In later sections, we will deal exclusively with linear systems which have a square coefficient matrix. We now justify this by showing that our system (4.1) cannot have exactly one solution for every right side unless the coefficient matrix is square.
Lemma 4.1 If x = x1 is a solution of the linear system Ax = b then any solution x = x2 of this system is of the form
where x = y is a solution of the homogeneous system Ax = 0.
Indeed, if both x1 and x2 solve Ax = b, then
i.e., then their difference y = x2 - x1, solves the homogeneous system A x = 0.
Example The linear system
has the solution xl = x2 = 1. The corresponding homogeneous system
has the solution x1 = of the original system
-is
2a, x2 = a, where a is an arbitrary scalar. Hence any solution of the form x1 = 1 - 2a, x2 = 1 + a for some number a.
The lemma implies the following theorem.
Theorem 4.1 The linear system Ax = b has at most one solution (i.e., the solution is unique if it exists) if and only if the corresponding homogeneous system Ax = 0 has only the “trivial” solution x = 0.
Next we prove that we cannot hope for a unique solution linear system has at least as many equations as unknowns.
unless our
Theorem 4.2 Any homogeneous linear system with fewer than unknowns has nontrivial (i.e., nonzero) solutions.
equations
We have to prove that if A is an m × n matrix with
then we can find such that Ay = 0. This we do by induction on n.
First, consider the case n = 2. In this case, we can have only one equation, and this equation has the nontrivial solution x1 = 0, x2 = 1, if a12 = 0;
otherwise, it has the nontrivial solution x1 = a12, x2 = - a11. This proves our statement for n = 2. Let now n > 2, and assume it proved that any homogeneous system with less equations than unknowns and with less than n unknowns has nontrivial solutions; further, let Ax = 0 be a homo-geneous linear system with m equations and n unknowns where m < n. We have to prove that this system has nontrivial solutions. This is certainly so if the nth column of A is zero, i.e., if an = 0; for then the nonzero n-vector x = in is a solution. Otherwise, some entry of an must be different from 0, say,
In this case, we consider the m × (n - 1) matrix B whose jth column is
4.1 PROPERTIES OF MATRICES 137
If we can show that the homogeneous system
has nontrivial solutions, then we are done. For if we can find numbers x1, . . . , xn-1 not all zero such that
then it follows from the definition of the bj’s that
thus providing a nontrivial solution to Ax = 0. Hence it remains only to show that Bx = 0 has nontrivial solutions. For this, note that for each j, the ith entry of bj is
so that the ith equation of Bx = 0 looks like
and is therefore satisfied by any choice of x1, . . . , xn-1. It follows that x = y solves Bx = 0 if and only if x = y solves the homogeneous system
which we get from Bx = 0 by merely omitting the ith equation. But now is a homogeneous linear system with m - 1 equations in n - 1 unknowns, hence with less equations than unknowns and with less than n unknowns. Therefore, by the induction hypothesis, has nontrivial solutions, which finishes the proof.
Example Consider the homogeneous linear system Ax = 0 given by
so that m = 2, n = 3. Following the argument for Theorem 4.2, nontrivial solution as follows: Since we pick i = 2 and get
The smaller homogeneous system Bx = 0 is therefore
w e construct a
We can ignore the last equation and get, then, the homogeneous system which consists of just one equation,
A nontrivial solution for this is x1 = 1, x2 = - 2. Hence, with
the 3-vector x = (xj) is a nontrivial solution of the original system.
Next we prove that we cannot expect to get a solution to our linear system (4.1) for all possible choices of the right side b unless we have no more equations than unknowns.
Lemma 4.2 If A is an m × n matrix and the linear system Ax = b has a solution for every m-vector b, then there exists an n × m matrix C such that
Such a matrix C can be constructed as follows: By assumption, we can find a solution to the system Ax = b no matter what b is. Hence, choosing b to be the jth column of I, we can find an n-vector cj, such that
But then, with C the n × m matrix whose jth column is cj, j = 1, . . . , m, we get
showing that the jth column of the product AC agrees with the j th column of I, j = 1, . . . , m. But that says that AC = I.
Lemma 4.3 If B and C are matrices such that
then the homogeneous system Cx = 0 has only the trivial solution x = 0.
Indeed, if Cx = 0, then
Theorem 4.3 If A is an m × n matrix and the linear system A x = b has a solution for every possible m-vector b, then m < n.
For the proof, we get from Lemma 4.2 that
for some n × m matrix C. But this implies by Lemma 4.3 that the homogeneous system Cx = 0 has only the trivial solution x = 0. Therefore, by Theorem 4.2, C must have at least as many rows as columns, that is, n > m, which finishes the proof.
4.1 PROPERTIES OF MATRICES 139 We now know that we cannot expect to get exactly one solution to our system (4.1) for every possible right side unless the system has exactly as many equations as unknowns, i.e.,unless the coefficient matrix is square.
We will therefore consider from now on only linear systems with a square coefficient matrix. For such square matrices, we prove a final theorem.
Theorem 4.4 Let A be an n × n matrix. Then the following are equivalent:
(i) The homogeneous system Ax = 0 has only the trivial solution x = 0.
(ii) For every right-side b, the system Ax = b has a solution.
(iii) A is invertible.
First we prove that (i) implies (ii). Let b be a given n-vector. We have to prove that Ax = b has a solution. For this, let D be the m × ( n + 1) matrix whose first n columns agree with those of A, while the ( n + 1)st column is b. Since D has more columns than rows, we can find, by Theorem 4.2, a nonzero (n + 1)-vector y such that Dy = 0, that is, such that
(4.12) Clearly, the number yn+1 cannot be zero. For if yn+1 were zero, then as
at least one of the numbers y1, . . . , yn would have to be nonzero, while at the same time
But this would say that Ax = 0 admits the nontrivial solution xi = yi, i = 1, . . . , n, which contradicts (i). Hence, since we can solve (4.12; for b to get that
But this says that Ax = b has a solution, viz., the solution xi = - (yi/yn+1), i = 1, . . . , n, which proves (ii).
Next we prove that (ii) implies (iii). Assuming (ii), it follows with Lemma 4.2 that there exists an n × n matrix C such that
Hence, by Lemma 4.3, the equation Cx = 0 has only the trivial solution x = 0. This says that the n × n matrix C satisfies (i); hence, by the argument we just went through, C satisfies (ii); therefore, by Lemma 4.2, there exists an n × n matrix D such that
But now we are done. For we showed earlier that if
with A, C, D square matrices, then C is invertible and
Hence A is the inverse of an invertible matrix, therefore invertible.
Finally, Lemma 4.3 shows that (iii) implies (i).
Example We showed in an earlier example that the 2 × 2 matrix
is not invertible and, in another example, that for this matrix the homogeneous system A x = 0 has nontrivial solutions. By Theorem 4.4, the linear system Ax = b should therefore not be solvable for some 2-vector b. Indeed, with b = i1, we get the system
which has no solution since the second equation demands that
while the first equation demands that
As a simple application of Theorem 4.4, we now prove that A square and AB = I implies B = A-1 and BA = I. Indeed, if A is of order n × n, then AB = I implies that B is of order n × n, and that, for all n -vectors b, A(B b) = b. But this says that we can solve A x = b for x no matter what b, hence A is invertible by Theorem 4.4, and that then x = B b is the solution, hence Bb = A-1b for all b, or B = A-1. But then, finally, BA = I.