Difference of harmonic number

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

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


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.

Read more

Determinant of nilpotent matrices

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

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


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.

Read more

Equality of fields

Prove that

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


Coming soon!

Read more

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

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


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 .

Read more


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



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,


because P(A+N)P^{-1}PAP^{-1} are both triangular and the determinant of a triangular matrix is the product of its diagonal entries. So,


The exercise along its solution have been migrated from here.

Read more

Who is Tolaso?

Find out more at his Encyclopedia Page.

Donate to Tolaso Network