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.

Problem 1 

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)

Problem 2

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.

Problem 3

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 P_S<--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

Problem 5 

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 A^T.  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)}

Problem 6

Let  S  =  {v₁, v₂, ..., v_n} 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 + ... + a_nv_n        and        v  =  b₁v₁ +b₂v₂ + ... + b_nv_n     

then

        a₁  =  b₂, a₂  =  b₂, ... , a_n  =  b_n 

Solution

Suppose that 

      v  =  a₁v₁ +a₂v₂ + ... + a_nv_n        and        v  =  b₁v₁ +b₂v₂ + ... + b_nv_n     

Then

      a₁v₁ +a₂v₂ + ... + a₂v_n   =  b₁v₁ +b₂v₂ + ... + b_nv_n      

Moving everything to the left hand side and combining like terms, we get

        (a₁ - b₁)v₁ + (a₂ - b₂)v₂ + ... + (a_n- b_n)v_n   =  0

Since the vectors are linearly independent, all of the coefficients are zero, that is

        (a₁ - b₁)  =  (a₂ - b₂)  =  ...  =  (a_n- b_n)  =  0

so that 

        a₁ = b₁ ,   a₂  =  b₂  ,  ... ,   a_n = b_n