The Big Theorem on Equivalencies
[A More Careful Statement of Theorem 8 page 120]
Let A be an m x n matrix [so that A has m rows and n columns] and let T: Rn ---> Rm be the corresponding linear transformation. That is, T is the function given by T(x) = Ax, for every x in Rn. [We also say that the function T is given by: x ----> Ax.]
Furthermore, consider the transpose of A, denoted by AT, and let S: Rm ---> Rn be the linear transformation given by S(y) = AT(y) for every y in Rm.
Below are listed three groups of statements: Group I, Group II and Group III.
Theorem: :
(1) The statements in Group I are equivalent to each other.
(2) The statements in Group II are equivalent to each other
(3) The statements in Group III are equivalent to each other
furthermore:
(4) When m = n, then all the statements in all the groups are
all equivalent to each other.
GROUP I -- [The "one-to-one" group.]
(a) The linear transformation T: Rn ---> Rm is one-to-one.
That is, if T(x1) = T(x2), then x1 = x2.
That is, if Ax1 = Ax2, then x1 = x2.(b) The linear system Ax = 0 has only the trivial solution.
That is, if Ax = 0, then x = 0.(c) The n columns of the matrix A are independent vectors in Rm.
(d) There exists an n x m matrix C such that CA = In.
(e) The matrix AT, the transpose of A, satisfies all the conditions of Group II.
GROUP II -- [The "onto" group.]
(a) The linear transformation T: Rn ---> Rm is onto.
That is, for any vector b in Rm, there exists at least
one vector, x, in Rn such that T(x) = b.(b) For any vector b in Rm, the linear system Ax = b always
has at least one solution.(c) The n columns of the matrix A span Rm.
(c') A has a pivot position in every row.[NEW!!-- from Thm 4, page 42]
(d) There exists an n x m matrix D such that AD = Im.
(e) The matrix AT, the transpose of A, satisfies all the conditions of Group I.
GROUP III -- [The "invertible" group.] [Assume: A is a square n x n matrix.]
(a) The matrix A is invertible. That is, there exists an n x n matrix Z
such that AZ = ZA = In.(b) A is row equivalent to In.
(c) A has n pivot positions (equivalently, n pivot columns).
(d) There exits elementary matrices, E1, E2, ...., Es
such that E1E2...EsA = In(e) The matrix AT, the transpose of A, is invertible and thus satisfies
all the conditions of Group III.(f) Det (A) is NOT equal to 0
(g) Det (AT) is NOT equal to 0
Remarks:
(1) If A satisfies any (hence all) of the conditions of Group I, then n < m.
(2) If A satisfies any (hence all) of the conditions of Group II, then n > m.
(3) The only "hard" part of the above theorem is the proof that any of (a), (b),
or (c) in Group I implies either (d) or (e) [of Group I]. However, with the extra
condtion that m=n, our text shows that (d) of Group I implies (c) of Group III, from which
all else may be derived.
(4) No p vectors in Rn can span if p < n.
(5) No p vectors in Rm can be independent if p > n.
(6) Any n vectors in Rn span Rn if and only if they are independent.
Some Important Reminders:
(1) If A is an m x n matrix with columns a1, a2, ...., an,
that is, if A = [a1 a2 .... an],
and if x is a column in Rn with
components x1, x2, ..., xn, then
the product Ax is given by the following
linear combination of the columns of A: x1a1
+ a2x2 + .... + xnan.
(2) If A is an m x n matrix and B is an n x p matrix, then the product
AB is
the m x p matrix given by AB = [Ab1 Ab2
.... Abp],
where b1, b2, ..., bp are
the p columns of B.
(3) A collection of vectors, v1, v2, ..., vn, is said
to be independent
if the only linear combination of them that is zero is the
trivial linear combination.
That is, if x1v1 + x2v2
+...+ xnvn = 0, then it must be that x1 = x2
= .... = xn = 0.
(4) [And thus..] A collection of vectors, v1, v2, ..., vn,
is said to be dependent
there exists a non trivial linear combination of them that is
zero. That is, if there
exists constants x1, x2, ..., xn
not all of which are zero (although some of them might be zero)
such that x1v1 + x2v2
+...+ xnvn = 0.