# Problems in Mathematics

Problems are listed in the increasing order of difficulty. When a problem is simply a mathematical statement, the readers are supposed to supply a proof. Answers are given (or will be given) to all of problems. This is mostly for the 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 readers are strongly discouraged to see the answers before they successfully solved the problems themselves.

## Commutative algebraEdit

Problem: A finite integral domain is a ﬁeld.

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

$x \mapsto ax: A \to A$

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

Problem: A polynomial has integer values for sufficiently large integer arguments if and only if it is a linear combination (over $\mathbf{Z}$) of binomial coefficients $\binom{t}{n}$.

Since $\deg \binom{t}{n} = n$, for dimensional reason, if $f \in R[t]$, then we can write:

$f(t) = a_0 + a_1 \binom{t}{1} + ... + a_d \binom{t}{d}, a_n \in \mathbf{R}.$

Applying finite differential operator $\Delta g(t) = g(t + 1) - g(t)$ to both sides d times, one finds that $a_d$ is an integer. By induction, $a_n$ are all integers then.$\square$

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 $\mathfrak{i} \in S$. We will reach a contradiction once we show $\mathfrak{i} \in \operatorname{Spec}(A)$. For that end, let $xy \in \mathfrak{i}$. If $x \not\in \mathfrak{i}$, then, by maximality, $(\mathfrak{i}, x) \not\in S$. That is, it is principal; say,

$(\mathfrak{i}, x) = (d)$.

Let $\mathfrak{j}$ be an ideal consisting of $a \in A$ such that $ax \in \mathfrak{i}$. It turns out that

$\mathfrak{j}d = \mathfrak{i}$.

Indeed, $\mathfrak{j}d = (\mathfrak{ji}, \mathfrak{j}x) \subset \mathfrak{i}$. Conversely, if $z \in \mathfrak{i}$, then $z = z' d$ and $z'x \in z'(d) \subset \mathfrak{i}$. Thus, $z \in \mathfrak{j}d$. We now conclude that $y \in \mathfrak{i}$, for, otherwise, $\mathfrak{j}d$ is principal. $\square$

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

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

$(\mathfrak{i}, x) = (i_1 + a_1 x, ..., i_n + a_n x)$.

Let $\mathfrak{j}$ be an ideal consisting of $a \in A$ such that $ax \in \mathfrak{i}$. It turns out that

$\mathfrak{i} = (i_1, ..., i_n, \mathfrak{j}x)$.

In fact, if $z \in \mathfrak{i}$, then

$z = b_1 (i_1 + a_1 x) + ... + b_n (i_n + a_n x)$

Here, $b_1 a_1 + ... + b_n a_n \in J$. We conclude that $y \in \mathfrak{i}$, for, otherwise, $\mathfrak{j}$ and thus $\mathfrak{i}$ are finitely generated.$\square$

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 $u \in A$. Then u is invertible in $F(u) \subset A$. $\square$

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

Problem: If $f \in A[X]$ is a unit, then $f - a_0$ is nilpotent, where $a_0 = f(0)$ is the constant term of f.

Let $g = b_0 + b_1x + ... + b_m x^m$ be the inverse of f. Then $a_n b_m = 0$. Since $a_n b_{m-1} + a_{n-1} b_m = 0$, it follows ${a_n}^2 b_{m-1} = 0$. By obvious induction, for some r, we see ${a_n}^r$ kills every coefficient of $g$; hence, g. Thus, ${a_n}^r = {a_n}^r f g = 0$, meaning $a_n$ is nilpotent. Recall that the sum of a unit and a nilpotent element is a unit. Since $a_n x^n - f$ is a unit, by applying the above argument, we see that $a_{n-1} x^{n-1}$ is nilpotent. In the end, we conclude that $a_0 - f$ is a sum of nilpotent elements; thus, nilpotent.

Problem: The nilradical and the Jacobson radical of $A[X]$ coincide.

