Notes preparatory to "rank"

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This note contemplates ak-by-nmatrixA.

Whenever ak-deep column matrixbis given, we may inquire whether there are anyn-deep columnsxfor which the matrix equation

(#)A•x=b

is valid.

This equation is just the matrix counterpart of a certain corresponding system of simultaneous linear equations whose coefficient matrix isA, and whose augmented matrixMhas the form

M= [A|b] .

What do we really know aboutAif we know that

(1) for each conceivablek-deep column matrixb, equation (#) always has one or more solution(s)x?

Certainly we know that the fully reduced row echelon form (frref) of the augmented matrixMnever has any "embarrassing" row starting withnzeros but ending with a non-zero entry in the last column, and this regardless what the originalk-deep column matrixb. But this means exactly that

(2) the frref ofAitself has no row of zeros whatsoever.

(From this condition onA, in turn, it is easy to deduce (1).)

(Of course (1) or (2) can happen ONLY if there are at least as many variables as equations, i.e.,k<n.)

And we know, in particular, using ask-deep column matrixbthe various columns of thek-by-kidentity matrixI_{k×k}, that

(3) there is ann-by-kmatrixXwithA•X=I_{k×k}

(as each successive column ofXjust use any solutionxto the instance of equation (#) in whichbis the corresponding column ofI_{k×k}).

From (3) in turn we can again deduce (1): for ifA•X=I_{k×k}, and if columnbis given, we may choose to setx=X•b. Thenxis ann-deep column and

A•x=A•(X•b) = (A•X)•b=I_{k×k}b=b,

as required.

---

A new question: what do we really know aboutAif we know that

whenbis thek-deep column matrix with zeros

(a) everywhere, equation (#) never has any solutionsx

OTHER THAN the "trivial" solution with eachx_{i}= 0 ?

The first thing to observe is that (a) is the case if, and only if, none of the variablesx_{i}is "free" -- that is, eachx_{i}is "bound" -- equivalently, in the frref ofA, there is, for each variablex_{i}, a row whose leading 1 appears in thei^{th}position -- which means:

(b) the frref ofAhasnnon-zero rows.

(In particular, we must haven<k.)

In fact, the frref ofAis then a matrix (with at leastnrows) whose firstnrows constitute the identity matrixI_{n×n}:

|I_{n×n}|

(c) frref(A) = | ----- | .

| 0 |

In consequence,

(d) there is ann-by-kmatrixYwithY•A=I_{n×n}.

(Indeed, ifEis the invertible matrix,k-by-k, product of elementary matrices, corresponding to any sequence of elementary row operations that reduceAto its frref, it's enough to set

Y= the firstnrows ofE,

|I_{n×n}|

for sinceE•A= | ----- | , it then follows thatY•A= the firstnrows ofE•A=I_{n×n}.)

| 0 |

Not only does (d) thus follow from (a) (via (b) and (c)), however, but also conversely, (a) follows from (d):

for, whatever the solutionxtoA•x= 0 , if we really have aYwithY•A=I_{n×n}, then

x=I_{n×n}•x= (Y•A)•x=Y•(A•x) =Y•0 = 0 .

---

Conditions (1) (2) and (3) thus all mean some same thing aboutA, and force the inequalityk<n; while, independently, conditions (a) (b) (c) and (d) also mean some (other) same thing aboutA, and force the opposite inequalityn<k.

This means, in particular, that an invertible matrix -- a matrixA(suppose itk-by-n, still) for which there is a matrixB(necessarilyn-by-kso that both products can be available) satisfying bothA•B=I_{k×k}andB•A=I_{n×n}-- must be square.

That is, we must haven=k.

In fact, a matrixAsatisfying BOTH any one of (1) (2) or (3) AND ALSO any one of (a) (b) (c) or (d) MUST SATISFY ALL OF THESE -- and be square -- and be invertible.

And for matricesAthat happen already to be square (n=k) the situation is even nicer: as soon asn-by-nAsatisfies ANY ONE of the seven conditions (1), (2), (3), (a), (b), (c), or (d), it will satisfy them ALL -- for its frref will beI_{n×n}.

--- Back --- First posted 13 Feb 2004.