Joy of Mathematics

Difference of harmonic number

August 15, 2019 / Leave a comment

Let $\mathcal{H}_n^{(s)}$ denote the $n$ – th harmonic number of weight $s$ . Prove that

$\sum_{n=1}^{\infty}\frac{\mathcal{H}_n^2- \mathcal{H}_n^{(2)}}{(n+1)(n+2)}=2$

Solution

Recall the generating function

(1) $\begin{equation*} \sum_{n=1}^{\infty}\left(\mathcal{H}_n^2-\mathcal{H}_n^{(2)}\right)x^n = \frac{\ln^2(1-x)}{1-x} \end{equation*}$

Thus $(1)$ from $0$ to $t$ we get that

(2) $\begin{equation*} \sum_{n=1}^{\infty}\left(\mathcal{H}_n^2-\mathcal{H}_n^{(2)}\right)\frac{t^{n+1}}{n+1}=-\frac{1}{3}\ln^3(1-t) \end{equation*}$

Integrating $(2)$ from $0$ to $1$ the result follows.

Determinant of nilpotent matrices

August 15, 2019 / Leave a comment

Let $P, Q$ be nilpotent matrices such that $PQ + P+Q=0$ . Evaluate the determinant

$\Delta = \det \left( \mathbb{I} + 2 P + 3 Q \right)$

Solution

Lemma: If $A$ and $B$ are nilpotent matrices that commute and $a,b$ are scalars, then $aA + bB$ is nilpotent.

Proof: Since $A$ and $B$ commute, they are simultaneously triangularizable. Let $S$ be an invertible matrix such that $A = STS^{-1}$ and $B = SUS^{-1}$ , where $T$ and $U$ are upper triangular. Note that since $A$ and $B$ are nilpotent, $T$ and $U$ must have zeros down the main diagonal. Hence $aT + bU$ is upper triangular with zeros along the main diagonal which means that it’s nilpotent. Finally $aA + bB = S(aT + bU)S^{-1}$ and so $aA + bB$ is nilpotent.

We have $\left(P+\mathbb{I}\right)\left(Q+\mathbb{I}\right) = \mathbb{I} \implies \left(Q+\mathbb{I} \right) \left(P+\mathbb{I}\right) = \mathbb{I}$ . Then equating the two left hand sides and simplifying gives us $PQ = QP$ . Thus by the lemma we know that $2P + 3Q$ is nilpotent, i.e., it’s eigenvalues are all zero. It follows that the eigenvalues of $\mathbb{I} + 2P + 3Q$ are all one and so $\det \left(\mathbb{I}+2P+3Q \right) = 1$ .

Equality of fields

August 12, 2019 / Leave a comment

Prove that

$\mathbb{Q} \left(\sqrt{2}+ \sqrt{5}+ \cdots +\sqrt{n^2+1} \right)=\mathbb{Q} \left(\sqrt{2}, \sqrt{5}, \cdots , \sqrt{n^2+1} \right)$

Solution

Coming soon!

Inequality on groups

August 12, 2019 / 1 Comment on Inequality on groups

Let $\mathcal{G}$ be a finite group and suppose that $\mathcal{H}$ , $\mathcal{K}$ are two subgroups of $\mathcal{G}$ such that $\mathcal{H} \neq \mathcal{G}$ and $\mathcal{K} \neq \mathcal{G}$ . Show that

$\left|\mathcal{H} \cup \mathcal{K} \right| \leq \frac{3}{4} \left| \mathcal{G} \right|$

Solution

Recall that $\displaystyle |\mathcal{H}\mathcal{K}|=\frac{|\mathcal{H}||\mathcal{K}|}{|\mathcal{H} \cap \mathcal{K}|}$ and thus $\displaystyle \frac{|\mathcal{H}||\mathcal{K}|}{|\mathcal{H} \cap \mathcal{K}|} \leq |\mathcal{G}|$ . Hence $\displaystyle |\mathcal{H} \cap \mathcal{K}| \ge \frac{|\mathcal{H}| |\mathcal{K}|}{|\mathcal{G}|}$ and so

(1) $\begin{align*} |\mathcal{H} \cup \mathcal{K}|&=|\mathcal{H}|+|\mathcal{K}|-|\mathcal{H} \cap \mathcal{K}| \\ &\leq |\mathcal{H}|+|\mathcal{K}|-\frac{|\mathcal{H}| |\mathcal{K}|}{|\mathcal{G}|} \\ & =(a+b-ab)|\mathcal{G}| \end{align*}$

where $\displaystyle a=\frac{|\mathcal{H}|}{|\mathcal{G}|}$ and $\displaystyle b=\frac{|\mathcal{\mathcal{K}}|}{|\mathcal{G}|}$ .

Now, since $\mathcal{H} \neq \mathcal{G}$ and $\mathcal{K} \neq \mathcal{G}$ we have $[\mathcal{G}: \mathcal{H}] \geq 2$ and $[\mathcal{G}:\mathcal{K}] \geq 2$ that is $a \leq \frac{1}{2}$ and $b \leq \frac{1}{2}$ . So if we let $a'=1-2a$ and $b'=1-2b$ then $a', b' \geq 0$ and thus

$a+b-ab=\frac{3}{4}-\frac{a'+b'+a'b'}{4} \leq \frac{3}{4}$

due to $(1)$ .

The exercise along its solution have been migrated from here .

Equality of determinants

August 8, 2019 / 1 Comment on Equality of determinants

Let $A,N \in \mathcal{M}_n(\mathbb{C})$ and suppose that $N$ is nilpotent. Show that if $A,N$ commute then

$\det(A+N)=\det(A)$

Solution

Since $\mathbb{C}$ is algebraically closed and $A,N$ commute this means that $A,N$ are simultaneously triangularizable, there exists an invertible element $P \in \mathcal{M}_n(\mathbb{C})$ such that both $PNP^{-1}$ and $PAP^{-1}$ are triangular. Since $PNP^{-1}$ is both nilpotent and triangular, all its diagonal entries are zero and so the diagonal entries of $P(A+N)P^{-1}=PAP^{-1}+PNP^{-1}$ are the same as the diagonal entries of $PAP^{-1}$ . Thus,