Elementary Matrices and Matrix Inverses

Mathematics 3A, Winter 1997

May 5, 1997

The following is a complete proof of some important facts about inverse matrices, and illustrates the role of the elementary matrices in Linear Algebra.

Fact. Every elementary row operation that converts a matrix A ∈ M(n×n) into another matrix B can be written as multiplication with an elementary matrix E:

EA = B.  (1)

All we need to know about elementary matrices is that they are invertible, which can be checked by explicit calculation. The following theorem states in a precise form that our method for checking invertibility of a matrix is indeed the correct one.

Theorem. Given a matrix A ∈ M(n × n), the following statements are equivalent:

(i) A is invertible.

(ii) There exists a matrix C such that CA = I.

(iii) A can be transformed into the identity matrix I by elementary row operations (i.e. A is row equivalent to the identity matrix).

Moreover, when the inverse of A exists, it is unique.

Proof. The best strategy here is to prove that (i) implies (iii), (iii) implies (ii), and (ii) implies (i). We start with the easiest part.

(iii) ⇒ (ii): A being row-equivalent to I_{n} means, by the above fact, that there are elementary matrices E_{1}, E_{2}, …, E_{i} so that

E_{1}E_{2} ⋯ E_{i}A = I.  (2)

By simply setting

C := E_{1}E_{2} ⋯ E_{i},  (3)

we obtain statement (ii).

(ii) ⇒ (i): What we really need to check here is that matrix C not only satisfies CA = I, but also AC = I (we stated this as a theorem in class, but never proved it).

First consider the particular matrix C as defined by equation (3). Leftmultiply equation (2) with E_{1}^{−1} and right-multiply with E_{1} to obtain

E_{2}E_{3} ⋯ E_{i}AE_{1} = E_{1}^{−1}E_{1} = I,  (4)

then do the same with E_{2}, E_{3}, and so on. We finally get

AE_{1}E_{2} ⋯ E_{i} = I.  (5)

Thus we obtain AC = I. We claim that any matrix D which satisfies DA = I must be equal to C. Let’s check:

D = DI = D(AC) = (DA)C = IC = C.  (6)

This proves at the same time that the inverse is unique, and that A^{−1} = C.

(i) ⇒ (iii): By contradiction. Suppose that A is not row-equivalent to I. Then there exist elementary matrices E_{1}, E_{2}, …, E_{i} so that the matrix

H := E_{1}E_{2} ⋯ E_{i}A  (7)

has zeros in the last row. Define a vector

b = E_{1}^{−1}E_{2}^{−1} ⋯ E_{i}^{−1}e_{n}  (8)

where e_{n} is the unit vector

e_{n} = (0 ⋮ 0 1) ,  (9)

and try to solve the linear system Ax = b. Left-multiplying this equation with E_{1}E_{2} ⋯ E_{i} yields the equivalent system

Hx = e_{n}.  (10)

This system does not have a solution because the last row of the matrix on the left is zero, but the last row of the vector on the right is one. So Ax = b does not have a solution for some vector b ∈ ℝ^{n}, so A cannot be invertible.