Solutions to
3.7.{33,34,35,44,45}

33
Show that Ax=b has a solution if and only if b is in CS(A).

ANSWER:
If A has columns [C1 C2 ... Cn], and x is the vector <x1,...,xn>,
then Ax is the (m x 1) column x1*C1 + ... + xn*Cn.
So if b is a linear combination of the columns (r1*C1+...+rn*Cn), then
b is a possible value of Ax (x1*C1+...+xn*Cn), and vice-versa.

34
Show that if the product of two matrices is the zero matrix, AB=0, then the
column space of B is contained in the null space of A.

ANSWER:
Assume y is in CS(B).
By #33, there is a solution x to the equation y=Bx.
So, Ay=A(Bx)=(AB)x=0x=0. This means y is in NS(A) by definition.

35
Show that Ax=b has a solution if and only if rk(A)=rk[A|b].

ANSWER:
Since the rank of a matrix is the dimension of its column space,
rk(A)=rk[A|b] if and only if
    the span of {columns of A}
and the span of {columns of A, b}
have the same dimension.
Since the former is a subspace of the latter, they have the same dimension
if and only if they're equal! (Come to O.H. if you want a more detailed
explanation of this step.)
Finally, the spaces being equal is the same as b already being in CS(A),
which by #33 is the same as Ax=b having a solution.

44
Suppose that a matrix A is formed by taking n vectors from R^m as its columns.
(a) If these vectors are linearly independent, what is the rank of A
    and what is the relationship between m and n?
(b) If these vectors span R^m instead, what is the rank of A
    and what is the relationship between m and n?
(c) If these vectors form a basis for R^m, what is the relationship between
    m and n then?

ANSWER
(a) rk(A)=n (as the vectors then form a basis for CS(A)), and n<=m
    (or else the vectors couldn't have been linearly independent).
(b) rk(A)=m (CS(A)=R^m), and n>=m
    (or else the vectors couldn't have spanned R^m).
(c) m=n.

45
Add more entries to the list in theorem (3.83) by showing that for an nxn
    matrix A,
        A is invertible <=> the    rows of A are linearly independent
                        <=> the columns of A are linearly independent
                        <=> the    rows of A span R^n
                        <=> the columns of A span R^n

ANSWER:
We know for an n x n matrix A, A is invertible <=> rk(A) = n.
rk(A) = dim RS(A): n vectors in R^n have an n-dimensional span
<=> they span R^n <=> they are linearly independent. (See (3.64), 3.6.)
Since rk(A) is also equal to dim CS(A), the same argument applies to the columns.