Tiberiu Popoviciu, Author at Tiberiu Popoviciu Institute of Numerical Analysis

by
Tiberiu Popoviciu
in Cernăuți.
Received February 26, 1937.

1.
- —
  
  Either $\Delta=\left\|a_{1k}\right\|$ a determinant with real elements and order $n_{\sigma}$ According to the well-known theorem of MJ Hadamard we have

|\Delta|^{2}\leq\prod_{=1}^{n}\left(\sum_{=1}^{n}a_{tk}^{2}\right)

so
(1)

|\Delta|\leq\sqrt{n^{n}}\mathbb{M}^{n}\text{ if }\left|a_{lk}\right|\leq\mathbb{M}\text{. }

Either $\mathrm{F}=\mathrm{F}\left(x_{1},x_{2},\ldots,x_{n}\right)=\sum_{i,k=1}^{n}c_{ik}x_{i}x_{k}$ a positive definite quadratic form. Considering the linear transformation which brings back $\mathrm{F}^{-}$ to its canonical form $\sum_{i=1}^{n}x_{i}^{2}$ , we immediately find that

|\Delta|^{2}\leq\frac{\prod_{i=1}^{n}\mathrm{\penalty 10000\ F}\left(a_{i1},a_{i2},\ldots,a_{in}\right)^{(1)}}{\delta},

(2)

Or $\delta$ is the determinant of the form F. Inequality (2) is only a consequence of that of MJ Hadamard. Simple examples
(1) We obtain this formula by multiplying the determinant line by line by any determinant of order $n$ And $\neq 0$ .

More generally, $\mathrm{F}\left(x_{1},x_{2},\ldots,x_{n}\right)$ being a positive definite Hermitian form, with determinant $\delta$ , we have
$\left(\text{ absolute value of }\left\|a_{i\mathrm{k}}\right\|\right)^{2}\leq\frac{\prod_{i=1}^{n}\mathrm{\penalty 10000\ F}\left(a_{i1},a_{i},\ldots,a_{in}\right)}{\delta}$
for a determinant $\left\|a_{ik}\right\|$ with any real or complex elements.
We show that a suitable choice of the form F allows, in certain cases, to give, by formula (2), a limitation better than that of MJ Hadamard, for the absolute value of the determinant.
2. - In particular, we focus our attention on formula (1). We will assume that the $a_{ik}$ be all $\geq 0$ and we will show that we can then lower the factor $\sqrt{n^{n}}$ in the second member of formula (1). We can obviously take $\mathrm{M}=1$ for the demonstrations and our problem can then be stated as follows:

Determine the form F which, under the hypothesis $0\leq a_{dk}\leq 1$ , generally gives the best limitation (2).

We can immediately see that this other problem must be solved:

Determine the quadratic form F , positive definite, of determinant $\delta$ , so that

\max_{0\leq x_{i}\leq 1}\frac{F\left(x_{1},x_{2},\ldots,x_{n}\right)}{\sqrt[n]{\delta}}

(3)

be as small as possible.
The function $F\left(x_{1},x_{2},\ldots,x_{n}\right)$ is convex in the Jensen sense, - we therefore have
$\mathrm{F}\left(\frac{x_{1}+x_{1}^{\prime}}{2},\frac{x_{2}+x_{2}^{\prime}}{2},\ldots,\frac{x_{n}+x_{n}^{\prime}}{2}\right)<\frac{1}{2}\left[\mathrm{\penalty 10000\ F}\left(x_{1},x_{2},\ldots,x_{n}\right)+\mathrm{F}\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{n}^{\prime}\right)\right]$ ,
provided that $\left|x_{1}-x_{1}^{\prime}\right|+\left|x_{2}-x_{2}^{\prime}\right|+\ldots+\left|x_{n}-x_{n}^{\prime}\right|>0$ . We then know that, in the (convex) domain $0\leq x_{i}\leq 1$ , it can only reach its maximum on the boundary. The function being a fortiori convex with respect to each group of variables, we conclude that

\max_{0\leq x_{i}\leq 1}F\left(x_{1},x_{2},\ldots,x_{n}\right)

:can only be achieved if $x_{1},x_{2},\ldots,x_{n}$ are all equal to 0 or 1. - This maximum is therefore equal to one of the $2^{n}$ numbers that we obtain by replacing all the variables in F $x_{i}$ by 0 or 1. The value is 0 for $x_{1}=x_{2}=\ldots=x_{n}=0$ and we can therefore leave it aside; the others are then divided into $n$ groups. The $h^{\text{th }}$ group of values is formed by the $\binom{n}{k}$ numbers that we obtain by giving to $n-k+1$ variables the value 1 and to the others $k-1$ variables the value 0. The arithmetic mean of the numbers in the $k^{\text{th }}$ group is equal to

(n-k+1)\frac{\sum_{i=1}^{n}c_{it}}{n}+(nk)(n-k+1)\frac{\sum_{i,k=1}^{n}c_{ik}}{n(n-1)}

by agreeing to designate by $\sum_{i,k=1}^{n}$ a summation where the values $i=E_{i}$ are excluded.
3. - To find the minimum of expression (3) it is sufficient to consider only forms F symmetric with respect to the variables… This property will result from the following lemma:

Lemma. If the quadratic form $\mathrm{F}=\sum_{i,k=1}^{n}c_{ik}x_{l}x_{k}$ is positive definite, if $\delta$ is its determinant and if $\delta_{1}$ is the determinant of the symmetric quadratic form

\mathrm{G}=\mathrm{C}\left(\sum_{i=1}^{n}x_{i}^{2}\right)+\mathrm{D}\left(\sum_{i,k=1}^{n}x_{i}x_{k}\right)

Or

\mathrm{C}=\frac{\sum_{i=1}^{n}c_{it}}{n},\quad\mathrm{D}=\frac{\sum_{i,k=1}^{n}c_{i,k}^{\prime}}{n(n-1)}

we $has$ :
10. The form G is positive definite.
$2^{0}.\quad\delta\leq\delta_{1}$ , equality being possible only if $\mathrm{F}\equiv\mathrm{G}$ , so if F is symmetric.

The determinant $\delta$ is a function of the coefficients $c_{ik}$ (it is a polynomial in $c_{ik}$ ). Suppose that these coefficients vary in such a way that the form remains positive definite and that

\sum_{i=1}^{n}c_{il}=\mathrm{A}=\text{ const },\quad\sum_{i,k=1}^{n}c_{ik}=\mathrm{B}=\text{ const. }

The domain of variation of the $c_{k}$ is then open and obviously limited. On the border of this domain $\delta$ becomes null $\left({}^{2}\right)$ . The maximum of $\delta$ . is therefore reached inside and we obtain it by applying the rules of differential calculus. If we denote by $\mathrm{C}_{1k}$ minors (with their

signs) of $\delta$ , it is necessary for the maximum that

\frac{\partial\delta}{\partial c_{d}}=\lambda,\quad\frac{\partial\delta}{\partial c_{ik}}=\mu,\quad i,j,k=1,2,\ldots,n,\quad j\neq k

so that

\mathrm{C}_{11}=\mathrm{C}_{22}=\ldots=\mathrm{C}_{mn},\quad\mathrm{C}_{ik}=\mathrm{C}_{r^{\prime}k^{\prime}}\quad\left(i\neq k,i^{\prime}\neq k^{\prime}\right)

(5)

It is therefore necessary that the adjoining form and, consequently, that the form itself be symmetrical.

Since the maximum must necessarily exist and since, on the other hand, system (5) has only the single solution G, the properties result ³ ).

Theorem I. If the form F is not symmetrical, we can construct another form for which the number (3) is smaller.