We only have to prove: if $f$ is in the Jacobson radical, then $f$ is nilpotent, since the converse is true for any ring. Recall that $1 - fg$ is a unit for every $g$. In particular, $1 - xf$ is unit. Now use the previous problem to conclude $f$ is nilpotent.

Problem: Let A be a ring such that every ideal not contained in its nilradical contains an element e such that $e^2 = e \ne 0$. Then the nilradical and the Jacobson radical of $A$ 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 $e(1-e) = 0$. Since $1-e$ is a unit, $e = 0$, a contradiction.

Problem: $f \in A[[X]]$ is a unit if and only if the constant term of f is a unit.

## Real analysisEdit

Problem: $\sqrt{3} + 2^{1/3}$ is irrational.

Let $x = \sqrt{3} + 2^{1/3}$. Then $2 = (x - \sqrt{3})^3$. The equation can then be solved for $\sqrt{3}$ $\square$

Problem: Is $\sqrt{2}^\sqrt{2}$ irrational?

Problem: Compute $\int_{-\infty}^\infty {\sin x \over x}$

Problem: If $\lim_{x \to c} f(x) + f'(x)$ exists, then $\lim_{x \to c} f(x)$ exists and $\lim_{x \to c} f'(x) = 0$

Apply L'Hospital's rule to $e^x f(x) \over e^x$

Problem: Let $f:\R \to [0,+\infty)$ nonvanishing and such that $f(x)f''(x)\geq0$, then $\int_{-\infty}^{+\infty}f(x)^2dx=+\infty$

Derive twice $g(x)=f(x)^2$, look at the asymptotic behaviour of $g'(x)$.

Problem Let $X$ be a complete metric space, and $f:X \to X$ be a function such that $f \circ f$ is a contraction. Then $f$ admits a fixed point.

By Banach's fixed point theorem, $f \circ f$ has a unique fixed point $x_0$; i.e., $x_0 = (f \circ f)(x_0)$. But then

$f(x_0) = (f \circ f)(f(x_0))$

In other words, $f(x_0)$ is also a fixed point of $f \circ f$. By uniqueness, $x_0 = f(x_0)$. $\square$

Problem Let $X$ be a compact metric space, and $f:X \to X$ be such that

$d(f(x), f(y)) < d(x, y)$

for all $x \ne y \in X$. Then $f$ admits a unique fixed point. (Do not use Banach's fixed point theorem.)

Let $c = \inf \{ d(x, f(x)) | x \in X \}$. By compactness, there is $x_0$ such that $c = d(x_0, f(x_0))$. If $x_0 \ne f(x_0)$, then, by hypothesis, we have:

$d(x_0, f(x_0)) \le d(f(x_0), f \circ f(x_0)) < d(x_0, f(x_0))$,

which is absurd. Thus, $d(x_0, f(x_0)) = 0$. For uniqueness, suppose $y_0 = f(y_0)$. If $x_0 \ne y_0$, then

$d(x_0, y_0) \le d(x_0, f(x_0)) + d(f(x_0), f(y_0)) + d(f(y_0), y_0) = d(f(x_0), f(y_0)) < d(x_0, y_0)$,

which is absurd. Hence, $x_0$ is the unique fixed point. $\square$

Problem Let $f:\R^2 \to \R^2$ be such that

$d(f(x), f(y)) \geq A d(x, y), \qquad A>1$

then $f$ admits a unique fixed point.

Problem Let $X$ be a compact metric space, and $f:X \to X$ be a contraction. Then

$\bigcap_n^\infty f^n(X)$

consists of exactly one point.

