Page 236 - Advanced engineering mathematics

216 CHAPTER 7 Matrices and Linear Systems

Now suppose that X = C is a solution of AX = O. Then AC = O,so
O = (AC) = ( A)C = A R C = O,

so C is also a solution of the reduced system.
Conversely, suppose K is a solution of the reduced system, so A R K = O. We want to show
that AK = O. Since A = A R ,wehave ( A)K = O,so
(E r E r−1 ···E 2 E 1 )AK = O.
Now, each E j is an elementary matrix, and we know that there is an elementary matrix E that
∗
j
reverses the effect of E j . From the last equation, we have
∗
E E ···E ∗ E (E r E r−1 ···E 2 E 1 )AK = O.
∗
∗
1 2 r−1 r
But E E r = I n , and E ∗ E r−1 = I n , and so on until E E 1 = I n , so in the last product all of the
∗
∗
r r−1 1
elementary matrices cancel in pairs, leaving AK = O. Therefore K is also a solution of the
original system, completing the proof.
We can therefore concentrate on solving a reduced system. As we have seen in the examples,
the solution of A R X=O is easily read from this matrix, and has the added dividend that it reveals
the structure of these solutions. The set of all solutions of the homogeneous system AX = O
m
form a vector space, which is a subspace of R if A is n × m. Furthermore, the dimension of this
solution space can read from A R , as we saw in the examples.
If A R has k nonzero rows (hence rank k), then k of the x i ’s are determined by the m − k free
variables, which can be assigned any values in writing solutions of the system. This means that
x 1 ,··· , x k are determined by x k+1 ,··· , x m , which can be chosen arbitrarily. The general solution
will have m − k arbitrary constants in it.
We will summarize these observations.

THEOREM 7.12 Solution Space of a Homogeneous System

Let A be n × m. Then
m
1. The set of all solutions of AX = O forms a subspace of R , called the solution space of
this system.
2. The dimension of this solution space is
m − number of nonzero rows of A R ,
which is the same as m − rank (A).

Proof Let S be the set of all solutions of the system. Since
x 1 = x 2 = ··· = x m = 0
is a solution, the zero m-vector is in S.
Now suppose X 1 and X 2 are solutions, and α and β are numbers. Then
A(αX 1 + βX 2 ) = αAX 1 + βAX 2 = O + O = O,
m
so linear combinations of solutions are solutions, and S is a subspace of R .
For the dimension of S, use the fact that the system has the same solution space as the
reduced system. As the examples suggest, the nonzero rows of A R enable us to express the
general solution as a linear combination of linearly independent solutions, one for each free
variable. Since the number of free variables is the number of columns of A R , minus the number
of nonzero rows, then the dimension of S is m − rank(A).

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October 14, 2010 14:23 THM/NEIL Page-216 27410_07_ch07_p187-246