# Complex matrix without a square root

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

# Intersection and union of interiors

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.

# Additive subgroups of vector spaces

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 $$\mathbb F_{p^2}$$.

# A differentiable real function with unbounded derivative around zero

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.