a note on systems of linear equations

A Note on Systems of Linear EquationsAuthor(s): David L. ElliottSource: SIAM Review, Vol. 3, No. 1 (Jan., 1961), pp. 66-69Published by: Society for Industrial and Applied MathematicsStable URL: http://www.jstor.org/stable/2027250 .

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SIAM REVIBW Vol. 3, No. 1, January, 1961

A NOTE ON SYSTEMS OF LINEAR EQUATIONS'

DAVID L. ELLIOTT2

THIS NOTE IS A COMMENT on reference [1] and a generalization of the method there presented. We consider a system of m linear equations in n unknowns Xl, X X2 X* *XXnX

n (1) aijxj = ci i = 1,2,**,m,aij,c,real

j=1

or A *x = c in matrix notation. We distinguish three cases: (I) There is no finite vector x satisfying (1) (inconsistent case);

(II) There is a unique vector x satisfying (1); (III) There are an infinity of vectors x satisfying (1), such that their end-

points lie on some line, plane, or higher-dimensional linear manifold. Semarne considers Case II for m = n, but the presentation in reference [1]

requires modification, as will be shown below. Semarne's method of solving (1) combines the two classical ideas of the aug-

mented-matrix equation and Gram-Schmidt orthogonalization. (The method does not involve computing a matrix inverse to A.) As modified in this report, the method will handle any system (1) whether or not A has an inverse, furnish- ing a solution if one exists and informing us if one does not exist.

We shall use vectors (or points)

X = (X1, X2 X * Xn)

in En (euclidean n-space),

y = (yo,Yi, i *, Yn)

in E . We extend En by introducing, in addition to the xi, an artificial variable t. Multiplying (1) by t and transposing, we obtain:

n

( 1') -cit + , aijtxj = 0 i = 1, 2, * * m, j=1

or n

(2) E bijyj = O i = 1,2, * m, jaO

where

bio = -ci , bij = aij j)=1 2, * n,

and

(3) y =txj,1 <ijn,andyo=t.

1 Received by the editors August 10, 1959. 2 U. S. Naval Ordnance Test Station, Pasadena Annex.

66



SYSTEMS OF LINEAR EQUATIONS 67

(This notation differs from that of reference [1].) If x is a solution of (1), then for arbitrary t (3) gives a solution of (2) lying on a line through the origin of Efn+l; (1, xi, X , * x n) is the intersection of this line with the hyperplane yo = 1. Conversely, if we find any point y* satisfying (2), then every point on the line y = sy*, - O0 < s < oc, also satisfies (2). If yo* # 0, let s* = l/yo*; for the

point y = st y, yo = 1. Then in (3) we have t = yo = 1,

(4) *Syj*= = j xjX 1 _ ji n,

and x satisfies (1 ) . If yo* = 0, the line y = sy* lies in the hyperplane yo = 0 and can have no

finite intersection with yo = 1, so no corresponding solution of (1) can be found. If yo = 0 for every y satisfying (2), we classify (1) as Case I, inconsistent. We shall give explicit forms for solutions of (2) and (1) below.

Equation (2),

B.y = 0,

where B is the m by n + 1 matrix (bij), can be interpreted geometrically in two ways. The first is that the vector y lies in the intersection of m hyperplanes (each passing through the origin). That is, the "solution" of (2) is a vector subspace, U, of E n+l: the origin itself, or some line, plane, etc., through the origin. De- fining the vector (bil , bi2 X , bin) = bt for any i = 1, 2, * * , m, each hyperplane has equation

(2') b.y = O i = 1 2, **,m.

Then the vector bi is perpendicular to the ith hyperplane. Our second interpretation of (2) is "any solution-vector y is perpendicular to the m vectors bl, * b* , and a "solution" is the subspace U which can be described as the orthogonal complement in En+l of the subspace V spanned by bl, * , bn. (That is, any point in V can be written as a linear combination of the bt.) Then we need to find a set of vectors which span U, in order to define the solution of (2) completely.

