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.
Let A =
Let B =
We can see that B is the inverse of A since:
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.
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.
- Gareth Williams, Linear Algebra with Applications, Wm. C. Brown Publishers, 1996.