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)


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


(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


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


(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.


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


(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.




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?