Problems in Mathematics

The problems are listed in increasing order of difficulty. When a problem is simply a mathematical statement, the reader is supposed to supply a proof. Answers are given (or will be given) to all of the problems. This is mostly for quality control; the answers allow contributors other than the initial writer of the problem to check the validity of the problems. In other words, the reader is strongly discouraged from seeing the answers before they successfully solve the problems themselves.

Commutative algebraEdit

Problem: A finite integral domain is a field.


Let   be an element in the finite integral domain A. Then the map


is injective (since A is an integral domain) and is surjective by finiteness. 

Problem: A polynomial has integer values for sufficiently large integer arguments if and only if it is a linear combination (over  ) of binomial coefficients  .


Since  , for dimensional reason, if  , then we can write:


Applying finite differential operator   to both sides d times, one finds that   is an integer. By induction,   are all integers then. 

Problem: An integral domain is a PID if its prime ideals are principal. (Hint: apply Zorn's lemma to the set S of all non-principal prime ideals.)


Suppose, on the contrary, that S is nonempty. Then there is a a maximal element  . We will reach a contradiction once we show  . For that end, let  . If  , then, by maximality,  . That is, it is principal; say,


Let   be an ideal consisting of   such that  . It turns out that


Indeed,  . Conversely, if  , then   and  . Thus,  . We now conclude that  , for, otherwise,   is principal.  

Problem: A ring is noetherian if and only if its prime ideals are finitely generated. (Hint: Zorn's lemma.)


The direction ( ) is obvious. For the converse, let   be the set of all proper ideals of   that are not finitely generated. We want to show   is empty then. Suppose not. Then, by Zorn's lemma,   contains a maximal element  . It follows that   is prime. To see that, let  . If  , then, by maximality,  . That is, it is finitely generated; say,


Let   be an ideal consisting of   such that  . It turns out that


In fact, if  , then


Here,  . We conclude that  , for, otherwise,   and thus   are finitely generated. 

Problem: Every nonempty set of prime ideals has a minimal element with respect to inclusion.

Problem: If an integral domain A is algebraic over a field F, then A is a field.


Since A is an integral domain, it is a subring of some field. Let  . Then u is invertible in  .  

Problem: Every two elements in a UFD have a gcd.

Problem: If   is a unit, then   is nilpotent, where   is the constant term of f.


Let   be the inverse of f. Then  . Since  , it follows  . By obvious induction, for some r, we see   kills every coefficient of  ; hence, g. Thus,  , meaning   is nilpotent. Recall that the sum of a unit and a nilpotent element is a unit. Since   is a unit, by applying the above argument, we see that   is nilpotent. In the end, we conclude that   is a sum of nilpotent elements; thus, nilpotent.

Problem: The nilradical and the Jacobson radical of   coincide.


We only have to prove: if   is in the Jacobson radical, then   is nilpotent, since the converse is true for any ring. Recall that   is a unit for every  . In particular,   is unit. Now use the previous problem to conclude   is nilpotent.

Problem: Let A be a ring such that every ideal not contained in its nilradical contains an element e such that  . Then the nilradical and the Jacobson radical of   coincide.


In general, the nilradical is contained in the Jacobson radical. Suppose this inclusion is strict. Then by hypothesis there is a nonzero e such that  . Since   is a unit,  , a contradiction.

Problem:   is a unit if and only if the constant term of f is a unit.

Real analysisEdit

Problem:   is irrational.


Let  . Then  . The equation can then be solved for    

Problem: Is   irrational?

Problem: Compute  

Problem: If   exists, then   exists and  


Apply L'Hospital's rule to  

Problem: Let   nonvanishing and such that  , then  


Derive twice  , look at the asymptotic behaviour of  .

Problem Let   be a complete metric space, and   be a function such that   is a contraction. Then   admits a fixed point.


By Banach's fixed point theorem,   has a unique fixed point  ; i.e.,  . But then


In other words,   is also a fixed point of  . By uniqueness,  .  

Problem Let   be a compact metric space, and   be such that


for all  . Then   admits a unique fixed point. (Do not use Banach's fixed point theorem.)


Let  . By compactness, there is   such that  . If  , then, by hypothesis, we have:


which is absurd. Thus,  . For uniqueness, suppose  . If  , then


which is absurd. Hence,   is the unique fixed point.  

Problem Let   be such that


then   admits a unique fixed point.

Problem Let   be a compact metric space, and   be a contraction. Then


consists of exactly one point.


Since f is a contraction, it admits a fixed point  . Thus,  . Let  . Then


for some sequence  . Let c be the Lipschitz constant of f. Now,


which goes to 0 as   since   is bounded and   and since for any     by Banach's fixed point theorem.  

Problem: Every closed subset of   is separable.


Let   be a countable dense subset of  , and let


Then  . In fact, since   is a subset of   for any  ,   and so  . Conversely, if  , then for some  ,



Problem: Any connected nonempty subset of   either consists of a single point or contains an irrational number.


Let   be a connected nonempty subset of  . Then   is an interval with end-points a, b. If   has more than one point, then   contains a nonempty interval (a, b), which contains an irrational.  

Problem: Let   be a bounded function.   is continuous if and only if   has closed graph.

Problem: Let   be a homeomorphism, then   is monotone.

