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MML learning notes (2) : Linear algebra of n-order determinant, commutation

preface

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1.3 The determinant of order n

The third-order determinant is: ∣ a11a12a13a21a22a23a31a32a33 ∣ = a11 ∗ a22 ∗ a33 + a12 ∗ a23 + a13 ∗ ∗ a31 a21 ∗ a32 – a11 ∗ a23 ∗ a32 – a12 ∗ a21 ∗ a33 – a13 ∗ a22 ∗ a31 \ begin {vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\\ \end{vmatrix}=a_{11}*a_{22}*a_{33}+a_{12}*a_{23}*a_{31}+a_{13}*a_{21}*a_{32}-a_{11}*a_{23}*a_{32}-a_{12}*a_{21}*a_{33}-a * a_ _ {13} {and} * a_ {31} ∣ ∣ ∣ ∣ ∣ ∣ ∣ a11a21a31a12a22a32a13a23a33 ∣ ∣ ∣ ∣ ∣ ∣ ∣ = a11 ∗ a22 ∗ a33 + a12 ∗ a23 + a13 ∗ ∗ a31 a21 ∗ a32 – a11 ∗ a23 ∗ a32 – a12 ∗ a21 ∗ a33 – a13 ∗ a22 ∗ a31

We can find the rules: ∣ a11a12a13a21a22a23a31a32a33 ∣ = ∑ (- 1) ta1p1a2p2a3p3 \ begin a_ {vmatrix} {11} & a_ {12} & a_ {13} \ \ a_ {21} & a_ {and} & a_ {23} \ \ a_{31} & a_{32} & a_{33}\\ {vmatrix} = \ \ end sum (1) ^ ta_ p_1 {1} a_ p_2 {2} a_ p_3 {3} ∣ ∣ ∣ ∣ ∣ ∣ ∣ a11a21a31a12a22a32a13a23a33 ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∑ ta1p1a2p2a3p3 (1 -)

Where t is the number of inversions of p1p2p3p_1p_2p_3p1p2p3

And then the determinants of order n: ∣ A11a12… a1na21a22… a2n…… an1an2… Ann ∣ = ∑ (1 -) ta1p1a2p2… anpn\begin{vmatrix} a_{11} & a_{12} &… & a_{1n}\\ a_{21} & a_{22} & … &a_{2n}\\ . & . & & . \\ . & . & & . \\ a_{n1} & a_{n2} &… & a_{nn}\\ \end{vmatrix}=\sum(-1)^ta_{1p_1}a_{2p_2}… A_ {np_n} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a11a21. J. an1a12a22.. an2……… a1na2n.. Ann ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∑ (1 -) ta1p1a2p2… anpn

Special cases 1: ∣λ1λ2.. Lambda n ∣ lambda = 1 lambda 2… Lambda n \ begin {vmatrix} \ lambda_1 & & & & \ \ & \ lambda_2 & & & \ \ & & & & \ \ & & & & \ \ & & & & \ lambda_n \end{vmatrix}=\lambda_1\lambda_2… \ lambda_n ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ lambda lambda. 2. 1. Lambda n ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ lambda lambda = 1 2… Lambda n

Special cases 2: ∣λ1λ2.. Lambda n ∣ = (1 -) n (n – 1) lambda lambda 1 2… λn (where (−1)n(n−1)2 is the arrangement of n, n−1… The reverse order of 3, 2, 1) \ begin {vmatrix} && && \ lambda_1 \ \ & && \ lambda_2 & \ \ && && \ \ & & & & \ \ \ lambda_n && && \ end = {vmatrix} (-1)^\frac{n(n-1)}{2}\lambda_1\lambda_2… \lambda_n (where (-1)^\frac{n(n-1)}{2} = n, n-1… | | | | | | | | | λn. Lambda 2 lambda 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = (1 -) 2 n (n – 1) lambda lambda 1 2… λn (where (−1)2n(n−1) denotes the array n, n−1… The number of inversions of 3, 2 and 1)

