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preface

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Self-introduction ଘ(੭, ᵕ)੭

Nickname: Haihong

Tag: programmer monkey | C++ contestant | student

Introduction: because of C language to get acquainted with programming, then transferred to the computer major, had the honor to get some national awards, provincial awards… Has been confirmed. Currently learning C++/Linux/Python

Learning experience: solid foundation + more notes + more code + more thinking + learn English well!

 

Machine learning little White stage

The article is only used as my own study notes for the establishment of knowledge system and review

Know what is, know why!

2.3 inverse matrix

define

For the n-order matrix AAA, if I have an n-order matrix BBB, such that AB equals BA equals EAB equals BA equals E

The matrix A is invertible and the matrix B is called the inverse matrix of A, or the inverse matrix for short

Keep in mind that

If A is invertible, then the inverse of A must be unique.

Proof: Suppose B and C are the inverse matrices of A, there are

$B=BE=B(AC)=(BA)C=EC=C$

It is concluded that B = C

So the inverse of A is unique.

writing

The inverse matrix of the AAA as A – 1 A ^ {1} A – 1 if AB = BA = E, the B = A – 1 AB = BA = E, the B = A ^ {1} AB = BA = E, B = A – 1

Theorem 1

content

If matrix A reversible, the ∣ A ∣ indicates 0 | | \ A neq 0 ∣ A ∣  = 0

prove

Because the matrix A is invertible, there must be A−1A^{-1}A−1, such that

$AA^{-1}=E$

launch

$|AA^{-1}|=|E| \\==> |A||A^{-1}|=|E|=1\\ ==> |A|\neq 0$

Theorem 2

content

If ∣ A ∣ indicates 0 | | \ A neq 0 ∣ A ∣  = 0, AAA reversible matrix, and A – 1 = 1 ∣ A ∣ A ∗ A ^ {1} = \ frac {1} {| A |} A ^ * A – 1 = ∣ A ∣ 1 A ∗, including A ∗ ∗ A ^ * A as AAA adjoint matrix

prove

Known AA ∗ = ∣ A ∣ EAA ^ * = | A | EAA ∗ = ∣ A ∣ E (| | A is A constant)

Because ∣ A ∣ indicates 0 | | \ A neq 0 ∣ A ∣  = 0

So ∣ A ∣ AA ∗ = 1 A1 ∣ A ∣ A ∗ = A (1 ∣ A ∣ A ∗) = E \ frac {1} {| |} A AA ^ * = A \ frac {1} {| A |} A ^ * = A (\ frac {1} {| A |} A ^ *) = E ∣ A ∣ 1 AA ∗ = A ∣ A ∣ 1 A ∗ = A (∣ A ∣ 1 A ∗) = E

And because A ∗ A = AA ∗ A * A = AA ^ ^ * AA ∗ ∗ A = A

So 1 ∣ ∣ A AA ∣ A ∣ A ∗ ∗ = 1 A = (1 ∣ A ∣ A ∗) A = E \ frac {1} {| |} A AA ^ * = \ frac {1} {| A |} ^ A * A = (\ frac {1} {| A |} A ^ *) ∣ A ∣ 1 AA ∗ = A = E ∣ A ∣ 1 A ∗ A = (∣ A ∣ 1 A ∗) A = E

by

We know that matrix A has an inverse matrix,

And A – 1 = 1 ∣ A ∣ A ∗ A ^ {1} = \ frac {1} {| A |} A ^ * A – 1 = ∣ ∣ A ∗ 1 A

Proof done!

inference

If AB=E (or BA=EAB=E (or BA=EAB=E (or BA=E) then B=A−1B=A^{-1}B=A−1

Proof:

Because the AB = the EAB = the EAB = E

So ∣ A ∣ ∣ B ∣ = ∣ ∣ E = 1 | | | | = | E | = 1 B ∣ A ∣ ∣ ∣ = B ∣ ∣ E = 1

So the ∣ A ∣ indicates 0 | A | \ neq0 ∣ A ∣  = 0, A – 1 A ^ {1} there – 1 A

$B=EB=(A^{-1}A)B=A^{-1}(AB)=A^{-1}E=A^{-1}$

Proof done!

Operation rule

Square matrix inverse matrix to meet operational rules (1) if the AAA reversible, then A – 1 A ^ {1} A – 1 is reversible, and (A) – 1-1 = A (A ^ {1}) ^ {1} = A (A) – 1-1 = A

Proof:

Because AAA is invertible

so

$AA^{-1}=A^{-1}A=E$

So let’s say that B is equal to AB is equal to AB is equal to A

$BA^{-1}=A^{-1}B=E$

launch

$A ^ {1} reversible$

and

$(A^{-1})^{-1}=B=A$

Proof done!

Number (2) if the AAA reversible, lambda indicates \ lambda \ lambda  neq 0 0 = 0, the lambda \ lambda A lambda A reversible, and lambda (A) – 1-1 = 1 lambda A \ lambda (A) ^ {1} = \ frac {1} {\ lambda} A ^ {1} (lambda A) = 1 A lambda – 1-1

Proof:

Because AAA is invertible

so

$AB=BA=E$

For λA\lambda AλA

There must be 1λB\frac{1}{\lambda} Bλ1B such that

Lambda (1 B) = (lambda A) lambda (1 B) lambda (A) = E \ lambda (A) (\ frac {1} {\ lambda} B) = (\ frac {1} {\ lambda} B) = E \ lambda (A) (lambda A) (1 B) lambda = (1 B) lambda lambda (A) = E so

λA\lambda AλA is also reversible

At the same time

$(\lambda A)^{-1}=\frac{1}{\lambda}B=\frac{1}{\lambda}A^{-1}$

(3) if A, BA, BA, B for the same order matrix and are reversible, the ABABAB are reversible, and (AB) – 1 = B – A – 1 (AB) ^ {1} = B ^ A ^ {1} {1} (AB) = B – 1-1 A – 1 to prove:

Because A, BA, BA, and B are matrices of the same order and invertible

so

• ​
  $AC=CA=E (C=A^{-1})$
• ​
  $BD=DB=E (D=B^{-1})$

because

• ​
  $(AB)(DC)=A(BD)C=AEC=AC=E$
• ​
  $(DC)(AB)=D(CA)B=DEB=DB=E$

so

ABABAB reversible and (AB) – 1 = DC = B – A – 1 (AB) ^ {1} = DC = B ^ A ^ {1} {1} (AB) – 1 = DC = B – 1-1 A

Proof done!

(4) if AAA reversible, ATA ^ {T} the AT is reversible, and (AT) – 1 = (A – 1) T (A ^ {T}) ^ {1} = (A ^ {1}) ^ {T} (AT) = (A – 1) T – 1

Proof:

Because AAA is invertible

the

$AA^{-1}=A^{-1}A=E$

I take the transpose, I get

$(AA^{-1})^T=(A^{-1}A)^T=E^T$

By (AB) T = BTAT (AB) ^ ^ T ^ T = B TA (AB) T = BTAT

$(A^{-1})^TA^T=A^T(A^{-1})^T=E^T=E$

so

ATA ^ TAT reversible

and

$(A^T)^{-1}=(A^{-1})^T$

Proof done!

conclusion

The essay is just a study note, recording a process from 0 to 1

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