Math Problems Vault

Problem 7.1

Let the $n$ vertices of a polyhedron be numbered in a definite order. Define a determinant of $n$ th order as follows:

If the $\lambda$ th and the $\mu$ th vertices are the two end-points of an edge of the polyhedron, let $a_{\lambda \mu } = a_{\mu \lambda } = 1$ .

If the line joining the $\lambda$ th and the $\mu$ th vertices is not an edge of the polyhedron, let $a_{\lambda \mu } = 0$ . In particular $a_{\lambda \lambda } = 0$ , $\lambda = 1, 2, \ldots , n$ .

Show that the value of this determinant is independent of the manner of numbering the vertices. Form and calculate the determinants for the regular tetrahedron, hexahedron and octahedron.

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Problem 7.1.1

(continued). If exactly $p$ edges end in each vertex the determinant is divisible by $p$ .

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Problem 7.2

Calculate

\begin{vmatrix} 1 & 1 & 1 & \ldots & 1 \\ b_{1} & a_{1} & a_{1} & \ldots & a_{1} \\ b_{1} & b_{2} & a_{2} & \ldots & a_{2} \\ \multicolumn {5}{c}{\dotfill } \\ b_{1} & b_{2} & b_{3} & \ldots & a_{n} \end{vmatrix}

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Problem 7.3

Prove the identity:

\left|\frac{1}{a_{\lambda } + b_{\mu }}\right|_{1}^{n} = \frac{\prod _{j > k}^{1, 2, \ldots , n} (a_{j} - a_{k})(b_{j} - b_{k})}{\prod _{\lambda , \mu }^{1, 2, \ldots , n} (a_{\lambda } + b_{\mu })}.

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Problem 7.4

Denoting the determinant of the quadratic form

\sum _{\lambda =1}^{n} \sum _{\mu =1}^{n} \frac{x_{\lambda } x_{\mu }}{\lambda + \mu }

by $D_{n}$ , we have

D_{n} = \frac{[1! 2! \ldots (n-1)!]^{3} n!}{(n+1)! (n+2)! \ldots (2n)!}.

Also calculate the determinant $D_{n}(\alpha )$ of the quadratic form

\sum _{\lambda =1}^{n} \sum _{\mu =1}^{n} \frac{x_{\lambda } x_{\mu }}{\lambda + \mu + \alpha }, \qquad \alpha > -2.

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Problem 7.5

\left|(a_{\lambda } - b_{\mu })^{n-1}\right|_{1}^{n} = \prod _{\nu =1}^{n-1} (n-\nu )^{n-2\nu } \prod _{j > k}^{1, 2, \ldots , n} (a_{j} - a_{k})(b_{j} - b_{k}).

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Problem 7.6

Derive Theorem 86 of Part V from Theorem 48 of Part V by means of a transformation of the determinant $\left|F(\alpha_{\lambda } \beta_{\mu })\right|$ .

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Problem 7.7

Setting $f(x) = (r_{1} - x)(r_{2} - x) \ldots (r_{n} - x)$ , we have

\begin{vmatrix} r_{1} & a & a & \ldots & a \\ b & r_{2} & a & \ldots & a \\ b & b & r_{3} & \ldots & a \\ \multicolumn {5}{c}{\dotfill } \\ b & b & b & \ldots & r_{n} \end{vmatrix} = \frac{a f(b) - b f(a)}{a - b}.

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Remark.

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Add to all the $n^{2}$ elements the variable $x$ . The determinant formed in this way is a linear function of $x$ and as such can be determined from two particular values.

Problem 7.8

Set $\Delta = ad - bc$ . Then the functional determinant

\frac{\partial (a\Delta , b\Delta , c\Delta , d\Delta )}{\partial (a, b, c, d)} = 3\Delta ^{4}.

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Problem 7.9

Prove that $\rho - 2$ is a divisor of the expression

\begin{vmatrix} \rho & \dfrac {l}{m} + \dfrac {m}{l} & \dfrac {n}{l} + \dfrac {l}{n} \\ ]1ex] \dfrac {l}{m} + \dfrac {m}{l} & \rho & \dfrac {m}{n} + \dfrac {n}{m} \\[1ex] \dfrac {n}{l} + \dfrac {l}{n} & \dfrac {m}{n} + \dfrac {n}{m} & \rho \end{vmatrix}, \qquad l \neq 0, \; m \neq 0, \; n \neq 0.