1.4 switch

1.4.1 Interchange of permutations

concept

• Transpose: In a permutation, to swap any two elements and leave the rest unchanged.
• Adjacent interchange: In a permutation, the interchange of two adjacent elements

Theorem 1

content

The parity changes when any two elements of a permutation change

prove

Let’s first prove the case of adjacent commutations

A permutation a1… aiabb1… bma_1… a_iabb_1… b_ma1… aiabb1… bm

A and B switch to a1… aibab1… bma_1… a_ibab_1… b_ma1… aibab1… bm

Obviously, a1… aia_1… a_ia1… Ai, b1… bmb_1… b_mb1… Bm the number of inversions of these elements has not changed

When a < b

• When we go from ab to BA, the number of inversions of A +1, the number of inversions of B stays the same

When a > b,

• From AB to BA, the number of inversions of A stays the same, and the number of inversions of B is -1.

so

If adjacent commutations occur in permutations, odd-even changes will be found (odd permutations -> even permutations or even permutations -> odd permutations)

Let’s prove the general case

a1… aiab1… bmbc1… cna_1… a_iab_1… b_mbc_1… c_na1… aiab1… bmbc1… Cn,a changes with B and becomes A1… aibb1… bmac1… cna_1… a_ibb_1… b_mac_1… c_na1… aibb1… bmac1… cn

We can first swap BBB with BMB_MBm to a1… aiab1… bbmc1… cna_1… a_iab_1… bb_mc_1… c_na1… aiab1… bbmc1… cn

Bm −1b_{m-1} BM −1; bm−1b_{m-1} BM −1; aiab1… BBM – 1 bmc1… cna_1… a_iab_1… bb_{m-1}b_mc_1… c_na1… aiab1… BBM – 1 bmc1… Finally, BBB and B1b_ {1} B1 make adjacent exchange and become A1… aiabb1… bmc1… cna_1… a_iabb_1… b_mc_1… c_na1… aiabb1… bmc1… cn

There are m adjacent swaps

And bm, bm – 1… B2, b1b_m, b_{m-1}… B_2, B_1BM, BM-1… B2, b1, that’s m times

And then we swap aaa with BBB to get a1… aibab1… bmc1… cna_1… a_ibab_1… b_mc_1… c_na1… aibab1… bmc1… cn

Then use AAA and b1b_1B1 for adjacent exchange to become A1… aibb1a… bmc1… cna_1… a_ibb_1a… b_mc_1… c_na1… aibb1a… bmc1… Bmb_mbm, bmB_MBM, bmB_MBM… aibb1… bmac1… cna_1… a_ibb_1… b_mac_1… c_na1… aibb1… bmac1… cn

We have m plus 1 adjacent swaps

To sum up

So we have m+ (m+1) =2m+1 adjacent swaps

From the very first proof

After 2m+1 adjacent transposition, the parity of permutations will still change (transposition once, parity will change; The parity does not change when you swap it twice –> the parity changes when you swap it an odd number of times. An even number of times does not. 2m+1 must be odd if m is a positive integer.

inference

Even permutations become standard permutations with an odd number of commutations, and even permutations become standard permutations with an even number of commutations.

instructions

First of all, the standard permutation is even permutation with the inverse number 0 and we know from theorem 1 that if you change it once, the parity changes if you change it once, odd -> even, even -> odd… If I do it odd times, I get an even permutation; You do it an even number of times, and you end up in an odd arrangement. So there must be an odd number of commutations in a uniform permutation to become a standard permutation. It is also true that even permutations become standard permutations with even commutations.

