Math 203 Practice Midterm 2   Please work out each of the given problems.  Credit will be based on the steps towards the final answer.  Show your work.   Let  L:  R2 --> R3   be a linear transformation such that          L (1,4) = (1,-1,3)   and     L (0,2) =  (2,1,4) Find L(1,0)  Solution We need to find constants c1 and c2 such that          c1(1,4) + c2(0,2)  =  (1,0) This gives          c1             =  1         4c1 + 2c2  =  0  we rref the matrix                hence         (1,0)  =  1(1,4) - 2(0,2) Now take the linear transformation          L(1,0)  =  L( (1,4) - 2(0,2) )  =   L((1,4)) - 2 L((0,2))         (1,-1,3) - 2(2,1,4)  =  (-3, -3, -5)          Of the following two subsets of the vector space of differentiable functions, determine which is a subspace.  For the one that is not a subspace, demonstrate why it is not.  For the one that is a subspace, prove that it is a subspace. A.  S  =  {f | f(3)  =  f '(3)}  B.  T  =  {f | f(0)f '(0)  =  0}   Solution A.  This is a subspace.  We need to show that the two closure properties hold.   If f and g are functions with          f(3)  =  f '(3)        and        g(3)  =  g'(3)  then         (f + g)(3)  =  f(3) + g(3)  =  f '(3) + g'(3)  =  (f + g)'(3) hence f + g is in the subset.  If c is a constant, then         (cf)(3)  =  cf(3)  =  cf '(3)  =  (cf)'(3) hence cf is in the subset.  Since both closure properties are true, we can conclude that S is a subspace. B.  This set is not a subspace.  Let f(t)  =  1 and g(t)  =  t.  Then          f(0)f '(0)  =  (1)(0)  =  0        and        g(0)g'(0)  =  (0)(1)  =  0 so both f and g are in T.  However         (f + g)(0)(f + g)'(0)  =  [f(0) + g(0)] [f '(0) + g'(0)]  =  (1 + 0)(0 + 1)  =  1 Hence T is not closed under addition.   Let  S  =  {t2, t2 + 2t, t2 + 3}  and T  =  {2t - 1, 5t - 3, t2} be subsets of P2  A.   Prove that S is a basis for P2. Solution Since the dimension of P2 is 3 and S has 3 vectors, we need only prove that the vectors in S are linearly independent.  If we let          c1t2 + c2(t2 + 2t) + c3(t2 + 3)  =  0 then we can combine like terms to obtain         (c1 + c2 + c3)t2 + 2c2t + 3c3  =  0 If a polynomial equals zero then each of the coefficients equal zero.  We get         c1 + c2 + c3  =  0                 2c2       =  0                      3c3  =  0 The matrix for this equation is         has determinant equal to 6.  Since the determinant is nonzero, the system has only the trivial solution, hence the three vectors are linearly independent.  We can conclude that S is a basis for P2.            B.  Find the transition matrix PS<--T . Solution Let E be a standard basis E  =  {t2, t, 1} and look at the following diagram         To go from T to S, we just multiply the matrices         Problem 4   Suppose that you want to change the graphic file from the one on the left to the one on the right.  What is the matrix needed to adjust the pixels appropriately?  Assume the center of the picture is the origin.                           Solution A close look at the picture shows that it has been reflected across the y-axis.  We have         Hence           Let         A.     Find the rank and the nullity of A.  Solution We find          Since there are three corners, the rank of A is 3.  From the equation         rank(A) + nullity(A)  =  n we get         nullity(A)  =  5 - 2  =  2 B.     Find a basis for the Null Space of A. We use the above rref matrix to get the basis          {(0,-1,-2,1,0), (0,1,-4,0,1)} C.     Find a basis for the Column Space of A using columns of A. We can just take the columns from the corresponding corner entries.  Since the corner entries appear in the first, second, and third columns, we take these columns.         {(1,2,0,3,2), (2,6,-4,1,4), (1,2,0,-2,2,2)} D.     Find a basis for the Row Space of A using rows of A. We find the basis for the column space of AT.  We have          Hence we take the first, second, and fourth rows of A.          {(1,2,1,4,2), (2,6,2,10,2), (3,1,-2,-1,-9)}         Let  S  =  {v1, v2, ..., vn} be a set of linearly independent vectors and let v be a vector in the span of S.  Prove that v can uniquely be written as a linear combination of elements of S.  That is that prove that if       v  =  a1v1 +a2v2 + ... + anvn        and        v  =  b1v1 +b2v2 + ... + bnvn      then         a1  =  b2, a2  =  b2, ... , an  =  bn  Solution Suppose that        v  =  a1v1 +a2v2 + ... + anvn        and        v  =  b1v1 +b2v2 + ... + bnvn      Then       a1v1 +a2v2 + ... + a2vn   =  b1v1 +b2v2 + ... + bnvn       Moving everything to the left hand side and combining like terms, we get         (a1 - b1)v1 + (a2 - b2)v2 + ... + (an- bn)vn   =  0 Since the vectors are linearly independent, all of the coefficients are zero, that is         (a1 - b1)  =  (a2 - b2)  =  ...  =  (an- bn)  =  0 so that          a1 = b1 ,   a2  =  b2  ,  ... ,   an = bn