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Exercise

1) Prove that Ax = b has solutions for any b ⇔ There is a matrix X, such that AX = I.

Answer Assume Ax = b has solutions for any b. By taking b to be vectors in the standard basis e1, e2, ..., en, we get vectors x1, x2, ..., xn, satisfying Axi = ei. Thus if we take X = [x1 x2 ... xn], then by this formula, AX = [Ax1 Ax2 ... Axn] = [e1 e2 ... en] = I.

Conversely, assume AX = I. Then for any b, we have A(Xb) = Ib = b. In other words, x = Xb is a solution of Ax = b.

2) Prove that solution of Ax = b is unique ⇔ There is a matrix X, such that XA = I.

Answer Assume A is an m by n matrix such that the solution of Ax = b is unique. Then all columns of A are pivot, which means that the reduced row echelon form of A is

J = [e1 e2 ... en] = [ In ]
O

where e1, e2, ..., en are the first n vectors in the standard basis of Rm (recall that the uniqueness implies nm). In other words, we can use row operations to change A to become J. The same row operations will change [In A] to [Y J], where Y is some n by n matrix. Then the reversed row operations change [Y J] = [Y e1 e2 ... en] to [In A] = [In a1 a2 ... an]. Consequently, [Y ei] can be row operated to become [In ai]. In other words, the systems Yx = ei and Inx = ai have the same solutions. Since x = AI is the unique solution of Inx = AI, we conclude that Yai = ei. Thus

YA = [Ya1 Ya2 ... Yan] = [e1 e2 ... en] = J

and if we take X = [In O]Y (i.e., X is the first n rows of Y), then

XA = [In O]YA = [In O]J = In.

Conversely, assume XA = I. Then Ax = 0 implies XAx = X0 = 0. Since XA = I, we conclude that x = 0. The whole implication is that the homogeneous system Ax = 0 has only the trivial solution.

3) Rephrase the first and second parts in terms of linear transformations.

Answer A linear transformation T is onto ⇔ There is a linear transformation S, such that TS = id.

A linear transformation T is one-to-one ⇔ There is a linear transformation S, such that ST = id.

We remark that the statements also hold for general transformations.