Download Section 1.6: Invertible Matrices One can show (exercise) that the

Section 1.6: Invertible Matrices
One can show (exercise) that the composition of finitely many invertible
functions is invertible. As a result, we have the following:
Theorem 6.1: Any admissible row operation is an invertible function.
Proof: If f : Fm×n → Fm×n is an admissible row operation, then by
definition, f = e1 ◦ . . . ◦ ek , where e1 , . . . , ek are elementary row operations on
Fm×n . Since each elementary row operation is an invertible function, the
theorem follows.
In fact, the result of this theorem is an important part of the reason for using
admissible row operations on the augmented matrix for a system of linear
equations in order to solve the system: It is equivalent to the symmetry
property of row- equivalence. Recall that any admissible row operation f is
effected by multiplication by an appropriate matrix, P. In particular, such a
matrix is the product of finitely many elementary matrices. In this context, we
make the following definition:
Definition: If f : Fm×n → Fm×n is an admissible row operation and
P ∈ Fm×m is such that f (A) = P A for all A ∈ Fm×n , then we say that the
matrix P represents the row operation f.
Lemma: The identity matrix I in Fm×m represents the identity operation,
1F , which is an admissible row operation.
Lemma: If I is the identity matrix in Fm×m ,, then I commutes with every
m × m matrix over F, i.e. we have IA = AI, for every matrix A in Fm×m .
Theorem 6.2: Each admissible row operation f is represented by a unique
matrix P.
Proof: Let f : Fm×n → Fm×n be an admissible row operation which is
represented by matrices P, Q ∈ Fm×m . If I is the m × m identity matrix, then
we have
P = 1F m(P ) = IP = P I = f (I) = QI = IQ = 1F m(P ) = Q,
so we are done.
The above results motivate the definitions of the following notions for
matrices, which are given in the text:
left inverse of a matrix, right inverse of a matrix, (two-sided) inverse of a
matrix, invertible matrix, square matrix
A useful result is
Theorem 6.3: If A ∈ Fn×n has a left (right) inverse, B, then B is also a right
(left) inverse for A. Thus, a square matrix is invertible iff it has either a left
inverse or a right inverse.
1
The proof of this theorem is essentially the ”marriage” of the proof of the
lemma for theorem 10 in the text (pg. 22) with the proof of the corollary to
theorem 13(pg.24). The remainder of the results of this section should be
familiar, but the reader is advised to review their proofs, which may differ
somewhat from those in more elementary texts. We will highlight the
remaining results of section 1.6, giving few proofs.
Theorem 6.4: If A, B ∈ Fn×n , then:
(i) If A is invertible, then its inverse is invertible, and (A−1 )−1 = A.
(ii) The product AB is invertible iff both A and B are invertible.
(iii) If A and B are both invertible, then (AB)−1 = B −1 A−1 .
Corollary: A product of finitely many n × n matrices is invertible iff each
factor is invertible. That is, if A1 , . . . , Ak ∈ Fn×n , and if A = A1 . . . Ak , then
A is invertible iff for each natural number j with 1 ≤ j ≤ k, Aj is invertible. In
this case, we have
A−1 = Ak . . . A1 .
Proof: We use induction on k. Note that the desired result holds trivially for
k = 1, and suppose we have shown the result for all products of kn × n
matrices. Let A1 , . . . , Ak+1 ∈ Fn×n , let A = A1 . . . Ak and let B = Ak+1 . By
theorem 6.4, AB is invertible iff A is invertible and B is invertible. But by the
inductive hypothesis, A is invertible iff A1 , . . . , Ak are all invertible, and since
B = Ak+1 , it follows that AB = A1 . . . Ak+1 is invertible iff A1 , . . . , Ak+1 are
all invertible. Also by theorem 6.4, if A and B are invertible, then
(AB)−1 = B −1 A−1 . By the induction hypothesis, we have in this case that the
inverse of A is given by
A−1 = Ak . . . A1 .
Since B = Ak+1 , we get the desired formula:
(A1 . . . Ak+1 )−1 = (AB)−1 = B −1 A−1
= Ak+1 (Ak . . . A1 ) = Ak+1 . . . A1 .
By mathematical induction, we are done.
Theorem 6.5: Any elementary matrix is invertible.
Corollary: Any admissible row operation is represented by an invertible
matrix.
Proof: If f : Fm×n → Fm×n is an admissible row operation which is
represented by the m × m matrix P, let E1 , . . . , Ek be m × m elementary
2
matrices such that P = E1 . . . Ek . Since each of the matrices E1 , . . . , Ek is
invertible, it follows that P is invertible and P −1 = Ek . . . E1 .
Theorem 6.6: Let A be an m × m matrix over a field F. Then the following
are equivalent:
(i) The matrix A is invertible.
(ii)A ≈R I.
(iii) A represents an admissible row operation.
(iv) A is a product of elementary matrices.
(v) The function f1 : Fm×1 → Fm×1 , defined by f1 (x) = Ax, is an invertible
function.
(vi) For each n ∈ N\{0}, the function fn : Fm×n → Fm×n , defined by
fn (B) = AB, is an invertible function.
Corollary: A square matrix A is invertible iff
A
I
≈R
I
A−1
.
n×n
That is, A ∈ F
is invertible iff there are admissible (or elementary)
operations f1 , . . . , fk on Fn×(2n) such that
fk ◦ . . . ◦ f1 ( A
I )= I
B
,
and in this case, B is the inverse of A.
Corollary: Let A and B be m × n matrices over a field F. Then we have that
A ≈R B iff B = P A for some invertible P ∈ Fm×m .
Theorem 6.7: If A is an n × n matrix over a field F, then the following are
equivalent:
(i) The matrix A is invertible.
(ii) The homogeneous system Ax = 0 has only the trivial solution, x = 0.
(iii) For every y ∈ Fn×1 , the system Ax = y has a solution.
(iv) For every y ∈ Fn×1 , the system Ax = y has exactly one solution.
(v) For every y ∈ Fn×1 , the system Ax = y has at most one solution.
Examples:
1. Compute the inverse of the following 3 × 3 complex matrix:

