Difference of harmonic number

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

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

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.

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Determinant of nilpotent matrices

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

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

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.

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Equality of fields

Prove that

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

Solution

Coming soon!

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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 .

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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,

    \[\det(P(A+N)P^{-1})=\det(PAP^{-1})\]

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,

    \[\det(A+N)=\det(P(A+N)P^{-1})=\det(PAP^{-1})=\det(A)\]

The exercise along its solution have been migrated from here.

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