(a) Let B be any matrix row equivalent to A. Let w1,...,wn be the vectors described by the columns of B. Prove that any linear relationx1v1+...+xnvn=0impliesx1w1+...+xnwn=0(b) Prove that the vectors given by the pivotal columns of A′ are linearly independent and the rest are linearly dependent on these.
(c) Prove v1,...,vn are linearly independent if and only if A′ has n nonzero rows.
(d) By (c), the vectors v1,...,vn are linearly dependent if and only if A′ has nonpivotal columns. The solutions to the linear dependence relations among v1,...,vn are given by the linear equations defined by A′. Show that the variables x1,...,xn corresponding to nonpivotal columns can be prescribed arbitrarily and the remaining variables are then uniquely defined to give the linear dependence relation.
(e) Prove that the subspace W spanned by v1,...,vn has dimension r where r is the number of nonzero rows of A′ and that a basis for W is given by the original vectors corresponding to the pivotal columns of A′.
Proof: (a) Since the other two row operations can be replicated by addition of scalar-multiplied rows to one another, it suffices to prove the relations are preserved for one operation. Viewing the equation in the form of the original basis, we can see that the other unmodified rows (i.e. basis vectors) result in zero in the sum, and since the scalar-multiplied row to be added when multiplied and summed with the xi is already known to be zero, its addition to any of the rows does not change the sum.
(b) These pivotal columns must be of the form ei, so are clearly independent. Every other vector is composed of a column whose nonzero entries have already been preceded by a pivotal element, and are thus generatable by these.
(c) Viewing A′ as A under a nonsingular matrix multiplication, we can see that v1,...,vn are linearly independent if and only if w1,...,wn are linearly independent if and only if A′ has n nonzero rows.
(d) After prescribing these variables arbitrarily, the partial sum represents a vector in basis vectors e1,...,en which are zero in coordinates not reached by a pivotal vector, so that there are unique scalars to the pivotal vectors to make this sum zero.
(e) Again viewing A′ as A under a nonsingular linear transformation, we can see that the vectors mapping to the pivotal column vectors of A′ precisely form the basis for W, of which pivotal columns there are exactly r.
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