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BUSI3119-E1 

A LEVEL 3 MODULE, AUTUMN SEMESTER 2020-2021

ADVANCED CALCULUS FOR BUSINESS, ECONOMICS AND FINANCE

1. (a) Prove that the intersection of any two subspaces of V is a subspace of V.                                   (13 Marks) 

Sol: Suppose that  and  are any two subspaces of .  Because  and  are subspaces of , so we have the additive identity 0 is in both  and , hence it is in the intersection of  and .

Suppose that . Then, it is clear that  and   That implies  and  as both U and W are subspaces of V with both being closed under addition.  Hence, we have . This shows that  is closed under addition.

Suppose that . Then, we have  and . Because U and W are subspaces of V, so we have  and  where . That means we have . This shows that  is closed under scalar multiplication.

Therefore, by showing that 0 is in , and  is closed under addition as well as under scalar multiplication, we proved that the intersection of any two subspaces of V is a subspace of V.  

        (b) Suppose  and  are subspaces of  such that , and               =. Prove that .           (7 Marks)

Sol:  By the fundamental theorem of linear mapping, we know that

Because , this implies that

 Hence, we have .

2. (a) Find all eigenvalues and eigenvectors of  defined by

                                                           (12 Marks) 

Sol: Suppose  is an eigenvalue of . For this particular, the eigenvalue-eigenvector equation becomes the system of equations

Based on the above equations, we see that  can be chosen freely and the other variables can be defined in terms of . So we have 

Thus, for each  being an eigenvalue of , we can find the corresponding eigenvector as

(b) Suppose  is invertible and  is a nonzero constant. Prove that  is an eigenvalue of  if and only if  is an eigenvalue of .                                                                (8 Marks)

Sol: Suppose  is an eigenvalue of . There exists a nonzero vector  such that

 . Multiplied both sides of the equation by , we obtain  Thus, we have . That is to say that  is an eigenvalue of .

To prove the implication in the other direction, replace  by , and  by , and then apply the result from the paragraph above.

3. Solve the following differential equation

                                              (20 Marks)

Sol: The general solution of the homogeneous equation is found by first solving its associated characteristic equation

.

The above equation has a repeated root which equals to  5. Hence, the general solution of the differential equation is of the following form

                                         where  are constant. 

While the particular solution of the given differential equation, when it is substituted into the differential equation it should produce the terms on the right hand side of the given differential equation. Hence, we have

  where , and  are constant. 

Substitute the PS into the differential equation, by equalizing the coefficients for each of the corresponding terms on both sides of the resulting equation, one can find the values for  , and  hence the particular solution

  for .

Therefore, for , the solution of the differential equation is given by

. 

When, the solution of the differential equation is given by

.

4. Solve the initial value problems

(a)                ,                                              (8 Marks)

Sol: This question can be solved by using separation of variables method, which is to rewrite the above equation so that each of two variables occurs on a different side of the equation. Next, one could integrate the terms on both side of the equation with the lower bound of the integrals set to be zero while its upper bound equals to t. Then, it is not difficult to find the solution equals to

(b)                       ,                                                            (12 Marks)

Sol: This question can be solved by using separation of variables method, which is to rewrite the above equation so that each of two variables occurs on a different side of the equation. Next, one could integrate the terms on both side of the equation with the lower bound of the integrals set to be 1 while its upper bound equals to t. Then, it requires bit more work than part a) to find the solution which equals

5. Find the solution of the variational problem

       (20 Marks) 

  Sol: