PROBLEM 4

(a) F  For example, the system  x=1, 2x=2  represented by  Ax = b where
	
          A = [1 2]     (a 2x1 matrix)

              [ 1 ]
          b = [   ]
              [ 2 ]

(b) F   The map does not take zero to zero.  (Reflection about
        a line _through the origin_ is a linear transformation.)

(c) T   For instance, use the Rank-Nullity Theorem:

          dim(im(T)) + dim(ker(T)) = n

        Here the image has dimension at most m < n,
        so the kernel's dimension is at least  n - m.
        Hence the kernel has positive dimension, and thus is infinite.
        (We have also shown this fact in other ways.)

(d) T   The inverse is AAAA.

(e) T   Multiply AB=0 from the left by the inverse of A to deduce B=0.

(f) F   We have

          (A+B)(A-B) = AA + BA - AB - BB = (AA-BB) + (AB-BA)

        which equals AA-BB only when AB-BA is zero, that is, when
        A and B commute.  But we know that in general matrix multiplication
        is not commutative, even for square matrices.  [Here the assumption
        that A,B are invertible is a red herring; already in dimension 2
        we saw examples of non-commuting invertible linear transformations.]

(g) T   Compare the row-reduced echelon forms.  (Conversely if the ranks
        are equal then the system is consistent.)

(h) F   For instance we may have  u=v  and then  w  can be any vector.

(i) T   A nontrivial linear relation   c2 v2  +  c3 v3  + c4 v4  =  0
        would yield a nontrivial relation on v1,v2,v3,v4 with coefficients
        0, c2, c3, c4.

(j) T   If (x,y) and (x',y') are in the set then so is (x+x',y+y')
        because the sum of the two integers x,x' is again an integer.
        [Of course this set is not a linear subspace; it is not closed under
        scalar multiplication -- consider (1/2) times the vector (1,1).]