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).]