1.4.2 Another representation of the determinant

The determinants of n are given a11a12… a1na21a22… a2n…… an1an2… Ann ∣ = ∑ (1 -) ta1p1… aipi… ajpj… anpn\begin{vmatrix} a_{11} & a_{12} &… & a_{1n}\\ a_{21} & a_{22} & … &a_{2n}\\ . & . & & . \\ . & . & & . \\ a_{n1} & a_{n2} &… & a_{nn}\\ \end{vmatrix}=\sum(-1)^ta_{1p_1}… a_{ip_i}… a_{jp_j}… A_ {np_n} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a11a21. J. an1a12a22.. an2……… a1na2n.. Ann ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∑ (1 -) ta1p1… aipi… ajpj… anpn

Let’s pick either one: A1p1… aipi… ajpj… anpna_{1p_1}… a_{ip_i}… a_{jp_j}… a_{np_n}a1p1… aipi… ajpj… Anpn, where 1… i… j… N is the natural arrangement, and t in (−1)t(-1)^t(−1)t is the inverse number

Then, the exchange of AIPI, AjpJA_ {IP_i}, a_{jp_j}aipi, ajpj, a1P1… ajpj… aipi… anpna_{1p_1}… a_{jp_j}… a_{ip_i}… a_{np_n}a1p1… ajpj… aipi… anpn

Let’s calculate the change in parity

First of all, we know that the value of this item does not change just by swapping the positions of the two elements.

Line marking from 1… i… j… N becomes 1… j… i… N, we can get the permutation 1… j… i… If the number of inversions of n is odd, let’s call it r

Because 1… i… j… N inverse number is 0, even permutation changes according to the permutation of any element, parity changes, 1… j… i… N becomes uniform, so the number of inversions must be odd

Again, let p1… pj… pi… pnp_1… p_j… p_i… pnp1… pj… pi… The inverse number of Pn (column standard) is T1T_1T1, then

a1p1… ajpj… aipi… anpna_{1p_1}… a_{jp_j}… a_{ip_i}… a_{np_n}a1p1… ajpj… aipi… The positive and negative signs before ANpn are (−1)r+ T1 (-1)^{r+t_1}(−1)r+ T1

because

$(-1)^{t_1}=(-1)(-1)^t=-(-1)^t$

p1… pi… pj… pnp_1… p_i… p_j… pnp1… pi… pj… The number of inversions of Pn is t a1P1… aipi… ajpj… anpna_{1p_1}… a_{ip_i}… a_{jp_j}… a_{np_n}a1p1… aipi… ajpj… The coefficient in front of anpn is (−1)t(-1)^t(−1)t changes once to p1… pj… pi… pnp_1… p_j… p_i… pnp1… pj… pi… Change the pn parity Is multiplied by (1) (exchange arrangement, any two elements, changes of parity, and is multiplied by (1)) so (1) – (1 -) t1 = t (1) (1) – (1) ^ {t_1} = (1) ^ t (1) – (1 -) t1 = (1 -) t

And since r is odd, there are

$(1) ^ r = 1$

Synthesize the following two expressions: T1 = {(1 -) (1) – (- 1) t = – (1) – (- 1) t r = 1 \ the begin – {cases} (1) ^ {t_1} = (1) (1) ^ t = – (1) ^ t \ \ ^ (1) r = 1 \ end {cases} {(−1) T1 =(−1)(−1) T =(−1) T (−1) r=−1

Get:

$(-1)^{r+t_1}=(-1)^r(-1)^{t_1}=(-1) * (-1)^{t_1}=(-1)*(-(-1)^t)=(-1)^t$

Introduction:

$(-1)^ta_{1p_1}… a_{ip_i}… a_{jp_j}… a_{np_n}=(-1)^{r+t1}a_{1p_1}… a_{jp_j}… a_{ip_i}… a_{np_n}$

instructions

The position of two elements of a certain term in the commutative determinant makes the row coordinates and column coordinates change simultaneously, but it does not change the parity of the term.