From the viewpoint of the second interpretation, above, a good solution method is evident, namely, the Gram-Schmidt orthogonalization procedure. We note these facts:

a. One complete basis of orthogonal vectors for En+l is the set

I = (1, 0,...,0,0),

= (0, 1, ... ,0, 0),

{u} = {U ....................

u n= ((0, 0, ... 0, 1).

b. We can form an orthogonal basis for V, consisting of M mutually orthogonal vectors (M < m)f',f2, * M , from the (not necessarily independent) vectors bl, * , bm by using the Gram-Schmidt process.



68 DAVID L. ELLIOTT

c. If we continue the Gram-Schmidt process by adjoining vectors-say, from {ul-which do not belong to V, we eventually will obtain a new complete orthogonal basis for En+l,

fl f2 I fM, fM+1 ... ffn+l,

and the vectors (if any) fM+1 **, fn +l will be a complete orthogonal basis for U, hence a "solution". If M = n + 1, U contains only the origin.

As Semarne points out, the beauty of this approach is that the Gram-Schmidt computation is quite simple; the basis vectors are:

f1 = bl,

f = 2 I(f ) f X

*................

1 bm \ {2 bmX - mn-1 bm

fm = - ( J) fi - J f2 f - . b fm-1 f

= bm ~~~fl .f 1 f f2 .f 2 f,;n-1 ff-

Now M ( m) of these vectors are non-zero; we may omit the others and re- number the non-zero ones as fl, * * *, fM; they span V. Now we continue computing until we obtain a total of n + 1 non-zero vectors:

fm - (f *) - (fM fM

= 0- (f/l) f' - (4o) fM

(5') =M+2 Ul - ( fi )+ - - I f+l /fTM1+

fl = - f - f) . - _( )

where again we have dropped all f's which are zero. The "solution" of (2) is then the space U consisting of all y such that:

(6) y= +lfM+l + ... + an+if8f, for any real a? .

We can obtain the solution of (1) as the intersection (if any) of U with the hyperplane yO = 1; that is, points of U having yO = 1 satisfy ( 1 ). We can now, from the vectors of (5'), decide which case (of the first paragraph) we have:

(I) IffoR = 0 for all thefR in (5'), U lies in the hyperplane yo = 0, and no finite solution of ( 1 ) exists.

(II) If it happens that M = n and that fo= l0+l , (6) becomes y =a we take St = 1/fol+l and write:

(I)xIf solves (1) uniqey. = (S t bl. i * * * = a)

x solves (1) uniquely. (See note below.)



SYSTEMS OF LINEAR EQUATIONS 69

(III) If M < n and some foK # 0, K > AM, (2) is satisfied, from (6), by n+1

E aR f n+l (7) Y n=+l whenever E aK fo # 0.

21 aKfo K-M+1

It is evident that go = 1, so y = (1, xl, * , xn) and x solves (1) if the conditions of (7) are satisfied. Thus, (1) is completely solved.

NOTE. Reference [1] considers only Case II, for m = n, but the author forms fn+l not from u?, but from his vector:

en+1 (1, ,**,1 )

This vector is unsatisfactory for the following reason. In reference [1], the matrix A of (1) is assumed n-by-n non-singular; then no combination of its row vectors at can satisfy:

n ~~~~n ,8?as = 0, E8 fl O.

Then no combination of the bt of the matrix B can equal u0, so fn+l $ 0 and we obtain our solution. But if we replace u? with efn+l in the orthogonalization it may occur that, for a non-singular A, some combination of the rows of B will equal e8+'; then we get a zero solution of (2), and the solution of (1) is inde- terminate, as in the example:

2x, + X2 = -2, (8)

X1 + 0 = -1.

In such a case, reference [1] givesf3 = (0, 0, 0), whereas ourf3 = ( T7, 0), giving the solution x1 =-1, X2 = 0.

REFERENCE

1. H. MANVEL SEMARNE, New direct method of solution of a system of simultaneous linear equations, SIAM Review, 1, 1(1959), pp. 53-54.



a note on systems of linear equations

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