Determine the remaining factors.

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Problem 7.10

The determinant

\left|1, \; x_{\nu }, \; x_{\nu }^{2}, \; \ldots , \; x_{\nu }^{n-q-1}, \; x_{\nu }^{n-q+1}, \; \ldots , \; x_{\nu }^{n-1}, \; x_{\nu }^{n}\right|, \qquad \nu = 1, 2, \ldots , n

is an alternating rational entire function of the $n$ numbers $x_{1}, x_{2}, \ldots , x_{n}$ and thus divisible by the product of differences

\prod _{j > k}^{1, 2, \ldots , n} (x_{j} - x_{k}).

Show that the quotient $S_{q}$ is equal to the $q$ th elementary symmetric function of the $n$ numbers $x_{1}, x_{2}, \ldots , x_{n}$ .

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Problem 7.11

Let the numbers $a_{0}, a_{1}, a_{2}, \ldots , a_{n}$ be different from $0$ . Then we have identically in $z$

a_{0} \begin{vmatrix} z + \dfrac {a_{1}}{a_{0}} & \dfrac {a_{2}}{a_{1}} & \dfrac {a_{3}}{a_{2}} & \ldots & \dfrac {a_{n-1}}{a_{n-2}} & \dfrac {a_{n}}{a_{n-1}} \\ ]1ex] -\dfrac {a_{1}}{a_{0}} & z & 0 & \ldots & 0 & 0 \\[1ex] 0 & -\dfrac {a_{2}}{a_{1}} & z & \ldots & 0 & 0 \\ \multicolumn {6}{c}{\dotfill } \\ 0 & 0 & 0 & \ldots & -\dfrac {a_{n-1}}{a_{n-2}} & z \end{vmatrix} = a_{0} z^{n} + a_{1} z^{n-1} + \cdots + a_{n}.

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Problem 7.11.1

Let $a_{0}, a_{1}, a_{2}, \ldots , b_{0}, b_{1}, b_{2}, \ldots$ denote constants, $z$ a variable. Suppose that $a_{0} \neq 0$ and that we have identically

(a_{0} + a_{1} z + a_{2} z^{2} + \cdots )(b_{0} + b_{1} z + b_{2} z^{2} + \cdots ) = 1.

Then

a_{0}^{n+1} b_{n} = (-1)^{n} \begin{vmatrix} a_{1} & a_{0} & 0 & 0 & \ldots & 0 \\ a_{2} & a_{1} & a_{0} & 0 & \ldots & 0 \\ a_{3} & a_{2} & a_{1} & a_{0} & \ldots & 0 \\ \multicolumn {6}{c}{\dotfill } \\ a_{n} & a_{n-1} & a_{n-2} & a_{n-3} & \ldots & a_{1} \end{vmatrix}.

(This serves as a method of computing the reciprocal of a given power series. The supposed identity may be merely formal; no considerations of convergence are involved.)

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Problem 7.11.2

Compute the coefficient of $z^{n}$ in the expansion of $[(1+z)^{a} - xz]^{-1}$ first from Problem 11.1 then directly.

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Problem 7.11.3

Analogously to Problem 11.2, compute the coefficient of $z^{n}$ in the expansion of $[e^{z} - xz]^{-1}$ .

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Problem 7.12

Let $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}, c$ be real numbers. The system

\begin{aligned} -a_{3} x_{2} + a_{2} x_{3} & = b_{1}, \\ a_{3} x_{1} \phantom{{} + a_{2} x_{2}} - a_{1} x_{3} & = b_{2}, \\ -a_{2} x_{1} + a_{1} x_{2} \phantom{{} - a_{1} x_{3}} & = b_{3}, \\ a_{1} x_{1} + a_{2} x_{2} + a_{3} x_{3} & = c \end{aligned}

can be consistent in two different ways. In one case the unknowns $x_{1}, x_{2}, x_{3}$ are completely undetermined, in the other case they are completely determined.