The form G constructed above answers the question. This follows immediately from the facts that (4) is an arithmetic mean, that this expression is the same for the form $G$ and that the lemma is proven.
4. - Let us now suppose that $\mathrm{F}=\sum_{i=1}^{n}x_{i}^{2}-\lambda\sum_{i,k=1}^{n}x_{i}x_{k}$ (
3) Property 10 of the lemma can also be established directly. We can always write $c_{ik}=\sum_{j=1}^{n}\alpha_{ij}\alpha_{kj}$ where the real numbers $\alpha_{ik}$ are such that the determinant $\left\|\alpha_{ik}\right\|\neq 0$ . We then have

\begin{gathered}\mathrm{G}=\frac{\sum_{j=1}^{n}\sum_{i=1}^{n}\alpha_{j}^{2}}{n}\sum_{i=1}^{n}x_{i}^{2}+\frac{\sum_{j=1}^{n}\sum_{\imath,k=1}^{n}\alpha_{ij}^{\prime}x_{kj}}{n(n-1)}\sum_{i,k=1}^{n}x_{i}x_{k}=\\ =\frac{1}{n(n-1)}\sum_{j=1}^{n}\left[(n-1)\left(\sum_{i=1}^{n}\alpha_{ij}^{2}\right)\left(\sum_{\imath=1}^{n}x_{i}^{2}\right)+\left(\sum_{\imath,k=1}^{n}\alpha_{ij}\alpha_{kj}\right)\left(\sum_{i,k=1}^{n}x_{i}x_{k}\right)\right]=\\ =\frac{1}{n!}\sum_{j=1}^{n}\left[\sum^{*}\left(\alpha_{1j}x_{1}+\alpha_{2j}x_{2}+\ldots+\alpha_{nj}x_{n}\right)^{2}\right]\end{gathered}

the sum $\Sigma^{*}$ being extended to $n!$ permutations of variables $x_{1},x_{2},\ldots,x_{n}$ . An analogous property holds for positive definite Hermitian forms.
symmetric. We have $\delta=[1-(n-1)\lambda](1+\lambda)^{n-1}$ and for the form to be positive definite it is necessary that $-1<\lambda<\frac{1}{n-1}$ .

Let's consider the numbers

\mathrm{N}_{\mathrm{k}}(\lambda)=\mathrm{N}_{\mathrm{k}}=(n-k+1)\frac{1-(n-k)\lambda}{\sqrt[n]{\delta}}

—

We need to determine $\lambda$ such as $\max_{1\leq k\leq n}\mathrm{\penalty 10000\ N}_{k}$ be as small as possible. We have

\mathrm{N}_{\mathrm{k}}-\mathrm{N}_{\mathrm{k}_{1}}=\left(k_{1}-k\right)\frac{1-\left(2n-k-k_{1}+1\right)\lambda}{\sqrt{\delta}}

(6)

and we see immediately that $\mathrm{N}_{\mathrm{k}}>\mathrm{N}_{\mathrm{k}_{1}}$ if $k_{1}>k,k+k_{1}=n+2$ ; so we just need to consider the numbers $\mathrm{N}_{i},i=1,2,\ldots,\left\lceil\frac{n}{z}\right\rceil+1$ , by designating by $[\alpha]$ the largest integer included in $\alpha$ . If we pose $\lambda_{9}-1$ , $\lambda_{1}=\frac{1}{2(n-i)},\lambda_{\left[\frac{n}{2}\right]+1}=\frac{1}{n-1}$ , formula (6) then shows us that $\max_{1\leq k\leq n}\mathrm{\penalty 10000\ N}_{\mathrm{k}}=\mathrm{N}_{i}$ in the meantime $\left(\lambda_{l-1},\lambda_{l}\right),i=1,2,\ldots,\left|\frac{n}{2}\right|+1$ .

It remains to be examined $\mathrm{N}_{i}$ in the meantime ( $\lambda_{i-1},\lambda_{i}$ ). Let us designate by $\mathrm{N}_{f}^{*}(\lambda)$ derivatives $\mathrm{d}\rightarrow\mathrm{N}_{i}$ compared to $\lambda$ , freed from a factor which is positive in the interval ( $\lambda_{1-1},\lambda_{i}$ ). We have

N_{i}^{*}(\lambda)=[(n-2)(n-i)+n-1]\lambda-(n-i)

and we see that $N_{i}^{n}\left(\lambda_{i}\right)<0$ For $i\leq\left|\frac{n}{2}\right|$ , therefore, for $i=1,2,\ldots,\left|\frac{n}{2}\right|$ ;
(7) $\min_{\left(\lambda_{i-1},\lambda l\right)}\mathrm{N}_{l}(\lambda)=\mathrm{N}_{i}\left(\lambda_{l}\right)=\frac{(n-i)(n-i+1)}{\sqrt[n]{(n-2i+1)(2n-2i+1)^{n-1}}}$ .

When $i=\left\lceil\frac{n}{2}\right\rfloor+1$ , we have $N_{\left[\frac{n}{2}\right]+1}^{*}\left(\lambda_{\left[\frac{n}{2}\right]}\right)>0$ if $n$ is even and $\mathrm{N}_{\left[\frac{n}{2}\right]+1}^{*}\left(\frac{1}{n}\right)=0$ if $n$ is odd, so in the interval $\left(\lambda_{\left[\frac{n}{2}\right]},\frac{1}{n-1}\right)$ ,

\min N_{\left[\frac{n}{2}\right]+1}=N_{\left[\frac{n}{2}\right]+1}\left(\frac{1}{n}\right)=\begin{cases}\frac{n(n+2)}{n}&n\text{ pair }\\ 4\sqrt{(n+1)^{n-1}}&\\ \frac{n}{(n+1)^{n+1}}&n\text{ impair. }\end{cases}

We easily find that (8) is smaller than the numbers (1)
and so we have the $\left({}^{4}\right)$
Theorem II. If $\mathrm{F}=\sum_{i=1}^{n}x^{2}-\lambda\sum_{i,k=1}^{n}x_{i}x_{k}$ is positive definite we have $\min_{-1<\lambda<\frac{1}{n-1}}\max_{0\leq x_{l}\leq 1}\frac{\mathrm{\penalty 10000\ F}\left(x_{1},x_{2},\ldots,x_{n}\right)}{\sqrt[n]{\delta}}=\begin{cases}\frac{n(n+2)}{n}&n\text{ pair }\\ 4\sqrt{(n+1)^{n-1}}&\\ \frac{n}{\sqrt{(n+1)^{n+1}}}&n\text{ impair }\end{cases}$ and this minimum is reached for $\lambda=\frac{1}{n}$ .
5. - Returning to the determinant $\Delta$ , we can state the

Theorem III. If all the elements of the determinant $\Delta=\left\|a_{ik}\right\|$ are non-negative and at most equal to M, we have

|\Delta|=\begin{cases}\sqrt{\frac{n^{n}(n+2)^{n}}{4^{n}(n+1)^{n-1}}}M^{n}&n\text{ pair }\\ \frac{\sqrt{(n+1)^{n+1}}}{2^{n}}M^{n}&n\text{ impair. }\end{cases}

For equality to hold in (9) it is necessary:
$1^{0}$ . that the equality takes place in (2), which comes from Mr. Hadamard's inequality.
20. that among the elements of a row (or a column) $a_{i1},a_{i2},\ldots,a_{in}n-\left\lceil\frac{n}{2}\right\rfloor$ are equal to 1 (to M) and the others are equal to 0.

The condition $1^{0}$ is written

\sum_{j=1}^{n}a_{ljj}\frac{\partial\mathrm{\penalty 10000\ F}\left(i1,a_{l2},\ldots,a_{ln}\right)}{\partial a_{ij}}=0,\quad i\neq k

which, taking into account the special form of the shape $F\left(\lambda=\frac{1}{n}\right){}^{\mathrm{e}}\mathrm{t}$ of the condition $2^{0}$ , becomes

(n+1)\sum_{j=1}^{n}a_{ij}a_{kj}=\left(n-\left\lceil\frac{n}{2}\right\rfloor\right)^{2},\quad i\neq k.

(10)

( ⁴ ) If we pose $n-2i=x$ in the second member of formula (7) we easily verify that the derivative with respect to a $x$ of this expression is positive for $x>0$ . For $n$ odd we still have to verify the inequality $-2(n+1)(n+2)^{n-1}<(n+3)^{n}(n>1)$ , which is immediate.

The first member of (10) is a multiple of $n+1$ , it is therefore necessary that $n$ either of the form $n=4p-1$ . In this case the condition $2^{0}$ and equalities (10) are necessary and sufficient for the determinant to be maximizing. For example for $n=3,7,11$ we have the maximizing determinants

		$\displaystyle\left\|\begin{array}[]{lll}1&1&0\\ 1&0&1\\ 0&1&1\end{array}\right\|$

		$\displaystyle\left\|\begin{array}[]{llllllllllll}1&1&1&1&1&1&0&0&0&0&0\\ 1&1&1&0&0&0&1&1&1&0&0\\ 1&1&0&1&0&0&1&0&0&1&1\\ 1&0&1&0&1&0&0&1&0&1&1\\ 1&0&0&1&0&1&0&1&1&1&0\\ 1&0&0&0&1&1&1&0&1&0&1\\ 0&1&1&0&0&1&0&0&1&1&1\\ 0&1&0&1&1&0&0&1&1&0&1\\ 0&1&0&0&1&1&1&1&0&1&0\\ 0&0&1&1&1&0&1&0&1&1&0\\ 1&0&1&0&1&0&1\\ 0&0&1&1&0&1&1&1&0&1&0&1\\ 0&1&1&0&0&1&1\\ 0&1&0&1&1&0&1\end{array}\right\|$

which are equal, in absolute value, to $2,\Sigma^{5},2.3^{6}$ respectively.
6. - We can pose the more general problem of seeking the best limitation by assuming that the elements $a_{1k}$ of the determinant are included between two numbers $m$ And $\mathrm{M},m<\mathrm{M}$ . We can assume- $\mathrm{M}>0,-\mathrm{M}\leq m<\mathrm{M}$ , without restricting the generality. We can solve this problem by following the same path as above, but it should be noted that the results are, in general, more complicated…

It is still sufficient to consider only forms $F$ symmetrical. with respect to the variables. The numbers $\mathbf{N}_{k}$ become

	$\displaystyle\mathrm{N}_{k}=$	$\displaystyle\frac{(n-k+1)\mathrm{M}^{2}+(k-1)m^{2}}{n}-$
		$\displaystyle\quad-\frac{\left[(n-k)(n-k+1)\mathrm{M}^{2}+2k-1\right)(n-k+1)\mathrm{M}m+(k-1)(k-2)m^{2}}{\sqrt{\delta}}$

Formula (6) becomes
(6') $\quad\mathrm{N}_{k}-\mathrm{N}_{k_{1}}=\left(k_{1}-k\right)(\mathrm{M}-m)\frac{\mathrm{M}+m-\left[\left(2n-k-k_{1}+1\right)\mathrm{M}+\left(k+k_{1}-3\right)m\right]\lambda}{\sqrt[n]{\delta}}$ and we still see that $\mathrm{N}_{k}>\mathrm{N}_{k_{1}}$ if $k_{1}>k,k+k_{1}=n+2$ .

In this case it is necessary to ask $\lambda_{0}=-1,\lambda_{l}=\frac{\mathrm{M}+m}{2(n-i)\mathrm{M}+2(i-1)m}$ ,
$\lambda_{\left[\begin{array}[]{l}n\\ i\end{array}\right]+1}=\frac{1}{n-1}$ and we will still have
$\max_{\leq k\leq n}\mathrm{\penalty 10000\ N}_{k}=\mathrm{N}_{i}$ in the meantime $\left(\lambda_{i-1},\lambda_{i}\right),i=1,2,\ldots,\left\lceil\frac{n}{2}\right\rceil+1$ .
The expression $\mathrm{N}_{i}^{*}(\lambda)$ becomes
$\mathrm{N}_{i}^{*}(\lambda)=\left\{(n-i+1)[(n-2)(n-i)+n-1]\mathrm{M}^{2}+2(i-1)(n-i+1)(n-2)\mathrm{M}m+\right.$

\begin{array}[]{r}\left.+(i-1)[(n-2)(i-2)+n-1]m^{2}\right\}\lambda-(n-i)(n-i+1)\mathrm{M}^{2}\\ -2(i-1)(n-i+1)\mathrm{M}m-(i-1)(i-2)m^{2}\end{array}

and it all depends on the sign of this function of $\lambda$ in the meantime. ( $\lambda_{i-1},\lambda_{i}$ ). We can write

\begin{array}[]{r}\mathrm{N}_{i}^{*}\left(\lambda_{i}\right)=\frac{\mathrm{M}-m}{2(n-i)\mathrm{M}+2(i-1)m}\left\{\left(n-i+1,\left[-2i^{2}+(3n+2)i-n^{2}-n-1\right](\mathrm{M}-m)+\right.\right.\\ \left.+2(n-i+1)\left[(2n-1)i-n^{2}\right](\mathrm{M}-m)m+n(n-1)(2i-n-1)m^{2}\right\}.\end{array}

7.
- —
  
  Suppose, in particular, that $m\geq 0$ . We then see that $\mathrm{N}_{i}^{*}\left(\lambda_{i}\right)<0$ For $i\leq\left\lfloor\frac{n}{2}\right\rfloor$ and it is now permissible to write the formulas (7) again, which become
  (7') $\min_{\left(\lambda_{i-1},\lambda_{i}\right)}\mathrm{N}_{i}(\lambda)=\mathrm{N}_{i}\left(\lambda_{i}\right)=\frac{(\mathrm{M}-m)\left[(n-i)(n-i+1)\mathrm{M}^{2}-i(i-1)m^{2}\right]}{\sqrt{}(\mathrm{M}-m)(n-2i+1)[(2n-2i+1)\mathrm{M}+(2i-1)m]^{n-1}}$ : For $i=1,2,\ldots,\left|\begin{array}[]{l}n\\ 2\end{array}\right|$ .

If we pose $n-2i=x$ , we easily verify that the derivative of the second member of $\left(7^{\prime}\right)$ is positive for $x>0$ . The smallest among. the numbers ( $7^{\prime}$ ) is therefore

\frac{n(\mathrm{M}-m)\left[(n+2)\mathrm{M}^{2}-(n-2)m^{2}\right]}{4\sqrt{(\mathrm{M}-m)[(n+1)\mathrm{M}+(n-1)m]^{n-1}}}\text{ pour }n\text{ pair }

(11)

\frac{(\mathrm{M}-m)\left[(n+1)(n+3)\mathrm{M}^{2}-(n-1)(n-3)m^{2}\right]}{4\sqrt[n]{2(\mathrm{M}-m)[(n+2)\mathrm{M}+(n-2)m]^{n-1}}}\text{ pour }n\text{ impair. }

8.
- —
  
  Let us first assume that $n$ is even. We then find that $\mathrm{N}_{\left[\frac{n}{2}\right]+1}^{*}\left(\lambda_{\left[\frac{n}{2}\right]}>0\right.$ and the minimum sought is equal to
  $\mathrm{N}_{\left[\frac{n}{2}\right]+1}\left(\frac{\mathrm{M}+m}{n\mathrm{M}+(n-2)m}\right)$ whose numerical value is (11), so
  Theorem IV. If all the elements of the determinant $\Delta=\left\|a_{ik}\right\|$ , of even order $n$ , are between two positive numbers $m$ And $M>m$ , with a

|\Delta|\leq\sqrt{\frac{n^{n}(\mathrm{M}-m)^{n-1}[(n+2)\mathrm{M}-(n-2)m]^{n}}{4^{n}[(n+1)\mathrm{M}+(n-1)m]^{n-1}}}

For equality to occur, it is necessary
that

\sum_{j=1}^{n}a_{ij}a_{kj}=\frac{n^{2}(\mathrm{M}+m)^{3}}{4[(n+1)\mathrm{M}+(n-1)m]}\quad i\neq k

2 than among the elements of a row (or column) $\frac{n}{2}$ are equal to $M$ And $\frac{n}{2}$ equal to $m$ .

—

The sum $\sum_{j=1}^{n}a_{ij}a_{kj}$ is of the form $\mu\mathrm{M}^{2}+(n-2\mu)\mathrm{M}m+\mu m^{2}==\mu(\mathrm{M}-m)^{2}+n\mathrm{M}m$ , Or $\mu$ is a positive integer (the case $\mu=0$ is obviously to be excluded if $n>2$ ). We easily find

\mu=\frac{n}{4}\left[1-\frac{(\mathrm{M}+m)^{2}}{(\mathrm{M}-m)[(n+1)\mathrm{M}+(n-1)m]}\right]

So we must have $\mu<\frac{n}{4}$ . On the other hand, we can always write the first two lines of a maximizing determinant in
wet form so that we can place a third line verifying the
conditions $1^{0}$ And $2^{0}$ it is necessary that $\mu>\frac{n}{6}$ . We can therefore affirm that for. $n=4,6,8$ there are surely no maximizing determinates.

Note. The case $n=2$ is an exception. In this case the determinant $\left|\begin{array}[]{cc}\mathrm{M}&m\\ m&\mathrm{M}\end{array}\right|$ can be maximizing. It is necessary and sufficient for this that $\mathrm{M}=(1+\sqrt{2})m$ .
9. - The case where $n$ is odd is more interesting. In this case $N_{\left[\frac{n}{2}\right]+1}^{b}\left(\lambda_{\left[\frac{n}{2}\right]}\right)<0$ and the minimum is given by the root of the equation, $\mathrm{N}_{\left[\begin{array}[]{l}n\\ 2\\ 2\end{array}\right]+\mathrm{i}}^{*}(\lambda)=0$ which is equal to

\lambda^{\prime}=\frac{(u+1)\mathrm{M}^{2}+2(n+1)\mathrm{M}m+(n-3)m^{2}}{n(n+1)\mathrm{M}^{2}+2(n+1)(n-2)\mathrm{M}m+\left(n^{2}-3n+4\right)m^{2}}

This minimum is equal to

\mathrm{N}_{\left[\begin{array}[]{l}n\\ 2\\ 2\end{array}\right]+1}\left(\lambda^{\prime}\right)=\frac{(n+1)(\mathrm{M}-m)^{2}[(n+1)\mathrm{M}+(n-1)m]^{2}}{4\sqrt[n]{(n+1)(\mathrm{M}-m)^{2}[(n+1)\mathrm{M}+(n-1)m]^{2n-2}}}

which is indeed smaller than (12) ( ⁵ ), so we have
Theorem V. If all elements of the determinant $\Delta=\left\|a_{ik}\right\|$ odd order $n$ , are between two positive numbers $m$ And $M>m$ , we $a$ .

|\Delta|\leq\frac{\sqrt{(n+1)^{n-1}}}{2^{n}}(M-m)^{n-1}[(n+1)M+(n-1)m]

For equality to take place it is necessary
$1^{0}$ that

\sum_{j=1}^{n}a_{ij}a_{lj}=\frac{(n+1)\mathrm{M}^{2}+2(n+1)\mathrm{M}m+(n-3)m^{2}}{4}

$2^{0}$ than among the elements of a row (or column) $\frac{n+1}{2}$ are equal to M and the others $\frac{n-1}{2}$ equal to $m$ .

In this case the sum $\sum_{j=1}^{n}a_{ij}a_{kj}$ is of the form $\mu\mathrm{M}^{2}++(n-2\mu+1)\mathrm{M}m+(\mu-1)m^{2}=\mu(\mathrm{M}-m)^{2}+(n+1)\mathrm{M}m-m^{2},\mu$ being an integer
$\left({}^{5}\right.$ ) In the form

\left|\frac{(n+1)(n+3)\mathrm{M}^{2}-(n-1)(n-3)n^{2}}{(n+1)(\mathrm{M}-m)[(n+2)\mathrm{M}+(n-2)m]}\right|^{n}>\frac{2[(n+1)\mathrm{M}+(n-1)m]^{2}}{(n+1)(\mathrm{M}-m)[(n+2)\mathrm{M}+(n-}

returns to the elementary inequality of Bernoulli.
positive. We easily find $\mu=\frac{n+1}{4}$ and it is therefore necessary that $n$ either of: the form $n=4p-1$ . The result is the same as in the case $m=0$ . Moreover, the fact that a determinant is maximizing does not depend on M and $m$ .

Final remark. We wanted to simply show some elementary consequences of formula (2). It would remain to demonstrate the existence of maximizing determinants of any order of the form $n=4p-1$ . When $n$ is not of this form the maximum ine problem can be solved by formula (2) (at least for $n=4p-3$ and for $n=4,6,8$ ).

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REMARKS ON THE MAXIMUM OF A DETERMINANT WHOSE ELEMENTS ARE ALL NON-NEGATED

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