Review Problem Solutions

(1)

• True, as ker A is the solution of the linear system Ax = 0.
• False, e.g.

            1 1
            1 1

• True, as im A^2 is a subset of im A.
• False, e.g. a line minus the origin is not a linear space.
• True, as can be no [0 0 0 0 0 0 | 1] row.

(2) A basis of the image are columns 1,3, and 5. A basis for the kernel is (2,1,0,0,0) and (-3,0,-2,1,0).

(3) Analyzing what PS does to e1 and e2 gives (PS) =

         1 2 
         0 0
 Squaring this matrix leaves it unchanged, and so raising it 
to the 50th power leaves it unchanged. So (PS)^50 of (22,17) is (56,0).

 (4) The image of T is of dim no more than 3 (as the output
	space for T is R^3.)  Sincce dim(im T)+dim(ker T)=7, we have
	that dim(ker T) is at least 4.  But ker T is a subset of ker(ST),
	so ker(ST) is of dim at least 4. Thus ST is not inverible,	
	as it would have to have a ker of just 0.

 (5)  T^5 = T^3 means (T^2 - I) T^3 = 0.  So (T^2 - I) T^3 x = 0
	for any vector x. In other words, any vector T^3 x in im(T^3)
	must be killed by T^2 - I.  Similarly we have T^3 (T^2 - I) = 0,
	which implies im(T^2-I) is in ker(T^3).

 (6)  The matrices are 
    Proj onto |  Refl in  |   Rotation about   |  Refl in the 
    xy-plane  |  xy-plane |       y-axis       |  plane y=z
    --------------------------------------------------------
     1 0 0       1  0 0       cos(a) 0 -sin(a)       1 0 0
     0 1 0       0 -1 0         0    1   0           0 0 1 
     0 0 0       0  0 1       sin(a) 0  cos(a)       0 1 0


 (7)  Performing Gauss-Jordan on the matrix yields an 
upper triangular matrix with diagonal entries 1,1, and k^2-3k+2.
So the matrix is invertible when k is not equal to 1 or 2.

 (8) proj_L(x) - proj_M(x) is refl_L(x)

 (9) The inverse of A is 
        4  6 -11
       -1 -2   4
        3  5  -9
 and the solution vector is (39,-12,32).

 (10) By thinking about what happens to e_1, e_2, and e_3,
you can determine that SR is a rotation of 120 degrees around the line
spanned by (1,1,1).