4.2 Algebraic Properties of Matrix Inverses
The operation of taking the inverse of a matrix has several algebraic properties that are
similar to taking the inverse of a number. However, with some of them one has to be
careful of the order that we multiply matrices.
Proposition 1.
(a) If A is invertible, then so is A-1 and (A-1)-1 = A.
(b) If A and B are invertible, then so is AB and (AB)-1 = B-1A-1.
(c) If E1, E2, …, En are invertible, then so is E1E2…En and
(E1E2…En)-1 = (En)-1…(E2)-1(E1)-1.
(d) If A is invertible, then so is An and (An)-1 = (A-1)n.
(e) If E1, E2, …, En are invertible and E1E2…EnA = I, then A is invertible and
A-1 = E1E2…En.
(f) If we can transform A by elementary row operations to the identity matrix then
A is invertible.
(g) If A is invertible, then so is AT and (AT)-1 = (A-1)T.
Proof. (a) (A-1)-1 is that matrix B that satisfies that A-1B = I and BA-1 = I. However since
A-1 is the inverse of A one has AA-1 = I and A-1A = I. So (A-1)-1 = A. (b) (AB)-1 is that
matrix C that satisfies that (AB)C = I and C(BA) = I. However (AB)(B-1A-1) = ABB-1A-1 =
AIA-1 = AA-1 = I and (B-1A-1)(AB) = B-1A-1AB = BIB-1 = BB-1 = I. So (AB)-1 = B-1A-1. (c)
This follows from (b). (d) This follows from (c) by taking E1 = E2 = … = En = A. (e)
From (c) it follows that E1E2…En is invertible. From Propostion 1 in section 4.1 it
follows that A = (E1E2…En)-1. From (a) it follows that A is invetible and A-1 = E1E2…En.
(f) One has I = En…E2E1A where the Ej are elementary matrices. We saw in section 3.1
that the elementary matrices were invertible. So A is invertible by (e) (g) (AT)-1 is that
matrix B that satisfies that ATB = I and BAT = I. Since A-1 is the inverse of A one has AA-1
= I and A-1A = I. Therefore (AA-1)T = IT = I and (A-1A)T = IT = I. Using (AB)T = BTAT we
get (A-1)TAT = I and AT(A-1)T = I. So (AT)-1 = (A-1)T. //
If A is invertible, then we can define the negative powers of A.
Definition 1. If A is invertible and n is a positive integer, let A-n = (A-1)n.
The powers of an invetible matrix obey some of the usual rules of exponents that are true
for numbers.
4.2 - 1
Proposition 2. Suppose A is invertible and n and m are integers
(a) An is invertible for all n and (An)-1 = A-n
(b) Am+n = AmAn
(c) (Am)n = Amn
(d) If B is also invertible and AB = BA then (AB)n = AnBn
Proof. (a) This follows from parts (a) and (d) of Proposition 1. (b) There are four cases.
1.
m  0 and n  0
2.
m  0 and n < 0 and m + n  0
3.
m < 0 and n  0 and m + n  0
4.
m+n<0
Case 1 has already been proved in our discussion of positive powers. To prove case 2 we
let n = - p where p > 0. So we must show Am-p = AmA-p. By Case 1 one has Am = Am-pAp
and by (a) one has A-p = (Ap)-1. So AmA-p = Am-pAp(Ap)-1 = Am-pI = Am-p which is what we
wanted to show. The proof of case 3 is similar. To prove case 4 note that by (a) one has
Am+n = (A-(m+n))-1. Using case 1, 2 or 3 one has A-(m+n) = A(-m)+(-n) = A(-n)A(-m). Using
Proposition 1b one has (A-(m+n))-1 = (A(-n)A(-m))-1 = (A(-m))-1(A(-m))-1. Using (a) again one has
(A(-m))-1(A(-m))-1 = AmAn. Combining proves case 4. We leave the proof of (c) and (d) to
the reader. //
Singular matrices.
Definition 2. Let A be a square matrix. A is singular if it is not invertible.
Proposition 3. Let A and B be a square matrices of the same size.
(a) If A is singular and B is invertible the AB and BA are singular.
(b) If any row of A consists entirely of zeros, the A is singular.
(c) If any column of A consists entirely of zeros, the A is singular.
(d) Suppose we transform A by elementary row operations to a matrix C. If C has
has either a row of zeros or a column of zeros then A is singular.
Proof. (a) If AB were invertible, then by Proposition 1a and 1b the matrix ABB-1 = A
would be invertible. Similarly if BA were invertible. (b) If the ith row of A consists
entirely of zeros, then the ith row of AB would also consist entirely of zeros and could not
be equal to the identity. (c) This follows form (b) and part (f) of Proposition 1. (d) One
has C = En…E2E1A where the Ej are elementary matrices. We saw in section 3.1 that the
elementary matrices were invertible. So En…E2E1 is invertible by Proposition 1c and
A = (En…E2E1)-1C. By (b) the matrix C is singular and by (a) the matrix A is singular. //
Corollary. Suppose we transform A by elementary row operations to a reduced row
echelon form matrix C. Then A is invertible or singular depending on whether or not C is
the identity.
Proof. This follow from Proposition 1f and Proposition 3g.
4.2 - 2