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 = 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_nB = 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 = I_n

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]

= I_nB [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]

I_nX₁ = 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*...*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' = M₁*...*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.