One swap is not going to change parity, so many swaps are not going to change parity

(1 -) ta1p1a2p2… anpn(-1)^ta_{1p_1}a_{2p_2}… A_ {np_n} (1 -) ta1p1a2p2… Anpn has undergone several changes of column standard permutations p1P2… pnp_1p_2… p_np1p2… Pn must change into a natural arrangement (1, 2, 3… N)

Let’s say that after a number of transformations the column array becomes a natural array and let’s say that the row array is q1q2… qnq_1q_2… q_nq1q2… Qn, have

$(-1)^ta_{1p_1}a_{2p_2}… a_{np_n}=(-1)^ta_{q_11}a_{q_22}… a_{q_nn}$

For any item aiJA_ {ij}aij, {AIj = AIPIAIj = aqJJ \begin{cases} a_{ij}=a_{IP_i}\\ a_{ij}=a_{q_JJ}\ end{cases}{aij= AIPIAIj =aqjj

Get {j = pii = qj \ begin j = {cases} p_i \ \ I = q_j \ end {cases} {j = pii = qj

Pip_ipi can determine a unique corresponding qjq_jqj, such as 2=p32=p_32=p3, q2=3q_2=3q2=3 and unique!

So by the p1p2… pnp_1p_2… p_np1p2… Pn can determine the unique Q1Q2… qnq_1q_2… q_nq1q2… qn

Theorem 2

content

The determinant of order n can also be defined as: ∑(−1)tap11ap22… apnn\sum(-1)^ta_{p_11}a_{p_22}… A_ {p_nn} ∑ (1 -) tap11ap22… apnn

prove

First of all, the determinants of order n are given a11a12… a1na21a22… a2n…… an1an2… Ann ∣ = ∑ (1 -) ta1p1a2p2… anpn\begin{vmatrix} a_{11} & a_{12} &… & a_{1n}\\ a_{21} & a_{22} & … &a_{2n}\\ . & . & & . \\ . & . & & . \\ a_{n1} & a_{n2} &… & a_{nn}\\ \end{vmatrix}=\sum(-1)^ta_{1p_1}a_{2p_2}… A_ {np_n} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a11a21. J. an1a12a22.. an2……… a1na2n.. Ann ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∑ (1 -) ta1p1a2p2… anpn

The {D = ∑ (1 -) ta1p1a2p2… AnpnD1 = ∑ (1 -) tap11ap22… apnn\begin{cases} D=\sum(-1)^ta_{1p_1}a_{2p_2}… a_{np_n}\\ D_1=\sum(-1)^ta_{p_11}a_{p_22}… A_ {p_nn} {cases} {\ end D = ∑ (1 -) ta1p1a2p2… AnpnD1 = ∑ (1 -) tap11ap22… apnn

It can be concluded from the last discussion of theorem 1:

Any of the terms of D (−1) ta1p1a2P2… anpn(-1)^ta_{1p_1}a_{2p_2}… A_ {np_n} (1 -) ta1p1a2p2… Anpn has one and only one item in D1 (−1) TAq11AQ22… aqnn(-1)^ta_{q_11}a_{q_22}… A_ {q_nn} (1 -) taq11aq22… Aqnn corresponds to ** (q can be determined by P); ** Similarly, any item in D1 (−1) TAP11AP22… apnn(-1)^ta_{p_11}a_{p_22}… A_ {p_nn} (1 -) tap11ap22… Apnn also has and only one of the terms of D (−1)ta1q1a2q2… anqn(-1)^ta_{1q_1}a_{2q_2}… A_ {nq_n} (1 -) ta1q1a2q2… Anqn corresponds to it, indicating that any term in D and D1 can be one-to-one corresponded

So D is equal to D1D is equal to D_1D is equal to D1

so

$\sum(-1)^ta_{1p_1}a_{2p_2}… a_{np_n}=\sum(-1)^ta_{p_11}a_{p_22}… a_{p_nn}$

conclusion

Description:

• Refer to textbook “linear algebra” fifth edition tongji University mathematics department
• With the book concept explanation combined with some of their own understanding and thinking

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