Since f is a contraction, it admits a fixed point $x_0$. Thus, $x_0 \in \cap f^n(X)$. Let $y \in \cap f^n(X)$. Then

$y = f(x_1) = f^2(x_2) = f^3(x_3) = ...$

for some sequence $x_1, x_2, ...$. Let c be the Lipschitz constant of f. Now,

$d(x_0, y) = d(x_0, f^n(x_n)) \le d(x_0, f^n(x_1)) + d(f^n(x_1), f^n(x_n)) \le d(x_0, f^n(x_1)) + c^n d(f(x_1), f(x_n))$

which goes to 0 as $n \to \infty$ since $d(f(x_1), f(x_n))$ is bounded and $c < 1$ and since for any $y \in X$ $f^n(y) \to x_0$ by Banach's fixed point theorem. $\square$

Problem: Every closed subset of $\mathbf{R}^n$ is separable.

Let $E_n$ be a countable dense subset of $A \cap \overline{B}(0, k)$, and let

$E = \bigcup_{k=1}^\infty E_k$

Then $A = \overline{E}$. In fact, since $E_k$ is a subset of $A$ for any $k$, $E \subset A$ and so $\overline{E} \subset \overline{A} = A$. Conversely, if $x \in A$, then for some $k$,

$x \in A \cap \overline{B}(0, k) = \overline{E_k} \subset \overline{E}$.

$\square$

Problem: Any connected nonempty subset of $\mathbf{R}$ either consists of a single point or contains an irrational number.

Let $E$ be a connected nonempty subset of $\mathbf{R}$. Then $E$ is an interval with end-points a, b. If $E$ has more than one point, then $E$ contains a nonempty interval (a, b), which contains an irrational. $\square$

Problem: Let $f: \mathbf{R} \to \mathbf{R}$ be a bounded function. $f$ is continuous if and only if $f$ has closed graph.

Problem: Let $f: \mathbf{R} \to \mathbf{R}$ be a homeomorphism, then $f$ is monotone.

Problem Let $f:[0,1]^2 \to \mathbf{R}$ be a continuous function. Then

$g(x) = \sup \{ f(x, y) | y \in [0, 1] \} \quad (x \in [0, 1])$

is continuous.

Let $\epsilon > 0$. Since $f$ is uniformly continuous, there is $\delta > 0$ so that

$|f(x', y') - f(x, y)| < \epsilon$ whenever $|(x', y') - (x, y)| < \delta$

It follows that $g(x) < \epsilon + g(x')$ as well as $g(x') < \epsilon + g(x)$ when $|x' - x| < \delta$. Hence,

$|g(x') - g(x)| < \epsilon$. $\square$

Problem Let $f, g: \mathbf{R} \to \mathbf{R}$ be continuous functions such that: $f(g(x)) = g(f(x))$ for every $x$. The equation $f(f(x)) = g(g(x))$ has a solution if and only if $f(x) = g(x)$ has one.

($\Rightarrow$) is trivial. For ($\Leftarrow$), suppose we have $x$ so that $f(f(x)) = g(g(x))$. Define $h(y) = f(y) - g(y)$ for $y \in \mathbf{R}$. Then

$h(f(x)) = f(f(x)) - g(f(x)) = g(g(x)) - f(g(x)) = -h(g(x))$.

Thus, $h(f(x)) + h(g(x)) = 0$. If $h(f(x)) = 0$, then we are done. If $h(f(x)) < 0$, then, since $h(g(x)) > 0$, by the intermediate value theorem, $h(z) = 0$ for some $z$. The same argument works for the case when $h(f(x)) > 0$. $\square$

Problem Suppose $f: \mathbf{R} \to \mathbf{R}$ is uniformly continuous. Then there are constants $a, b$ such that:

$|f(x)| \le a|x| + b$

for all $x \in \mathbf{R}$.

There exists $\delta > 0$ such that

$|f(x) - f(y)| < 1$ whenever $|x-y| < \delta$.