Problem Let   be a continuous function. Then


is continuous.


Let  . Since   is uniformly continuous, there is   so that


It follows that   as well as   when  . Hence,


Problem Let   be continuous functions such that:   for every  . The equation   has a solution if and only if   has one.


( ) is trivial. For ( ), suppose we have   so that  . Define   for  . Then


Thus,  . If  , then we are done. If  , then, since  , by the intermediate value theorem,   for some  . The same argument works for the case when  .  

Problem Suppose   is uniformly continuous. Then there are constants   such that:


for all  .


There exists   such that

  whenever  .

Let  . Then   for some integer  . It follows:


Here,  . The estimate for   is analogous.  

Problem Let X be a compact metric space, and   be an isometry: i.e.,  . Then f is a bijection.


f is clearly injective. To show f surjective, let  . Since X is compact,   contains a convergent subsequence, say,  . Then


In other words,   is in the closure of  . Since the image of a closed set under an isometry is closed, we conclude:  .  

Problem Let   be a sequence of polynomials with degree ≤ some fixed D. If   converges pointwise to 0 on [0, 1], then   converges uniformly on [0, 1].


We first prove a weaker statement:

If   converges pointwise on all but finitely many points in [0,1], then   converges uniformly on all but finitely many points in [0,1].

We proceed by inducting on D. If D = 1, then the claim is obvious. Suppose it is true for D - 1. We write:


where   is a point such that  . Since the degree of   is strictly less than that of  , by inductive hypothesis,   converges uniformly on the complement of some finite subset F of [0, 1]. Thus,   converges on the complement of  . This completes the proof of the claim. By the claim,   converges uniformly except at some finitely many points. But since   converges pointwise everywhere, it converges uniformly everywhere.  

Problem On a closed interval a monotone function has at most countably many discontinuous points.

Problem Prove that in Rn the relation   implies r > s and find a metric space when the implication doesn't hold.

Linear algebraEdit

Throughout the section   denotes a finite-dimensional vector space over the field of complex numbers.

Problem Given an  , find a matrix with integer entries such that   but  


A permutation matrix.  

Problem Let A be a real symmetric positive-definite matrix and b some fixed vector. Let  . Then   if and only if  


Note that A is invertible. Fix   and let  . Then


Since  ,   is the minimum of f.  

Problem If   for all square matrices  , then  


Take  .  

Problem Let x be a square matrix over a field of characteristic zero. If   for all  , then   is nilpotent.


We may assume the field is algebraically closed. Suppose   has nonzero distinct eigenvalues  . The hypothesis then means that we have the system of linear equations:


Computing the determinant, we see the system has no nonzero solution, a contradiction.

Problem Let   be square matrices of the same size. Then   and   have the same eigenvalues.


Let   be an eigenvalue of  . If  , then  . Thus,   is an eigenvalue of  . If  , then   for some nonzero  . Thus,  . Since   implies  , a contradiction,   is an eigenvector. Hence,   is an eigenvalue of  . We thus proved that every eigenvalue of   is an eigenvalue of  . By the same argument, every eigenvalue of   is an eigenvalue of  .  

Problem Let   be square matrices of the same size. Then   and   have the same eigenvalues with same multiplicity.


If S is invertible, then


and thus


If S is not invertible, then   is invertible when   is sufficiently small. Thus,


and letting   gives the same identity. In any case, TS and ST share the same eigenvalues with same multiplicity. 

Problem Let   be a square matrix over complex numbers. A is a real symmetric matrix if and only if


is real for every x.


( ) is obvious. ( ) By hypothesis


Now recall that the numerical radius


is a norm.  

Problem Suppose the square matrix   satisfies:


for all  . Then   is invertible.


Suppose  . Then, in particular, each component of   is zero; i.e.,


The inequality   thus gives:


for all  . Pick   such that  . Then, by hypothesis,


which is absurd, unless  . Hence,  .  

Problem Let  . If   is finite-dimensional, then prove   is invertible if and only if   is invertible. Is this also true when   is infinite-dimensional?


For the first part, use determinant. For the second, consider a shift operator.  

Problem: Let   be linear operators on  . Then


The map


is well-defined. Hence,


Problem Every matrix (over an arbitrary field) is similar to its transpose.


The shortest proof would be to use a Smith normal form: a matrix A is similar to another matrix "B if and only if   and   have the same Smith normal form. Evidently, this is the case if B is the transpose of A.

Problem Every nonzero eigenvalue of a skew-symmetric matrix is pure imaginary.

Problem If the transpose of a matrix   is zero, then   is similar to a matrix with the main diagonal consisting of only zeros.

Problem   for any square matrix  .

Problem: Every square matrix is similar to an upper-triangular matrix.


Jordan form or Schur form.

Problem: Let A be a normal matrix. Then   is a polynomial in A.

Problem: Let A be a normal matrix. Then:


Problem: Let A be a square matrix. Then   (in operator norm) if and only if the spectral radius of  


The Spectral radius formula.

Problem: Let A be a square matrix. Then  

Problem:   is a norm for bounded operators T on a "complex" Hilbert space.


It is clear that the map is a seminorm. To see it is a norm, suppose   for all x. In particular,


Combing the two we get:


for all x and y. Take   and we get   for all x.