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发帖数: 1311 | 1
(1) rk(A) = rk(QR) =< min(rk(Q), rk(R)) . You can use linearly dependence of
row or column vector to prove this.
Also, any invertible matrix can be presented as products of fundamental
transformation matrixes. so if diagonal entries of R are nonzero, it's just
doing some basic transformation on column vectors of Q. so it's rk n.
(2) rk(A) = rk(Q) >=Q . first we can know rk(A) =< min(rk(Q),rk(R))=rk(Q) .
second, Let Q' be the transpose of Q. rk(A) >= rk(Q'A) = rk(Q'QR) = rk(DR)=
rk(Q), D is a nx |
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