Let $x \ge 0$. Then $n - 1 \le {x \over \delta} \le n$ for some integer $n \ge 1$. It follows:

$|f(x)| \le |f(0)| + |f(0) - f(\delta)| + ... + |f((n-1) \delta) - f(x)| \le |f(0)| + n$

Here, $n < 1 + {1 \over \delta} |x|$. The estimate for $x < 0$ is analogous. $\square$

Problem Let X be a compact metric space, and $f: X \to X$ be an isometry: i.e., $d(f(x), f(y)) = d(x, y)$. Then f is a bijection.

f is clearly injective. To show f surjective, let $x \in X$. Since X is compact, $f^n(x)$ contains a convergent subsequence, say, $f^{n_j}(x)$. Then

$d(x, f^{n_j}(x)) = d(f^{n_k}(x), f^{n_{j+k}}(x)) \to 0$

In other words, $x$ is in the closure of $f(X)$. Since the image of a closed set under an isometry is closed, we conclude: $x \in f(X)$. $\square$

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

We first prove a weaker statement:

If $p_n$ converges pointwise on all but finitely many points in [0,1], then $p_n$ 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:

$p_n(x) = p_n(x_0) + (x - x_0) q_n(x)$

where $x_0$ is a point such that $p_n(x_0) \to 0$. Since the degree of $q_n$ is strictly less than that of $p_n$, by inductive hypothesis, $q_n$ converges uniformly on the complement of some finite subset F of [0, 1]. Thus, $p_n$ converges on the complement of $F \cup \{x_0\}$. This completes the proof of the claim. By the claim, $p_n$ converges uniformly except at some finitely many points. But since $p_n$ converges pointwise everywhere, it converges uniformly everywhere. $\square$

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

Problem Prove that in Rn the relation $B_r(x)\supset B_s(y)$ implies r > s and find a metric space when the implication doesn't hold.

## Linear algebraEdit

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

Problem Given an $n$, find a matrix with integer entries such that $A \ne I$ but $A^n = I$

A permutation matrix. $\square$

Problem Let A be a real symmetric positive-definite matrix and b some fixed vector. Let $\phi(x) = \langle Ax, x \rangle - 2 \langle x, b \rangle$. Then $Az = b$ if and only if $\phi(z) \le \phi(x)$

Note that A is invertible. Fix $x \ne 0$ and let $f(t) = \phi(A^{-1} b + tx)$. Then

$f'(t) = 2t\langle Ax, x \rangle$

Since $f''(0) > 0$, $t = 0$ is the minimum of f. $\square$

Problem If $\operatorname{tr}(AB) = 0$ for all square matrices $B$, then $A = 0$

Take $B = A^T$. $\square$

Problem Let x be a square matrix over a field of characteristic zero. If $\operatorname{tr}(x^k) = 0$ for all $k > 0$, then $x$ is nilpotent.

We may assume the field is algebraically closed. Suppose $x$ has nonzero distinct eigenvalues $\lambda_1, \lambda_2, ..., \lambda_n$. The hypothesis then means that we have the system of linear equations:

$\{ \lambda_1^k + \lambda_2^k + ... + \lambda_n^k = 0 \}_{1 \le k \le n}$

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

Problem Let $S, T$ be square matrices of the same size. Then $ST$ and $TS$ have the same eigenvalues.

Let $\lambda$ be an eigenvalue of $ST$. If $\lambda = 0$, then $0 = \det(ST) = \det(TS)$. Thus, $\lambda$ is an eigenvalue of $TS$. If $\lambda \ne 0$, then $STx = \lambda x$ for some nonzero $x$. Thus, $(TS)Tx = \lambda Tx$. Since $Tx = 0$ implies $\lambda x = 0$, a contradiction, $Tx$ is an eigenvector. Hence, $\lambda$ is an eigenvalue of $TS$. We thus proved that every eigenvalue of $ST$ is an eigenvalue of $TS$. By the same argument, every eigenvalue of $TS$ is an eigenvalue of $ST$. $\square$

Problem Let $S, T$ be square matrices of the same size. Then $ST$ and $TS$ have the same eigenvalues with same multiplicity.

If S is invertible, then

$ST = S(TS)S^{-1}$

and thus

$\operatorname{det}(TS - \lambda I) = \operatorname{det}(ST - \lambda I)$.

