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I → H

Given: x → Ax maps onto R^n
WTS: spans R^n

Mapping onto R^n means that there's a solution for all R^n, so there's a solution for Ax=b  for all b. Therefore, b is a linear combination of the columns of A. The definition of span is the set of all linear combinations of the columns of A, so each b in R^n is equal to a linear combination, meaning A encompasses all linear combinations. Also, all columns of A are in R^n, so by closure (+ and *), the linear combination is also in R^n. Therefore A spans R^n.

B → I

Given: A is row equivalent to the n x n identity matrix
WTS: x → Ax maps onto R^n

If A is row equivalent to the n x n identity matrix, A has a pivot in each row. That means that there are no pivots in the augmented matrix [Ab], so the linear combination is consistent for all b in R^n. By the definition of Ax, that means that for all b in R^n, the equation Ax = b has a solution. Therefore, Ax maps onto R^n.

E → F

Given: The columns of A form a linearly independent set.
WTS: The linear combination x → Ax is one-to-one.

If the columns of A form a linearly independent set, then Ax = 0 only has the trivial solution. That's equivalent to saying a corresponding linear transformation T(0) only has the trivial solution. If T is one-to-one, T(0) has at most one solution, hence only the trivial solution. If T isn't one-to-one, let T(u) = b and T(v) = b, so T(u - v) = T(u) - T(v) = b - b = 0. So, T(x) = 0 doesn't only have the trivial solution. Therefore, if T(x) = 0 only has the trivial solution, then it's one-to-one. Because there is a corresponding standard matrix, the columns of matrix A are one-to-one.

G → B

Given: The equation Ax = b has at least one solution for each b in R^n
WTS: A is row equivalent to the n x n identity matrix.

If the equation Ax = b has at least one solution for each b in R^n, then the augmented matrix [Ab] is always consistent, meaning it doesn't have any pivots in the augmented column, which means the matrix A has n pivots. If the matrix A has n pivots, there are pivots along the main diagonal, which can be reduced to the n x n identity matrix.

A → D

Given: A is an invertible matrix.
WTS: The equation Ax = 0 has only the trivial solution.

Let Ax = 0. If A is invertible we can left multiply both sides by A inverse. So, A^-1 * Ax = A^-1 * 0, so x = 0. If A is an invertible matrix, Ax = 0 only when x = 0.

J → D

Given: There exists an n x n matrix C such that CA = I
WTS: The equation Ax = 0 has only the trivial solution.

Let Ax = 0. Left multiply by C. By our premise, CA = I, so Ix = C * 0 = 0, so x = 0. So, if there exists an n x n matrix C such that CA = I, then Ax = 0 has only the trivial solution.

F → B

Given: The linear transformation x → Ax is one-to-one.
WTS: A is row equivalent to the n x n identity matrix.

If a linear transformation is one-to-one Ax = 0 must have the trivial solution. By the existence and uniqueness theorem in 1.2, Ax = 0 has nontrivial solutions iff it has one free variable. Ax = 0 has no nontrivial solutions, so it has no free variables. Because it has no free variables, there is a pivot in each column, meaning there is a pivot along the diagonal. That can be row reduced to the n x n identity matrix.

E → I

Given: The columns of A form a linearly independent set.
WTS: The linear transformation x → Ax maps R^n onto R^n.

By the definition of linearly independent, the linear combination of the columns of A equals A only when all the coefficients are 0. So, Ax = 0 only when x = 0. By the existence and uniqueness theorem, it has no free variables. That means that A has n pivots, meaning there are no pivots in the augmented column of the augmented matrix [Ab], so Ax = b has a solution for all b in R^n, meaning it's onto.