1
A= 1
0
0
1
1

i
i .
1
Solution: We perform admissible row operations on the 3×6 matrix
A I .. According to the corollary
iff this
to theorem
6.6, A is invertible
procedure can succed in computing I A−1 from A I :
3
A
I

1
= 1
0
0
1
1


0 i
1 0 0
1 0
−1 1 0 
1 1
0 0 1

(1−i) i −i
−1
1 0 
0
0 1

i −i
1
0  = I A−1 .
−1 1


1 0 0
0 1 0 
0 0 1
i
i
1

1
⇔ 0
0
0
1
1
0
0
1

1
⇔ 0
0
1 0 0
(1−i)
−1
⇔ 0 1 0
0 0 1
1

(1−i) i −i
1
0 .
Thus we have A−1 =  −1
1
−1
1
2. Let f : Z5 → Z5 be defined by
⇔ f (A).
A
That is, If A = [aij ], then

a1,1
f (A) =  a2,1 + a1,1
2a3,1 + 4a1,1

...
a1,5
...
a2,5 + a1,5  .
. . . 2a3,5 + 4a1,5
Show that f is an admissible row operation.
Solution: We will find an invertible 3×3 matrix P which represents f. In fact,
if we let

1
P = 1
4
0
1
0


0
1
0  , and Q =  4
2
3

0 0
1 0 ,
0 3
then QP = I, and for any matrix A ∈ Z5 , we have f (A) = P A. Since f is
represented by P and P is invertible (Q = P −1 ), it follows that f is an
admissible row operation, as desired.
3. Define f : R2×3 → R2×3 by
⇔ f (A).
A
Show that f is not an admissible row operation.
Solution: Let Q ∈ R2×2 be given by
Q=
1
1
4
1
1
.
Then we claim that f (A) = QA for all A ∈ R2×3 , but that Q is not invertible.
We leave it to the reader to verify that f (A) = QA for all A ∈ R2×3 , and note
that the assumption of invertibility for Q would lead to a system of four
equations in two variables with a solution which would happen to imply 1 = 0.
Since in R(and in fact in any field), 1 6= 0, this contradiction proves the claim.
Now suppose by way of contradiction that f is an admissible row
operation. Then f is represented by an invertible 2 × 2 matrix P. But then we
would have
Q 0
=Q
I
0
= f( I
0 )=P I
0
=
P
0
,
where I is the 2 × 2 identity matrix over R, and 0 is the 2 × 1 zero matrix over
R. It follows that P = Q. But Q was shown to be noninvertible, so this is a
contradiction. Thus f is not an admissible row operation.
Homework Assignment for Section 1.6:
1. Let A ∈ Z13 be given by


1
2 1 0
0 3 5 .
A =  12
1 11 1 1
Find a row-reduced echelon matrix R which is row-equivalent to A, an admissible
row-operation f on 3×5 matrices over Z13 such that f (A) = R, and the invertible
matrix P which represents f.
2. Do problem #2 on pg. 26 of the text, following the instructions given in 1
above.
3. Let F be any field, and define f : Fm×n → Fm×n by
A
f(A),
where j ∈ {1, . . . , m} and a1 , . . . , am , b1 , . . . , bm ∈ F satisfy:
(i)0 6∈ {a1 , . . . , am }\{aj }
(ii)aj + bj 6= 0.
Prove that f is an admissible operation. Is the converse true?
4. Consider the matrix A y ∈ Z5 in example
1 on pg.
44 of these notes,
and the operations used to reduce this matrix to I x to solve the system
Ax = y. Call these operations f1 , f2 , f3 , f4 and f5 , respectively, so that they are
defined on Z5 by
X
⇔ f1 (X), X
X
⇔ f2 (X), X
⇔ f4 (X), X
5
⇔ f5 (X),
⇔ f3 (X),
Prove that these are admissible row operations by finding invertible matrices
P1 , P2 , P3 , P4 , P5 ∈ Z5 which represent f1 , f2 , f3 , f4 , and f5 , respectively.
5. Explain what is ”wrong” with

1
 1
A=
 1
0
the following computation in R4×4 :



2 2 2 1
1 1 1
 2 1 2 1 
1 1 0 


⇔
 1 1 2 2  = B. (*)

0 1 1
1 2 2 2
1 1 1
In particular, note that the systems Ax = 0 and Bx = 0 do not have the same
solutions, and explain why the computation (*) has ”caused” this undesirable
situation.
6. Do problem #6 on pg. 27 of the text.
7. Do problem #7 on pg. 27 of the text.
8. Do problem #10 on pg. 27 of the text.
9. Do problem #11 on pg. 27 of the text.
10. Do problem #12 on pg. 27 of the text. (Hint: First consider row
operations which involve only the last two rows of the matrix.)
11. Prove that the composition of finitely many invertible functions is
invertible.
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