If S is not invertible, then $S + tI$ is invertible when $t > 0$ is sufficiently small. Thus,

$\operatorname{det}(T(S + tI) - \lambda I) = \operatorname{det}((S + tI)T - \lambda I)$

and letting $t \to 0$ gives the same identity. In any case, TS and ST share the same eigenvalues with same multiplicity.$\square$

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

$\langle Ax, x \rangle$

is real for every x.

($\Rightarrow$) is obvious. ($\Leftarrow$) By hypothesis

$\langle Ax, x \rangle = \langle A^*x, x \rangle$

Now recall that the numerical radius

$w(T) = \sup_{\|x\|=1} |\langle T x, x \rangle|$

is a norm. $\square$

Problem Suppose the square matrix $a_{ij}$ satisfies:

$|a_{ii}| > \sum_{j \ne i} |a_{ij}|$

for all $i$. Then $A$ is invertible.

Suppose $Ax = 0$. Then, in particular, each component of $Ax$ is zero; i.e.,

$0 = \sum_j a_{ij}x_j = a_{ii}x_i + \sum_{j \ne i} a_{ij}x_j$

The inequality $| |a| - |b| | \le |a + b|$ thus gives:

$|a_{ii}| |x_i| = |\sum_{j \ne i} a_{ij}x_j| \le \sum_{j \ne i} |a_{ij}| |x_j|$

for all $i$. Pick $k$ such that $\max \{ |x_1|, |x_2|, ... |x_n| \} = |x_k|$. Then, by hypothesis,

$|a_{kk}| |x_k| \le (\sum_{j \ne i} |a_{ij}|) |x_k| < |a_{kk}||x_k|$,

which is absurd, unless $|x_k| = 0$. Hence, $x = 0$. $\square$

Problem Let $T, S \in \operatorname{End}(V)$. If $V$ is finite-dimensional, then prove $TS$ is invertible if and only if $ST$ is invertible. Is this also true when $V$ is infinite-dimensional?

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

Problem: Let $T, S$ be linear operators on $V$. Then

$\operatorname{dim}\operatorname{ker}(TS) \le \operatorname{dim}\operatorname{ker}(S) + \operatorname{dim}\operatorname{ker}(T)$

The map

$S: \operatorname{ker}(TS) \to \operatorname{ker}T$

is well-defined. Hence,

$\operatorname{dim}\operatorname{ker}(T) \ge \operatorname{dim}\operatorname{ker}(TS) - \operatorname{dim}\operatorname{ker}(S|_{\operatorname{ker}(TS)})$ $\square$

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 $XI - A$ and $XI - B$ 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 $A$ is zero, then $A$ is similar to a matrix with the main diagonal consisting of only zeros.

Problem $\operatorname{rank}(A^n) - \operatorname{rank}(A^{n-1}) \le \operatorname{rank}(A^{n+1}) - \operatorname{rank}(A^n)$ for any square matrix $A$.

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

Jordan form or Schur form.

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

Problem: Let A be a normal matrix. Then:

$\|A\| = \max_{ |x|=1 } |(Ax \mid x)| = \sup_{\lambda \in \operatorname{Sp}(A)} |\lambda|$

Problem: Let A be a square matrix. Then $A \to 0$ (in operator norm) if and only if the spectral radius of $A < 1$

Problem: Let A be a square matrix. Then $\|A\| = \|A^*A\|^{1/2}$

Problem: $T \mapsto \sup_{\|x\|=1} (Tx \mid x)$ 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 $(Tx \mid x) = 0$ for all x. In particular,
$0 = (Tx + y \mid x+y) = (Tx \mid y) + (Ty \mid x)$
$0 = (Tx + iy \mid x+ iy) = -i (Tx \mid y) + i (Ty \mid x)$
$(Tx \mid y) = 0$
for all x and y. Take $y = Tx$ and we get $Tx = 0$ for all x.