A Note on the Equality of the Column and Row Rank of a Matrix Author(s): George Mackiw Reviewed work(s): Source: Mathematics Magazine, Vol. 68, No. 4 (Oct., 1995), pp. 285-286 Published by: Mathematical Association of America Stable URL: http://www.jstor.org/stable/2690576 . Accessed: 27/03/2012 11:08
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http://www.jstor.org VOL. 68, NO. 4, OCTOBER 1995 285 A Note on the Equalityof the Column and Row Rankof a Matrix
GEORGE MAC KIW Loyola College in Maryland Baltimore,MD 21210
A basic resultin linearalgebra is thatthe row and columnspaces of a matrixhave dimensionsthat are equal. In thisnote we derivethis result using an approachthat is elementaryyet different from that appearing in mostcurrent texts. Moreover, we do not relyon the echelonform in our arguments. The columnand rowranks of a matrixA are the dimensionsof the columnand row spaces,respectively, of A. As is oftennoted, it is enoughto establishthat
row rank A < column rank A (1) forany matrixA. For, applyingthe resultin (1) to At, the transposeof A, would producethe desiredequality, since the rowspace of At is preciselythe columnspace of A. We presentthe argumentfor real matrices,indicating at the end the modifica- tionsnecessary in the complexcase and the case of matricesover arbitrary fields. We takefull advantage of the followingtwo elementaryobservations: 1) forany vector x in R', Ax is a linearcombination of the columnsof A, and 2) vectorsin the null space of A are orthogonalto vectorsin the row space of A, relativeto the usual Euclideanproduct. Bothof these remarks follow easily from the very definition of matrixmultiplication. For example,if the vector x is in the nullspace of A then Ax = 0, and thusthe inner productof x withthe rowsof A mustbe zero. Since theserows span the rowspace of A, remark2) follows.An immediateconsequence is thatonly the zero vectorcan be commonto boththe nullspace and rowspace of A, since2) requiressuch a vectorto be orthogonalto itself. Given an m by n matrixA, let the vectors xI, x2. Xr in Rn forma basis forthe row space of A. Then the r vectors Ax,, Ax2,..., Ax, are in the column space of A and, further,we claim they are linearly independent. For, if cl AxI + c2 Ax2 + +cr Axr = O forsome real scalars cl, c2) ... . cr, then A(c1xI + C2X2 ? +crxr)=0 and the vector v=clx X+c2x2 + +Cr xr would be in the nullspace of A. But,v is also in the rowspace of A, sinceit is a linearcombination of basis elements.So, v is the zero vectorand the linearindependence of xI, x2,..., Xr guaranteesthat cl = c2 = .. = Cr = 0. The existenceof r linearlyindependent vec- torsin the columnspace requiresthat r ? columnrank A. Since r is the rowrank of A, we have arrivedat the desiredinequality (1). This approachalso yieldsadditional information. Let YI, Y2 ... Yk forma basis for the null space of A. Since the row space of A and the null space of A intersect trivially,it followsthat the set {xI, x2., Xr, YI, Y2,. Yk} is linearlyindependent. Further,if z is a vectorin R' then Az is in the columnspace of A and hence expressible as a linear combinationof basis elements. Thus, we can write Az = E_ 1 d Ax., forscalars d . But thenthe vector z - Er d xj is in the nullspace of A and can be writtenas a linearcombination of YI, Y2, Yk Thus z can be expressed as a linear combinationof the vectors in the set {xI, x2,. .. Xr) 286 MATHEMATICS MAGAZINE
YI, Y2, Yk) and this set then formsa basis for Rn. A dimensioncount yields r + k = n, givingthe rankand nullitytheorem without use of the echelonform. Note thatthe relianceon orthogonalityin the argumentsabove is elementaryand does not requirethe Gram-Schmidtprocess. The ideas used here can be readily appliedto complexmatrices with minor modifications. The hermitianinner product is used insteadin the complexvector space Cn, as is the hermitiantranspose. Observa- tion2) wouldnote then that vectors in the null space of A are orthogonalto thosein the rowspace A. The rowand columnrank theorem is a well-knownresult that is validfor matrices over arbitraryfields. The notionof orthogonalcomplement can be generalizedusing linearfunctionals and dual spaces, and the generalstructure of the argumentshere can thenbe carriedover to arbitraryfields. The text[1, pp. 97 ff.]contains such an approach. Manyauthors base theirdiscussion of rankon the echelonform. The factthat the non-zerorows of the echelon formare a basis forthe rowspace, or thatcolumns in the echelonform containing lead ones can be used to identifya basis forthe column space in the originalmatrix A, are centralto such a developmentof rank.Results such as thesefollow easily if it is establishedindependently that row rank and column rankmust be equal. The authorwould liketo thankthe refereefor several valuable suggestions.
REFERENCE
1. KennethHoffman and Ray Kunze, Linear Algebra,2nd edition,Prentice-Hall, Englewood Cliffs, NJ, 1971.
A One-SentenceProof That V2 Is Irrational
DAVID M. BLOOM BrooklynCollege of CUNY Brooklyn,NY 11210
If V2 were rational,say F = mn/nin lowest terms,then also ? =(2n -m)/ (m - n) in lower terms,giving a contradiction. (The threeverifications needed-that the second denominatoris less thanthe first and stillpositive, and thatthe two fractionsare equal-are straightforward.) The argumentis not original;it's the algebraicversion of a geometricargument givenin [1, p. 84], and it was presented(in slightlydifferent form) by Ivan Nivenat a lecturein 1985. Note, though,that the algebraicargument, unlike the geometric, easilyadapts to an arbitraryVk where k is anypositive integer that is not a perfect square. Indeed, let j be the integersuch that j < xk REFERENCE 1. H. Eves, An Introductionto theHistory of Mathematics,6th edition, W. B. SaundersCo., Philadelphia, 1990.