Sunday, April 22, 2007

The Inverse of a Matrix

In today's blog, I review the basic properties of the inverse of a matrix.

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 = In, 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 = In

AC = CA = In

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

C(AB) = CIn

(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)

InB = 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-1A = In

We can generalize this definition to account for A-n:

Definition 3: A-n

A-n = (A-1)n = A-1A-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-1B

Proof:

(1) Assume that A-1 exists such that X = A-1B

(2) AX = B since:

AX = A(A-1B) =

= (AA-1)B [Associativity, see Property 1, here]

= InB [Definition of Inverse, see Definition 1, above]

= B [Identity, see Property 2, here]

(3) This proves that A-1B is a solution. Now, we need to show that the solution is unique.

(4) Let X1 be a solution such that AX1 = B.

(5) Then:

A-1AX1 = A-1B [Multiplying A-1 to both sides of step #4]

InX1 = A-1 B [Definition of A-1 above]

X1 = A-1B [Identity, see Property 2, here]

(6) Thus, we have shown that if X1 is a solution, then X1 = A-1B

QED

Lemma 3: If matrices A and B are invertible, then the product AB is invertible and (AB)-1 = B-1A-1.

Proof:

(1) (B-1A-1)(AB) = B-1(A-1A)B [Associative Property of Multiplication, see Property 1, here]

(2) B-1(A-1A)B = B-1(I)B [See Definition 1 above]

(3) B-1(I)B = (B-1I)B = B-1B [see Property 3, here]

(4) B-1B = I.

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

(AB)(B-1A-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-1A-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 = M1*...*Mn

(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' = M1*...*Mn-1

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

(6) So, A'*Mn is invertible by Lemma 3.

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

QED

References

3 comments :

Arie said...

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?

nagenahiru said...

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.

Judy said...

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