Monthly Archives: December 2016

Algebra

Complex matrix without a square root

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Consider for \(n \ge 2\) the linear space \(\mathcal M_n(\mathbb C)\) of complex matrices of dimension \(n \times n\). Is a matrix \(T \in \mathcal M_n(\mathbb C)\) always having a square root \(S \in \mathcal M_n(\mathbb C)\), i.e. a matrix such that \(S^2=T\)? is the question we deal with.

First, one can note that if \(T\) is similar to \(V\) with \(T = P^{-1} V P\) and \(V\) has a square root \(U\) then \(T\) also has a square root as \(V=U^2\) implies \(T=\left(P^{-1} U P\right)^2\).

Diagonalizable matrices

Suppose that \(T\) is similar to a diagonal matrix \[
D=\begin{bmatrix}
d_1 & 0 & \dots & 0 \\
0 & d_2 & \dots & 0 \\
\vdots & \vdots & \ddots & 0 \\
0 & 0 & \dots & d_n
\end{bmatrix}\] Any complex number has two square roots, except \(0\) which has only one. Therefore, each \(d_i\) has at least one square root \(d_i^\prime\) and the matrix \[
D^\prime=\begin{bmatrix}
d_1^\prime & 0 & \dots & 0 \\
0 & d_2^\prime & \dots & 0 \\
\vdots & \vdots & \ddots & 0 \\
0 & 0 & \dots & d_n^\prime
\end{bmatrix}\] is a square root of \(D\). Continue reading Complex matrix without a square root →

Topology

Intersection and union of interiors

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Consider a topological space \(E\). For subsets \(A, B \subseteq E\) we have the equality \[
A^\circ \cap B^\circ = (A \cap B)^\circ\] and the inclusion \[
A^\circ \cup B^\circ \subseteq (A \cup B)^\circ\] where \(A^\circ\) and \(B^\circ\) denote the interiors of \(A\) and \(B\).

Let’s prove that \(A^\circ \cap B^\circ = (A \cap B)^\circ\).

We have \(A^\circ \subseteq A\) and \(B^\circ \subseteq B\) and therefore \(A^\circ \cap B^\circ \subseteq A \cap B\). As \(A^\circ \cap B^\circ\) is open we then have \(A^\circ \cap B^\circ \subseteq (A \cap B)^\circ\) because \(A^\circ \cap B^\circ\) is open and \((A \cap B)^\circ\) is the largest open subset of \(A \cap B\).

Conversely, \(A \cap B \subseteq A\) implies \((A \cap B)^\circ \subseteq A^\circ\) and similarly \((A \cap B)^\circ \subseteq B^\circ\). Therefore we have \((A \cap B)^\circ \subseteq A^\circ \cap B^\circ\) which concludes the proof of the equality \(A^\circ \cap B^\circ = (A \cap B)^\circ\).

One can also prove the inclusion \(A^\circ \cup B^\circ \subseteq (A \cup B)^\circ\). However, the equality \(A^\circ \cup B^\circ = (A \cup B)^\circ\) doesn’t always hold. Let’s provide a couple of counterexamples.

For the first one, let’s take for \(E\) the plane \(\mathbb R^2\) endowed with usual topology. For \(A\), we take the unit close disk and for \(B\) the plane minus the open unit disk. \(A^\circ\) is the unit open disk and \(B^\circ\) the plane minus the unit closed disk. Therefore \(A^\circ \cup B^\circ = \mathbb R^2 \setminus C\) is equal to the plane minus the unit circle \(C\). While we have \[A \cup B = (A \cup B)^\circ = \mathbb R^2.\]

For our second counterexample, we take \(E=\mathbb R\) endowed with usual topology and \(A = \mathbb R \setminus \mathbb Q\), \(B = \mathbb Q\). Here we have \(A^\circ = B^\circ = \emptyset\) thus \(A^\circ \cup B^\circ = \emptyset\) while \(A \cup B = (A \cup B)^\circ = \mathbb R\).

The union of the interiors of two subsets is not always equal to the interior of the union.

Algebra

Additive subgroups of vector spaces

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Consider a vector space \(V\) over a field \(F\). A subspace \(W \subseteq V\) is an additive subgroup of \((V,+)\). The converse might not be true.

If the characteristic of the field is zero, then a subgroup \(W\) of \(V\) might not be an additive subgroup. For example \(\mathbb R\) is a vector space over \(\mathbb R\) itself. \(\mathbb Q\) is an additive subgroup of \(\mathbb R\). However \(\sqrt{2}= \sqrt{2}.1 \notin \mathbb Q\) proving that \(\mathbb Q\) is not a subspace of \(\mathbb R\).

Another example is \(\mathbb Q\) which is a vector space over itself. \(\mathbb Z\) is an additive subgroup of \(\mathbb Q\), which is not a subspace as \(\frac{1}{2} \notin \mathbb Z\).

Yet, an additive subgroup of a vector space over a prime field \(\mathbb F_p\) with \(p\) prime is a subspace. To prove it, consider an additive subgroup \(W\) of \((V,+)\) and \(x \in W\). For \(\lambda \in F\), we can write \(\lambda = \underbrace{1 + \dots + 1}_{\lambda \text{ times}}\). Consequently \[
\lambda \cdot x = (1 + \dots + 1) \cdot x= \underbrace{x + \dots + x}_{\lambda \text{ times}} \in W.\]

Finally an additive subgroup of a vector space over any finite field is not always a subspace. For a counterexample, take the non-prime finite field \(\mathbb F_{p^2}\) (also named \(\text{GF}(p^2)\)). \(\mathbb F_{p^2}\) is also a vector space over itself. The prime finite field \(\mathbb F_p \subset \mathbb F_{p^2}\) is an additive subgroup that is not a subspace of \(\mathbb F_{p^2}\).

Analysis

A differentiable real function with unbounded derivative around zero

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Consider the real function defined on \(\mathbb R\)\[
f(x)=\begin{cases}
0 &\text{for } x = 0\\
x^2 \sin \frac{1}{x^2} &\text{for } x \neq 0
\end{cases}\]

\(f\) is continuous and differentiable on \(\mathbb R\setminus \{0\}\). For \(x \in \mathbb R\) we have \(\vert f(x) \vert \le x^2\), which implies that \(f\) is continuous at \(0\). Also \[
\left\vert \frac{f(x)-f(0)}{x} \right\vert = \left\vert x \sin \frac{1}{x^2} \right\vert \le \vert x \vert\] proving that \(f\) is differentiable at zero with \(f^\prime(0) = 0\). The derivative of \(f\) for \(x \neq 0\) is \[
f^\prime(x) = \underbrace{2x \sin \frac{1}{x^2}}_{=g(x)}-\underbrace{\frac{2}{x} \cos \frac{1}{x^2}}_{=h(x)}\] On the interval \((-1,1)\), \(g(x)\) is bounded by \(2\). However, for \(a_k=\frac{1}{\sqrt{k \pi}}\) with \(k \in \mathbb N\) we have \(h(a_k)=2 \sqrt{k \pi} (-1)^k\) which is unbounded while \(\lim\limits_{k \to \infty} a_k = 0\). Therefore \(f^\prime\) is unbounded in all neighborhood of the origin.