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Problem 7.13

Let the entire function

1 + c_{1} z + c_{2} z^{2} + c_{3} z^{3} + \cdots

have all distinct zeros $a_{1}, a_{2}, \ldots , a_{n}, \ldots ,$

0 < \left|a_{1}\right| < \left|a_{2}\right| < \cdots < \left|a_{n}\right| < \cdots .

Consider the system of $n$ equations

\begin{aligned} 1 + a_{1} u_{1}^{(n)} + a_{1}^{2} u_{2}^{(n)} + \cdots + a_{1}^{n} u_{n}^{(n)} & = 0, \\ 1 + a_{2} u_{1}^{(n)} + a_{2}^{2} u_{2}^{(n)} + \cdots + a_{2}^{n} u_{n}^{(n)} & = 0, \\ & \dotfill \\ 1 + a_{n} u_{1}^{(n)} + a_{n}^{2} u_{2}^{(n)} + \cdots + a_{n}^{n} u_{n}^{(n)} & = 0 \end{aligned}

that completely determine the $u_{k}^{(n)}$ . If

\frac{1}{\left|a_{1}\right|} + \frac{1}{\left|a_{2}\right|} + \cdots + \frac{1}{\left|a_{n}\right|} + \cdots

is convergent, then $\lim_{n \rightarrow \infty } u_{k}^{(n)}$ exists, but it is not necessarily true that

\lim _{n \rightarrow \infty } u_{k}^{(n)} = c_{k}.

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Remark.

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One solution of the infinite system

1 + a_{\nu } u_{1} + a_{\nu }^{2} u_{2} + \cdots = 0, \qquad \nu = 1, 2, 3, \ldots

is: $u_{1} = c_{1}, u_{2} = c_{2}, u_{3} = c_{3}, \ldots$ .

Problem 7.14

In the equations

\begin{aligned} c_{11} z_{1} + c_{12} z_{2} + \cdots + c_{1n} z_{n} & = 0, \\ c_{21} z_{1} + c_{22} z_{2} + \cdots + c_{2n} z_{n} & = 0, \\ & \dotfill \\ c_{n1} z_{1} + c_{n2} z_{2} + \cdots + c_{nn} z_{n} & = 0 \end{aligned}

let the coefficients and the unknowns be complex:

c_{\lambda \mu } = a_{\lambda \mu } + i b_{\lambda \mu }, \qquad z_{\mu } = x_{\mu } + i y_{\mu },

$a_{\lambda \mu }, b_{\lambda \mu }, x_{\mu }, y_{\mu }$ real. For the equations to admit not only the identically zero solution

z_{1} = z_{2} = \cdots = z_{n} = 0, \qquad \text{i.e.} \quad x_{1} = x_{2} = \cdots = x_{n} = y_{1} = y_{2} = \cdots = y_{n} = 0,

it is necessary and sufficient that the determinant $\left|c_{\lambda \mu }\right|_{1}^{n}$ vanishes. This yields two equations in the $2n^{2}$ real numbers $a_{\lambda \mu }, b_{\lambda \mu }$ . On the other hand the given equations may be written as $2n$ linear homogeneous equations for $2n$ real unknowns. The necessary and sufficient condition for the existence of a not identically vanishing solution now consists in the vanishing of a real determinant, i.e. in one equation in the $a_{\lambda \mu }, b_{\lambda \mu }$ . How can this be correct?

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Problem 7.15

The six terms in the expansion of a third-order determinant cannot all be positive.

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Problem 7.16

The rule for the expansion of a determinant consists of two parts. The first part specifies what products are to be formed from the elements, while the second part determines the sign of these products.

The second part may be simplified in the case of second-order determinants in the following manner:

Associate with the elements

\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \text{ the signs } \begin{pmatrix} + & + \\ - & + \end{pmatrix} \text{ respectively.}

Then form the products prescribed in the first part from the elements, each with its fixed associated sign.

Now prove that a corresponding simplification is impossible in the case of a determinant of order higher than two, i.e. it is impossible to associate with the $n^{2}$ elements $n^{2}$ fixed signs in such a way that, if we form the products prescribed in the first part of the expansion rule from the elements each with its associated sign, we automatically obtain the correct sign in all the products.

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