The content in today's blog is taken from Linear Algebra with Applications by Gareth Williams.

Definition 1: Inverse of a Matrix

Let A be an n x n matrix. If a matrix B can be found such that AB = BA = I

_{n}, then A is said to be invertible and B is called the inverse of A. If such a matrix B does not exist, then A has no inverse.

Example 1:

Let A =

Let B =

We can see that B is the inverse of A since:

AB =

BA =

Now, let's look at two properties:

Lemma 1: The inverse of an invertible matrix is unique.

(1) Let B and C be inverses of a matrix A

(2) From the definition of an inverse, we have:

AB = BA = I

_{n}

AC = CA = I

_{n}

(3) Multiplying C to both sides of the first equation to get:

C(AB) = CI

_{n}

(4) Using properties of matrices, we get:

C(AB)=C (Matrix Identity, see Property 1, here)

(CA)B = C (Associativity of Matrix Multiplication, see Property 1, here)

I

_{n}B = C [From step #2 above]

B = C (Matrix Identity, see Property 1, here)

QED

Definition 2: A

^{-1}

Let A be an invertible matrix. Then A

^{-1}signifies the inverse of A so that AA

^{-1}= A

^{-1}A = I

_{n}

We can generalize this definition to account for A

^{-n}:

Definition 3: A

^{-n}

A

^{-n}= (A

^{-1})

^{n}= A

^{-1}A

^{-1}... A

^{-1}

Using these notations, we can use the inverse to solve a given equation:

Lemma 2: Let AX = B be a system of linear equations in n variables. If A

^{-1}exists, the solution is unique and is given by X = A

^{-1}B

Proof:

(1) Assume that A

^{-1}exists such that X = A

^{-1}B

(2) AX = B since:

AX = A(A

^{-1}B) =

= (AA

^{-1})B [Associativity, see Property 1, here]

= I

_{n}B [Definition of Inverse, see Definition 1, above]

= B [Identity, see Property 2, here]

(3) This proves that A

^{-1}B is a solution. Now, we need to show that the solution is unique.

(4) Let X

_{1}be a solution such that AX

_{1}= B.

(5) Then:

A

^{-1}AX

_{1}= A

^{-1}B [Multiplying A

^{-1}to both sides of step #4]

I

_{n}X

_{1}= A

^{-1}B [Definition of A

^{-1}above]

X

_{1}= A

^{-1}B [Identity, see Property 2, here]

(6) Thus, we have shown that if X

_{1}is a solution, then X

_{1}= A

^{-1}B

QED

Lemma 3: If matrices A and B are invertible, then the product AB is invertible and (AB)

^{-1}= B

^{-1}A

^{-1}.

Proof:

(1) (B

^{-1}A

^{-1})(AB) = B

^{-1}(A

^{-1}A)B [Associative Property of Multiplication, see Property 1, here]

(2) B

^{-1}(A

^{-1}A)B = B

^{-1}(I)B [See Definition 1 above]

(3) B

^{-1}(I)B = (B

^{-1}I)B = B

^{-1}B [see Property 3, here]

(4) B

^{-1}B = I.

(5) Using the same argument above, we can show the following:

(AB)(B

^{-1}A

^{-1}) =A(BB

^{-1})A

^{-1}= A(I)A

^{-1}= AA

^{-1}= I.

(6) Using Lemma 2 above, we know that if (AB)

^{-1}exists, it is the unique matrix X such that X(AB)=I=(AB)X.

(7) Hence, (AB)

^{-1}exists and (AB)

^{-1}= B

^{-1}A

^{-1}.

QED

Corollary 3.1: If A is a product of invertible matrices, then A is invertible.

Proof:

(1) Let A = the product of n matrices such that A = M

_{1}*...*M

_{n}

(2) Now, if n=2, by Lemma 3, A is invertible.

(3) Assume that A is invertible if up to n-1 invertible products only.

(4) Let A' = M

_{1}*...*M

_{n-1}

(5) A' is invertible by the inductive hypothesis.

(6) So, A'*M

_{n}is invertible by Lemma 3.

(7) Thus, using mathematical induction, we can conclude that any product of invertible matrices is itself invertible.

QED

References

- Gareth Williams, Linear Algebra with Applications, Wm. C. Brown Publishers, 1996.

## 3 comments :

Hello. Thank you.

Does the corollary 3.1 work the other way, too? That is, is this true?: If A=m1*m2 is invertible, then m1 and m2 are invertible.

And how would you prove or disprove it?

Hey, Larry thanks a lot for maintaining such a blog. I am an overage student following an IT course in grate difficulty. Your blog is a grate help to me. Thanks again and may god give you enrgy to continue you good work.

Can you give an example where AB = I, but BA is not I?

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