Zero determinant of a matrix

Suppose that for the complex square matrices A, B it holds

(1)   \begin{equation*} AB - BA =A \end{equation*}

Prove that \det A =0.


If \det A\ne0, then A is invertible. So we get


But this equality is impossible: taking trace on both sides, we get


We get a contradiction, which shows that A is singular, i.e. \det A=0.

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On a power of matrix

Let n be a natural number such that n \geq 2. Evaluate the power

    \[\mathcal{P} = \begin{pmatrix} 1 &1 \\ 1&0 \end{pmatrix}^n\]


This is a very standard exercise in diagonalisation of matrices and there would be no reason to post it here , if it did not include the Fibonacci result. We are proving that


where F_n denotes the n – th Fibonacci number. The proof now follows with an induction on n.

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Multiple integral on a zero measured set

Let A \subseteq \mathbb{R}^n be a Jordan measurable set of zero measure. Prove that \bigintsss \limits_{A} 1 \, {\rm d} \bar{x}= 0.


Since \mu(A)=0 there exists a sequence U_n of closed rectangles of \mathbb{R}^n such that A \subseteq \bigcup \limits_{n} U_n and \forall \epsilon>0 it is \sum \limits_{n} \mathcal{V} \left( U_n \right)< \epsilon where \mathcal{V}(U_n) is the volume of the rectangle U_n. Then foreach \epsilon>0

    \begin{align*} 0 &\leq \int \limits_{A} 1 \, {\rm d} \bar{x} \\ &\leq \int \limits_{\bigcup \limits_{n} U_n} 1 \, {\rm d} \bar{x} \\ &\leq \sum_{n} \int \limits_{U_n} 1\, {\rm d}\bar{x} \\ &= \sum_{n} \mathcal{V} \left ( U_n \right )\\ &< \epsilon \end{align*}


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Irreducible function on Q[z]

Let f(z) \in \mathbb{Q}[z] be irreducible with degree n>1. If f has a root on the unit circle then n is even and

    \[z^n f\left( \frac{1}{z} \right) = f(z)\]


Let \alpha be a root of f with |\alpha|=1. Since f has real coefficients \bar{\alpha}= \frac{1}{\alpha} is also a root of f. The product z^n f\left( \frac{1}{z} \right) is a polynomial in \mathbb{Q}[z] of degree n ( its leading coefficient is f(0) ) with root \alpha. By the irreducibility of f we have

(1)   \begin{equation*} z^n f\left( \frac{1}{z} \right) =  c f(z) \end{equation*}

for some non zero rational number c. Setting z=1 we have that f(1)=c f(1). Since f(1) \neq 0  , by our hypotheses , c=1 hence z^n f\left( \frac{1}{z} \right) = f(z) . Setting z=-1 we get that f(-1)=(-1)^n f(-1) and because f(-1) \neq 0 we deduce that n is even.

Note: The above tells us that f(z) can be expressed in terms of z + \frac{1}{z}.

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Inequality with roots

Let a, b, c be positive real numbers. Prove that



We apply the AM – GM inequality, thus:

    \begin{align*} \frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}} &=\frac{a}{b}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}} \\ &\geq 6\sqrt[6]{\frac{a}{b}\frac{b}{4c}\frac{c}{27a}} \\ &=\frac{6}{\sqrt[6]{4\cdot 27}} \end{align*}

Hence it suffices to prove that 3>\sqrt[6]{4\cdot 27} which holds because it is equivalent to 3^6> 4\cdot 27.

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