4) and 1),
(AB) =(B 0 A 0 ) ≤ ³ B 0´ = (B)
• A square, full rank matrix has an inverse.
Proof: We are to show that if A × has full
rank then there is an ‘inverse’ B with the property
that AB = BA = I . The columns of A are
independent, hence form a basis of R (why?).
Thus they span: the equations
A [b 1 ···b ]=[e 1 ···e ]=I