Numerical Methods Qualification Exam Problems and Solutions (University of Maryland)/August 2007

Problem 1Edit

A set of functions   is a Chebyshev system if

(i) The set is linearly independent.

(ii) If   is a linear combination of   which is not identically zero,   has at most   distinct zeros in  .

Problem 1aEdit

Show that   is a Chebyshev system if and only if for any   distinct points   the matrix   with   is non-singular.

Solution 1aEdit

Forward DirectionEdit

We want to prove the following statement:

If   is a Chebyshev system, then for any   distinct points   the matrix   with   is non-singular.

Writing out the matrix   yields:


Since the set   is linearly independent, there does not exist any non-zero sets of constants of   such that   for any  . Hence, the columns of the matrix   are linearly independent which implies that   is non-singular.

Reverse DirectionEdit

Proof of (i)Edit

Assume   is non-singular.

This implies the columns of


are linearly independent. Since   is non-singular for any choice of  ,   is a linearly independent set and we have shown  .

Proof of (ii)Edit

By hypothesis,   is a linear combination of   i.e.


for   not all zero.

Assume for the sake of contradiction that   has   zeros at  

This implies the following   equations:


Rewriting the equations in matrix form yields


Since   are not all zero, this implies that the columns of  ,  , are linearly dependent, a contradiction.

Hence,   has at most   zeros, and we have shown  .

Problem 1bEdit

Let   be such that   for all  . Let  . Show that   is a Chebyshev system. For this, you may use results from polynomial interpolation without proof.

Solution 1bEdit

We have to prove:

(i)   is a linearly independent set

(ii)any linear combination of this set has at most   zeros.

Proof of (i)Edit

If we evaluate   at   distinct points,  , we have the following Van Der Monde matrix whose determinant is non-zero.


Hence,   are linearly independent.

Assume for the sake of contradiction that   is a linear combination of  , that is


Hence,   is a polynomial of degree  . Taking   derivatives of   yields


which contradicts the hypothesis. Therefore   is a linearly independent set.

Proof of (ii)Edit

Let  .

Suppose   has   (or more) zeros in  . By Rolle's Theorem, the (n+1)st derivative of f vanishes on the given interval, a contradiction.

(i) and (ii) show that   is a Chebyshev system.

Problem 2Edit



be a sequence of integration rules.

Problem 2aEdit





for some constant  . Show that


for all  

Solution 2aEdit

By the Weierstrass Approximation Theorem, given  , there exists a polynomial   such that




Adding and subtracting   and  , yields,


By the triangle inequality and equation (2) and (3),


By equation (1), when   ,


Hence for arbitrary small   as  ,




Problem 2bEdit

Show that if all   then (1) implies (2).

Solution 2bEdit

For  , equation (1) yields,


Letting   in equation (0) yields,


Combining the two above results yields,


Since   is finite, there exists some number   such that  .

Since  ,


i.e. equation (2).

Problem 3Edit

Consider the real system of linear equations


where   is non singular and satisfies


for all real  , where the Euclidean inner product is used here.

Problem 3aEdit

Show that   for all real   where   is the symmetric part of  .

Solution 3aEdit

Substituting for   in   and expanding the inner product we have,


From properties of inner products we have,




Problem 3bEdit

Prove that


where   is the minimum eigenvalue of  .

Solution 3bEdit

First InequalityEdit


Since   is symmetric, it has a eigenvalue decomposition


where   is orthogonal and   is a diagonal matrix containing all the eigenvalues.

Substitution yields,




This implies the following three relationships:




Expanding the numerator we have,


Expanding the denominator yield




Second InequalityEdit

From part(a)


for all real  .

Therefore   is positive definite which implies all its eigenvalues are positive. In particular,


Problem 3cEdit

Now consider the iteration for computing a series of approximate solutions to (1),


where   and   is chosen to minimize   as a function of  . Prove that


Solution 3cEdit

First, we want to write   as a function of   i.e.


Changing the indices of equation   from   to  ,substituting for  , and applying the definition of norm yields,


From a property of inner products and since   is symmetric,




Next we want to minimize   as a function of  . Taking its derivative with respect to   yields,


Setting   and solving for   gives


Substituting for   into   gives,


By definition of norm,




Multiplying and dividing the second term on the right hand side of the above equation by   and applying a property of inner product yields,


From the result of part (b), we have our desired result: