The canonical form of a determinant and its applications

5 months ago

D.V. Ionescu

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D.V. Ionescu
Institutul de Calcul

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D.V. Ionescu, The canonical form of a determinant and its applications. (Romanian) Acad. R. P. Romîne Fil. Cluj Stud. Cerc. Mat. 10 1959 33–44.
[Forma canonică a unui determinant și aplicațiile sale]

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Studii si Cercetari Matematice Cluj

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Academia Republicii S.R.

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https://ictp.acad.ro/scm/journal/article/view/41

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THE CANONICAL FORM OF A DETERMINANT AND ITS APPLICATIONS

OF

DV IONESCU

It is known that any matrix with elements from a given field can be reduced to a canonical form by conveniently repeated elementary transformations. The elementary transformations are the following:

T_{1}

: one line is replaced by another line;

T_{2}

: multiply the elements of one line by a factor and add them to the elements of another line.

By changing the word "line" in the previous definitions with the word "column", we obtain the transformations

T_{1}^{'}, T_{2}^{'}

.

In particular, if the matrix is square

A = ‖ \begin{array}{cccc} A_{11} & A_{12} & \dots & A_{1 n} \\ A_{21} & A_{22} & \dots & A_{2 n} \\ \cdot & \cdot & \dots & \cdot \\ A_{n 1} & A_{n 2} & \dots & A_{n n} \end{array} ‖

its canonical form is

A^{*} = ‖ \begin{array}{lllll} A_{1} \\ A_{2} \\ ⋱ \\ A_{j} \\ 0 \\ 0 & ⋱ \\ 0 \end{array}

with all elements, except the diagonal, equal to zero.
We will call the canonical form of the determinant

D = | A |

, the determinant

D^{*} = | A^{*} |

.

In general we have

D = D^{*}

or

D = - D^{*}

, from which it follows:

1^{\circ}

If

D \neq 0

, then the canonical form of

D

is

3 - Mathematics studies and research

D^{*} = | \begin{array}{cccc} A_{1} & 0 \\ A_{2} \\ ⋱ \\ 0 & A_{n} \end{array} |

with all diagonal elements nonzero, and vice versa.

2^{\circ}

If

D = 0

, then the canonical form of

D

is

D^{*} = ‖ \begin{array}{cccccc} A_{1} & 0 \\ A_{2} & 0 \\ ⋱ \\ A_{j} \\ 0 & 0 \\ 0 \end{array}

with at least one element of the diagonal equal to zero, and vice versa.
In this paper we will make applications of the canonical form of a determinant. Although we have done several, we will limit ourselves to only two, namely:

1^{\circ}

. proving the well-known formula that relates the reciprocal determinant to a given order of the determinant

D

, by the determinant

D

;

2^{\circ}

. proving the well-known Sylvester identity.
The proofs we will make will be based on the following idea:
Let us assume that we have to calculate a determinant

Δ

which corresponds to a determinant

D

If an invariant can be highlighted

f (D, Δ)

for any elementary transformation

T_{1}, T_{2}, T_{1}^{'}, T_{2}^{'}

, then we will have

f (D, Δ) = f (D^{*}, Δ^{*})

where

Δ^{*}

is the determinant that corresponds to the canonical form

D^{*}

of the determinant D. If the determinant

Δ^{*}

is calculated directly and easily from

D^{*}

, then the previous formula will give us

Δ

.

§ 1. The reciprocal determinant of the order $j$ of a determinant

Whether

\begin{matrix} (1) & D = | \begin{array}{cccc} A_{11} & A_{12} & \dots & A_{1 n} \\ A_{21} & A_{22} & \dots & A_{2 n} \\ \cdot & \cdot & \dots & \cdot \\ A_{n 1} & A_{n 2} & \dots & A_{n n} \end{array} | \end{matrix}

an arbitrary determinant and consider its minors of the order

j

.

\begin{matrix} (2) & A (\begin{array}{ccc} {and}_{1}, & {and}_{2}, & \dots, \\ k_{1}, & k_{2}, & \dots, \\ k_{j} \end{array}) \end{matrix}

formed with the elements of the determinant

D

, common to the lines of rank

{and}_{1}, {and}_{2}, \dots \dots, {and}_{j}

and rank columns

k_{1}, k_{2}, \dots, k_{j}

. The groups (

{and}_{1}, {and}_{2}, \dots, {and}_{j}

),

(k_{1}, k_{2}, \dots, k_{j})

are made with

j

indices among indices

1, 2, \dots, n

The number of all
minors of the form (2) is

{(\binom{n}{j})}^{2}

We will arrange these minors in a determinant

Δ_{j}

with

(\binom{n}{j})

lines and

(\binom{n}{j})

columns in the following way. We arrange all the groupings of

j

INDICATORS

(α_{1}, α_{2}, \dots, α_{j})

take from the clues

1, 2, \dots, n

in a row

S

, so that

(α_{1}, α_{2}, \dots, α_{j}), (α_{1}^{'}, α_{2}^{'}, \dots, α_{j}^{'})

being two consecutive terms of the sequence, let us have

α_{h} ≦ α_{h}^{'}

for

h = 1, 2, \dots, j

.

We will form the determinant

Δ_{j}

, making the groups (

{and}_{1}, {and}_{2}, \dots, {and}_{j}

) and

(k_{1}, k_{2}, \dots, k_{j})

to go through all the terms of the sequence

S

.

It is noted

\begin{matrix} (3) & Δ_{j} = | A (\begin{array}{ccc} {and}_{1}, & {and}_{2}, & \dots, \\ k_{1}, & k_{2}, & \dots, \\ k_{j} \end{array}) | \end{matrix}

reciprocal determinant of the order

j

, his

D

These determinants were first considered by Cauchy. It was shown that

\begin{matrix} (4) & Δ_{j} = D^{(\binom{n - 1}{j - 1})} \end{matrix}

We will give the proof of this formula using the canonical form of the determinant

D

.

Paying attention to the first two lines of the determinant

D

, we will write the determinant

Δ_{j}

in the form of

Δ_{j} = \begin{aligned} \dots \dots \dots \dots \dots \dots \\ A (\binom{1, 2, {and}_{3}, \dots, {and}_{j}}{k_{1}, k_{2}, k_{3}, \dots, k_{j}}) \\ \dots \dots \dots \dots \dots \dots \dots \\ A (\binom{1, {and}_{2}, {and}_{3}, \dots, {and}_{j}}{k_{1}, k_{2}, k_{3}, \dots, k_{j}}) \\ \dots \dots \dots \dots \dots \dots, \\ A (\binom{2, {and}_{2}, {and}_{3}, \dots,}{k_{1}, k_{2}, k_{3}, \dots, k_{j}}) \\ \dots \dots \dots \dots \dots \dots \dots \\ A (\begin{array}{c} {and}_{1}, {and}_{2}, {and}_{3}, \dots, \\ k_{1}, k_{2}, k_{3}, \dots, k_{j} \\ \dots \dots \dots \dots \dots \dots, \dots \end{array}) \end{aligned}

in which the elements of the column (

k_{1}, k_{2}, \dots, k_{j}

).
In the first lines of the determinant

Δ_{j} s

-they highlighted the lines with minors of the order

j

of the determinant

D

containing the first two lines;

(i_{3}, i_{4}, \dots, i_{j})

being a group of some kind with

j - 2

indices taken from indices

3, 4, \dots, n

, the number of these lines is

(\binom{n - 2}{j - 2})

.

In the following lines, the minors in the determinant were highlighted.

D

, with

j

lines of which the first is the line

1 - a

and then the line a

2 - a

from the determinant

D; (i_{2}, \dots, i_{j})

is a group with

j - 1

indices taken from
indices

(3, 4, \dots, n)

, the number of these lines

ϵ

you are

(\binom{n - 2}{j - 1})

. Finally, in the following lines (

i_{1}, i_{2} \dots, i_{j}

) is a grouping with

j

indices taken from indices

3, 4, \dots, n

The number of these lines is

(\binom{n - 2}{j})

It is easy to verify that the number of all lines is

(\binom{n - 2}{j - 2}) + (\binom{n - 2}{j - 1}) + (\binom{n - 2}{j - 1}) + (\binom{n - 2}{j}) = (\binom{n}{j}) .

If in the determinant

D

change the first line with the second, we obtain the determinant

\bar{D} = | \begin{array}{cccc} a_{21} & a_{22} & \dots & a_{2 n} \\ a_{11} & a_{12} & \dots & a_{1 n} \\ \dots & \dots & \dots & \dots \\ a_{n 1} & a_{n 2} & \dots & a_{n n} \end{array} | = - D

and the reciprocal determinant of the order

j

corresponding to

\bar{D}

is

Taking into account that

a (\begin{array}{cccc} 2, & 1, i_{3}, & \dots, & i_{j} \\ k_{1}, & k_{2}, & k_{3}, & \dots, \\ k_{j} \end{array}) = - a (\begin{matrix} 1, \\ k_{1}, \\ k_{2}, \\ k_{3}, \end{matrix}, \dots, i_{j}, \dots, k_{j})

and permuting the line marked by

a (\binom{1, i_{2}, \dots, i_{j}}{k_{1}, k_{2}, \dots, k_{j}})

with the line marked by

a (\begin{array}{ccc} 2, & i_{2}, & \dots, \\ k_{1}, & k_{2} & \dots, \end{array}), {\bar{Δ}}_{j}

changing the sign at each permutation, we deduce that

{\bar{Δ}}_{j} = (- 1)^{(\binom{n - 1}{j - 1})} Δ_{j}

because

\overset{―}{Δ_{j}}

change the sign of

(\binom{n - 2}{j - 1})

times, you can take out -1 factor on the first

n - 2

lines and

(\binom{n - 2}{j - 2}) + (\binom{n - 2}{j - 1}) = (\binom{n - 1}{j - 1})

It is shown analogously that if in the determinant

D

two random rows or two random columns are swapped, we have the formulas

\begin{matrix} (5) & \bar{D} = - D, {\bar{Δ}}_{j} = (- 1)^{(\binom{n - 1}{j - 1})} Δ_{j} \end{matrix}

urd was noted with

\bar{D}

what becomes

D

through the change made and then through

\bar{Δ}

, the reciprocal determinant of the order

j

his/her

\bar{D}

.

In the determinant

D

, let's add to the elements of the 1st line, the elements of the line a

2 -

multiplied by

λ

, that is, let us consider the determinant

\overset{―}{\bar{D}} = | \begin{array}{cccc} a_{11} + λ a_{21} & a_{12} + λ a_{22} & \dots & a_{1 n} + λ a_{2 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ a_{n 1} & a_{n 2} & \dots & . \\ . & \dots & a_{n n} \end{array} |

The reciprocal determinant of the order

j

his/her

\overset{―}{\bar{D}}

is

When 1 gives the elements of any line of

D

, the elements of another row multiplied by a certain number are added, or to the elements of a certain column the elements of another column multiplied by a certain number are added, we have the formulas

\begin{matrix} (6) & \overset{―}{\bar{D}} = D, {\overset{―}{\bar{Δ}}}_{j} = Δ_{j} \end{matrix}

where was it noted with

\overset{―}{\bar{D}}

what becomes

D

through the transformation made and then through

\overset{―}{\bar{Δ}}

, the reciprocal determinant of the order

j

his/her

\overset{―}{\bar{D}}

.

Whether

D^{*}

the canonical determinant of

D

and

Δ_{j}^{*}

reciprocal determinant of the order

j

his/her

D^{*}

.

Because the determinant

D^{*}

is obtained from

D

through transformations

T_{1}

,

T_{2}, T_{1}^{'}, T_{2}^{'}

conveniently repeated and we have formulas (4), (5), the determinant

Δ_{j}^{*}

will be equal to

Δ_{j}

or will it differ from

Δ_{j}

by sign.

If

D = 0

, in canonical form

D^{*}

there will be at least one zero on the main diagonal, and

Δ_{\overset{⃛}{j}}^{j}

will also have at least one zero element on the main diagonal, that is, we will have

Δ_{j}^{*} = 0

It follows therefore that if

D = 0

, then we will also have

Δ_{j} = 0

.

If

D \neq 0

, its canonical form is

D^{*} = | \begin{array}{ccc} a_{1} & 0 \\ a_{2} \\ 0 & ⋱ \\ 0 & a_{n} \end{array} | = a_{1} a_{2} \dots a_{n}

determinant

Δ_{3}^{\overset{⃛}{j}}

, corresponding to 1ui

D^{*}

, it is

\begin{matrix} (7) & Δ_{j}^{*} = | \begin{array}{cccc} a_{1} a_{2} \dots a_{j} \\ a_{1} a_{3} \dots a_{j + 1} \\ ⋱ & 0 \\ 0 & ⋱ \\ ⋱ & a_{n - j + 1} a_{n - j + 2} \dots a_{n} \end{array} | = {(a_{1} a_{2} \dots a_{n})}^{(\binom{n - 1}{j - 1})} = {(D^{*})}^{(\binom{n - 1}{j - 1})} \end{matrix}

Formulas (5) and (6) show that

\frac{{\bar{Δ}}_{j}}{(\bar{D})^{(\binom{n - 1}{j - 1})}} = \frac{Δ_{j}}{D^{(\binom{n - 1}{j - 1})}}; \frac{{\overset{―}{\bar{Δ}}}_{j}}{(\overset{―}{\bar{D}})^{(\binom{n - 1}{j - 1})}} = \frac{Δ_{j}}{D^{(\binom{n - 1}{j - 1})}}

and therefore how much

\frac{Δ_{j}}{D^{(\binom{n - 1}{j - 1})}}

is invariant for transformations

T_{1}, T_{2}

,

T_{1}^{'}, T_{2}^{'}

. So we will have

\begin{matrix} (8) & \frac{Δ_{j}}{D^{(\binom{n - 1}{j - 1})}} = \frac{Δ_{j}^{*}}{{(D^{*})}^{(\binom{n - 1}{j - 1})}} \end{matrix}

But formula (7) shows that the second member of formula (8) is 1, which results in formula (4), which is also valid for

D = 0

, as shown above.

§ 2. Sylvester's Identity

Let's consider the determinants

\begin{matrix} (9) & D = | \begin{array}{cccc} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ \cdot & \cdot & \dots \\ a_{n 1} & a_{n 2} & \dots & a_{n n} \end{array} |, B = | \begin{array}{cccc} b_{11} & b_{12} & \dots & b_{1 m} \\ b_{21} & b_{22} & \dots & b_{2 m} \\ \cdot & \cdot & \dots & \cdot \\ b_{m 1} & b_{m 2} & \dots & b_{m m} \end{array} | \end{matrix}

and the elements

\begin{matrix} (10) & C_{i k} = | \begin{array}{cccc} \begin{matrix} q_{1 k} \\ . \\ . \\ D \end{matrix} \\ p_{i 1} & \dots & p_{i n} & b_{i k} \end{array} | \end{matrix}

where

i, k = 1, 2, \dots, m

We want to calculate the determinant

\begin{matrix} (11) & C = | \begin{array}{cccc} c_{11} & c_{12} & \dots & c_{1 m} \\ c_{21} & c_{22} & \dots & c_{2 m} \\ \cdot & \cdot & \cdot & \cdot \\ c_{m 1} & c_{m 2} & \cdot & \cdot \\ \cdot & c_{m m} \end{array} | \end{matrix}

and prove Sylvester's identity.

\begin{matrix} (12) & C = D^{m - 1} | \begin{array}{cccccc} a_{11} & \dots & a_{1 n} & q_{11} & \dots & q_{1 m} \\ \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot \\ a_{n 1} & \dots & a_{n n} & q_{n 1} & \dots & q_{n m} \\ p_{11} & \dots & p_{1 n} & b_{11} & \dots & b_{1 m} \\ \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot \\ p_{m_{1}} & \dots & p_{m n} & b_{m_{1}} & \dots & b_{m m} \end{array} | \end{matrix}

We will give a proof based on reducing a determinant to its canonical form.

Whether

\bar{D}

the determinant obtained from

D

by swapping two lines between them, and

\bar{C}

the determinant formed with the elements

\overset{―}{c_{i k}}

obtained by making in the determinant

c_{i k}

the same line changes as in the determinant

D

We have

\begin{matrix} (13) & \bar{D} = - D, \overset{―}{c_{i k}} = - c_{i k}, \bar{C} = (- 1)^{m} C \end{matrix}

and these formulas are also valid if in the determinant

D

two columns are swapped with each other.

Whether

\overset{―}{\bar{D}}

the determinant obtained

din D

, adding to the elements of one line the elements of another line multiplied by the factor

λ

We denote by

\overset{―}{\bar{C}}

the determinant formed with the elements

\overset{―}{c_{i k}}

obtained by making in the determinant

c_{i k}

the same transformation as in the determinant

D

We will have

\begin{matrix} (14) & \overset{―}{\bar{D}} = D, \overset{―}{\overset{―}{c_{i k}}} = c_{i k}, \overset{―}{\bar{C}} = C \end{matrix}

these formulas being valid even if in the determinant

D

are added to the elements of one column, the elements of another column multiplied by a factor

μ

.

Whether

D^{*}

canonical form of the determinant

D

, that is,
and

\begin{matrix} D^{*} = | \begin{array}{cccccc} a_{1} & 0 \\ a_{2} & 0 \\ ⋱ \\ a_{j} \\ 0 & ⋱ \\ 0 \end{array} | \\ c_{i k}^{*} = | \begin{array}{llllc} | \begin{matrix} q_{1 k}^{*} \\ ⋮ \end{matrix} \\ D^{*} \\ q_{n k}^{*} \\ p_{11}^{*} & \dots & . & p_{i n}^{*} & b_{i k} \end{array} | \end{matrix}

hate

p_{\ddot{i j}}^{*}

and

q_{\ddot{k}}^{*}

are the elements deduced from the elements

p_{i 1}, \dots, p_{i n}

and

q_{1 k}, \dots, q_{k, k}

the transformations that bring

D

to its canonical form

D^{*}

We have

C^{*} = | \begin{array}{cccc} c_{i 1}^{*} & \dots & c_{1, n}^{*} \\ \cdot & \cdot \\ \cdot & \cdot \\ \cdot & \cdot \\ c_{m 1}^{*} & \dots & c_{m m}^{*} \end{array} |

If

j ≦ n - 2

, then it is immediately seen - expanding the determinant

c_{c_{k}}^{*}

by the elements of the penultimate column - that

c_{i_{k}^{*}}^{*} = 0

, and therefore

C^{*} = 0

.

But if

j = n - 1

, but

m > 1

, then expanding the determinant sik by the elements of the penultimate column, we obtain
and therefore

c_{i k}^{*} = - a_{1} a_{2} \dots a_{n - 1} p_{i n}^{*} q_{n k}^{*}

C^{*} = {(- a_{1} a_{2} \dots a_{n - 1})}^{m} | \begin{array}{ccccc} p_{1 n}^{*} q_{n 1}^{*} & p_{1 n}^{*} q_{n 2}^{*} & \dots & p_{1 n}^{*} q_{n m}^{*} \\ p_{2 n}^{*} q_{n 1}^{*} & p_{2 n}^{*} q_{n 2}^{*} & \dots & p_{n 2}^{*} q_{n m}^{*} \\ \cdot & \cdot & \dots & \cdot \\ p_{m n}^{*} q_{n 1}^{*} & p_{m n}^{*} q_{n 2}^{*} & \dots & p_{m n}^{*} q_{n m}^{*} \end{array} | = 0 .

So we have shown that if

D = 0

, we also have

C^{*} = 0

, provided that in the case when

j = n - 1

to have

m > 1

From formulas (13) and (14) it follows that the determinant

C^{*}

is equal to

C

or differs from

C

by the sign. It follows that if we have

D = 0

, we also have

C = 0

(the case when

j = n - 1

and

m = 1

, reserved for later).

Let's assume

D \neq 0

, and let

D^{*}

its canonical form

D^{*} = | \begin{array}{cccc} a_{1} & 0 \\ a_{2} \\ ⋱ \\ 0 & a_{n} \end{array} |;

ELEMENTS

c_{i k}^{*}

are appropriate

c_{i k}^{*} = | \begin{array}{cccc} a_{1} & 0 \\ a_{2} \\ ⋮ \\ ⋱ \\ 0 & a_{n}^{*} \\ q_{k i}^{*} & p_{k i}^{*} & \dots & q_{i n}^{*} \\ q_{k i k}^{*} \end{array} |

Developing this determinant, we take the elements of its last column, and continuing on

a_{1} a_{2} \dots a_{n} = D^{*}

in the factor, we have

c_{i k}^{*} = D^{*} (b_{i k} - \frac{p_{i 1}^{*} q_{1 k}^{*}}{a_{1}} - \frac{p_{i 2}^{*}}{a_{2}} \frac{q_{2 k}^{*}}{a_{2}^{*}} - \dots - \frac{p_{i n}^{*}}{a_{n}} q_{n k}^{*}) = D^{*} b_{i k}^{'} .

determinant

C^{*}

can be written in the form

C^{*} = {(D^{*})}^{m} | \begin{array}{cccccccc} 1 & 0 & \dots & 0 & \frac{q_{11}^{*}}{a_{1}} & \frac{q_{12}^{*}}{a_{1}} & \dots & \frac{q_{1 m}^{*}}{a_{1}} \\ 0 & 1 & \dots & 0 & \frac{q_{21}^{*}}{a_{2}} & \frac{q_{22}^{*}}{a_{2}} & \dots & \frac{q_{2 m}^{*}}{a_{2}} \\ . & . & \dots & . & . & . & . & \dots \\ 0 & 0 & \dots & 1 & \frac{q_{n 1}^{*}}{a_{n}} & \frac{q_{n 2}^{*}}{a_{n}} & \dots & \frac{q_{n m}^{*}}{a_{n}} \\ 0 & 0 & \dots & 0 & 0 & b_{11}^{'} & b_{12}^{'} & \dots \\ 0 & 0 & \dots & 0 & b_{21}^{'} & b_{22}^{'} & \dots & b_{1 m}^{'} \\ . & . & \dots & . & . & b_{2 m}^{'} & . & . \end{array}

Multiplying the elements of the 1st, 2nd lines,

\dots

, of

n

-a with

p_{11}^{*}, p_{12}^{*}, \dots, p_{1 n}^{*}

and adding to the elements of the line a

(n + 1)

-a, these become

p_{11}^{*}, p_{12}^{*}, \dots, p_{1 n}^{*}, b_{11}, b_{12}, \dots, b_{1 m} .

Doing analogous operations for rank lines

n + 2, n + 3, \dots, n + m

get

C^{*} = {(D^{*})}^{m} | \begin{array}{cccccccc} 1 & 0 & \dots & 0 & \frac{q_{11}^{*}}{a_{1}} & \frac{q_{12}^{*}}{a_{1}} & \dots & \frac{q_{1 m}^{*}}{a_{1}} \\ 0 & 1 & \dots & 0 & \frac{q_{21}^{*}}{a_{2}} & \frac{q_{22}^{*}}{a_{2}} & \dots & \frac{q_{2 m}^{*}}{a_{2}} \\ . & . & \dots & . & . & . & . & \dots \\ 0 & 0 & \dots & 1 & \frac{q_{n 1}^{*}}{a_{n}} & \frac{q_{n 2}^{*}}{a_{n}} & \dots & . \\ p_{11}^{*} & p_{12}^{*} & \dots & p_{1 n}^{*} & b_{11} & b_{12} & \dots & \dots \\ p_{21}^{*} & p_{22}^{*} & \dots & p_{2 n}^{*} & b_{21}^{*} & b_{22} & \dots & b_{1 m} \\ . & . & \dots & . & . & . & \dots & b_{2 m} \\ p_{m 1}^{*} & p_{m 2}^{*} & \dots & p_{m n}^{*} & b_{m 1} & b_{m 2} & \dots & b_{m m} \end{array} |

or

\begin{matrix} (15) & C^{*} = {(D^{*})}^{m - 1} | \begin{array}{ccccccccc} a_{1} & 0 & \dots & 0 & q_{11}^{*} & q_{12}^{*} & \dots & q_{1 m}^{*} \\ 0 & a_{2} & \dots & 0 & q_{21}^{*} & q_{22}^{*} & \dots & q_{2 m}^{*} \\ . & . & \dots & . & . & . & \dots & . \\ 0 & 0 & \dots & a & q_{n 1}^{*} & q_{n 2}^{*} & \dots & q_{n m}^{*} \\ p_{11}^{*} & p_{12}^{*} & \dots & p_{1 n}^{*} & b_{11} & b_{12} & \dots & b_{1 m} \\ p_{21}^{*} & p_{22}^{*} & \dots & p_{2 n}^{*} & b_{21} & b_{22} & \dots & b_{2 m} \\ . & . & \dots & . & . & . & \dots & . \\ p_{m 1}^{*} & p_{m 2}^{*} & \dots & p_{m n}^{*} & b_{m 1} & b_{m 2} & \dots & b_{m m} \end{array} | \end{matrix}

From formulas (13) and (14) it follows that

\frac{\bar{C}}{(\bar{D})^{m}} = \frac{C}{D^{m}} şi \frac{\overset{―}{\bar{C}}}{(\overset{―}{\bar{D}})^{m}} = \frac{C}{D^{m}}

MEAN

\frac{C}{D^{m}}

is an invariant for the transformations

T_{1}, T_{2}, T_{1}^{'}, T_{2}^{'}

We will then have

\frac{C}{D^{m}} = \frac{C^{*}}{{(D^{*})}^{m}}

and taking into account formula (15), we deduce that

\begin{matrix} (16) & \frac{C}{D^{m}} = \frac{Δ^{*}}{D^{*}} \end{matrix}

where

Δ^{*}

is the determinant in the second member of formula (15).

Let us now consider the determinant

Δ = | \begin{array}{cccccccc} a_{11} & a_{12} & \dots & a_{1 n} & q_{11} & q_{12} & \dots & q_{1 m} \\ a_{21} & a_{22} & \dots & a_{2 n} & q_{21} & q_{22} & \dots & q_{2 m} \\ . & . & \dots & . & . & . & \dots & . \\ a_{n 1} & a_{n 2} & \dots & a_{n n} & q_{n 1} & q_{n 2} & \dots & q_{n m} \\ p_{11} & p_{12} & \dots & p_{1 n} & b_{11} & b_{12} & \dots & b_{1 m} \\ p_{21} & p_{22} & \dots & p_{2 n} & b_{21} & b_{22} & \dots & b_{2 m} \\ . & . & \dots & . & . & . & \dots & . \\ p_{m 1} & p_{m 2} & \dots & p_{m n} & b_{m 1} & b_{m 2} & \dots & b_{m m} \end{array} |

Whether

\bar{D}

the determinant which is obtained by changing to

D

two lines or two columns between them. We denote by

\bar{Δ}

the determinant which is obtained by making the same transformation as in

D

on rows or columns. We will have

\bar{D} = - D, \bar{Δ} = - Δ

Either now

\overset{―}{\bar{D}}

the determinant obtained by adding to the elements of a line of

D

, the elements of another line multiplied by a factor

λ

, or which is obtained by adding to the elements of a column of

D

, the elements of another column multiplied by a factor

μ

We note

cu \overset{―}{\bar{Δ}}

the determinant that is obtained from

Δ

making the same transformation as in

D

, on rows or on columns. We will have

\bar{D} = D, \overset{―}{\bar{Δ}} = Δ

From these formulas it follows that

\frac{\bar{Δ}}{\bar{D}} = \frac{Δ}{D}, \frac{\bar{Δ}}{\bar{D}} = \frac{Δ}{D}

MEAN

\frac{Δ}{D}

is an invariant for the transformations

T_{1}, T_{2}, T_{1}^{'}, T_{2}^{'}

It
follows that

\begin{matrix} (17) & \frac{Δ}{D} = \frac{Δ^{*}}{D^{*}} \end{matrix}

where

D^{*}

is the canonical form of

D

, and

Δ^{*}

is the determinant in the second member of formula (15).

Formulas (16) and (17) then show us that

\frac{C}{D^{m}} = \frac{Δ}{D}

from which it follows that

C = D^{m - 1} Δ

and with this, Sylvester's identity (12) is proven.
The case

j = n - 1, m = 1

is trivial, the determinant

C

is reduced to a single element

c_{11}

Sylvester's formula (12) is in this case a trivial identity, the factor

D^{m - 1}

from the second member which appears as

0^{\circ}

should be considered equal to 1.

THE CANONICAL FORM OF ONE DETERMINER AND ITS APPLICATIONS
(Brief summary)

It is known that elementary transformations

T_{1}, T_{2}

and

T_{1}^{'}, T_{2}^{'}

, produced over lines and columns, matrix

A = {‖ a_{i}^{k} ‖}_{1}^{n}

it is reduced to the canonical form

A^{*}

. We call it the canonical form of the determinant

D = | A |

determiner

D^{*} = | A^{*} |

.

In this work, the application of the canonical form of the determinant is given, and the formula (4) for the mutual determinant is proved

Δ

(order)

j

) determinant

D

. Доказываться также тождество (12) Sylvester.

Proofs are based on the following idea:
Let it be required to calculate the determinant

Δ

, corresponding to some determinant

D

. If the invariant is found

f (D, Δ)

transformed

T_{1}, T_{2}, T_{1}^{'}, T_{2}^{'}

, then from equality

f (D, Δ) = f (D^{*}, Δ^{*})

is displayed

Δ

, because

Δ^{*}

it's easy to calculate.

LA FORME CANONIQUE D'UN DÉTERMINANT FIT SES APPLICATIONS
(Résumé)
On sait que par des transformations élémentaires

T_{1}, T_{2}

and

T_{1}^{'}, T_{2}^{'}

, effected on the lines and columns of a matrix

A = {‖ a_{i k} ‖}_{1}^{n}

, on ramène celle-ci à la forme canonique

A^{*}

. On appelle forme canonique du determinant

D = | A |

, determining them

D^{*} = | A^{*} |

.

Dans ce travail, on fait des applications de la forme canonique d'un determinant, en démontrant la formulae (4) pour le réciproque determinant

Δ

, by order

j

, of a determinant

D

, et en démontrant l'identité (12) by Sylvester.

The demonstrations given are based on the following idea:
Supposons que nous ayons à calculer un determinant

Δ

qui correspond à un determinant

D

any And l'on peut mettre en évidence an invariant

f (D, Δ)

for the transformations

T_{1}, T_{2}, T^{'}_{1}, T^{'}_{2}

, then equality

f (D, Δ) = f (D^{*}, Δ^{*})

we can shoot

Δ

, easily calculating

Δ^{*}

.

1959

The canonical form of a determinant and its applications

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THE CANONICAL FORM OF A DETERMINANT AND ITS APPLICATIONS

§ 1. The reciprocal determinant of the order $j$ of a determinant

§ 2. Sylvester's Identity

THE CANONICAL FORM OF ONE DETERMINER AND ITS APPLICATIONS
(Brief summary)

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The canonical form of a determinant and its applications

Abstract

Authors

Keywords

Paper coordinates

PDF

About this paper

Journal

Publisher Name

DOI

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Online ISSN

References

Paper (preprint) in HTML form

THE CANONICAL FORM OF A DETERMINANT AND ITS APPLICATIONS

§ 1. The reciprocal determinant of the order j j jjof a determinant

§ 2. Sylvester's Identity

THE CANONICAL FORM OF ONE DETERMINER AND ITS APPLICATIONS (Brief summary)

Related Posts

§ 1. The reciprocal determinant of the order $j$ of a determinant

THE CANONICAL FORM OF ONE DETERMINER AND ITS APPLICATIONS
(Brief summary)