Math 1b Prac — Bases for row spaces, null spaces; orthogonal spacesJanuary 18, 2011 — s lightly revised January 21, 2011Given a matrix M,therow space of M is the span of its rows (the set of all linearcombinations of its rows). The null space of M is the set of all column vectors x so thatMx = 0. In other words, the null space of M is the set of all vectors orthogonal to the rowsof M. In yet other words, the null space of M is the “solution s pace” of the homogeneoussystem of linear equations corresponding to M. We will check (it is not hard) that theseare really subs paces. Row oper ations on a matrix M do not change the row space or thenullspace of M.These notes are very concise. Proofs of the following theorems will be given in class,or p ossibly in an expanded version of this handout.We use the matrixM =⎛⎝123423453456⎞⎠in our examples.∗∗∗Two ways to find a basis for the row space of ATheorem 1. Let E be a basic form of A. Then the nonzero rows of E form a basis fortherowspaceofA.Example: The reduced echelon form of M isE =⎛⎝10−1 −201 2 300 0 0⎞⎠.So one basis for the row space of M is, or consists of, (1, 0, −1, −2) and (0, 1, 2, 3).Theorem 2. Let E b e a basic form of A. Then the the columns of A that correspond tothe special columns of E form a basis for the column space of A.Example: A basic form ofN =2 0157−11381is E =2015 7−710−7 −20.The linear relations between the columns of E and N are the same. Clearly the secondand third columns of E form a basis for the column space of E, and so the second andthird columns of N form a basis for the column space of N. We also see that the firstcolumn of N is twice the third m inus 7 times the second, because this same r elation holdsamong the columns of E.Theorem 2. Let F be a basic form of A . Then the the rows of A that correspond tothe special columns of F form a basis for the row space of A.The reduced echelon form of M isF =⎛⎜⎝10−101 200 000 0⎞⎟⎠.This means that the first and second rows of M provide a basis for the r ow space of M.It also tells us that the third row of M is 2 times the second minus the second.∗∗∗Two ways to find a basis for the null space of ATheorem 3. SupposeA Ihas reduced echelon formEOFGwhere the rows of E ar e nonzero. Then the nonzero rows of G form a basis for the n ullspace of A.Example: The reduced echelon form for [M ,I]is⎛⎜⎝10−100−54012004−300010−3200001−21⎞⎟⎠.Here E =10−101 2and G =10−3201−21.That this works follows from the statement that (x, y) is a linear combination of therows of [A ,I] if and only if yA = x. Can you fill in the details? The procedure isrelated to what we d id in the handout on linear dep endence and also reminds us of whatwe did to find an inverse (or decide that there is none) for a square matrix A.Theorem 4. Suppose the reduced echelon form of A,withrowsof0’s deleted, is [I, B].Then a basis for the nullspace of A is given by the rows of [−B ,I].Example: The reduced echelon form of M isE =⎛⎝10−1 −201 2 300 0 0⎞⎠.Drop the last row to apply the theorem. Here B =−1 −223. So one basis for the nullspace of M is (1, −2, 1, 0) and (2, −3, 0, 1).Remark: Cf. the last paragraph of the handout on matrix mu ltiplication. We haveused reduced echelon form in the two theorems above in order to avoid technical detailsin the statements, but basic form is really what is important.∗∗∗Orthogonal spacesGiven a subspace U of Rn, we define the orthogonal space U⊥as the set of vectorsorthogonal to every vector of U.Ifu1,...,ukspan (generate) U,thenU is the row spaceof the matrix A with rows u1,...,ukand U⊥is the null space of A.The dim ension of the null s pace of A is called the nullity of A.Both procedures of Theorems 3 and 4, when explained fully, provide proofs of thefollowing theorems.Theorem 5. For any subspace U of Rn, dim(U)+dim(U⊥)=n.This theorem may b e stated “The r ank of A and the nullity of A sum to n”.Theorem 6. For any subspace U of Rn, (U⊥)⊥= U .Example: Let U be the row space of M.IsU the solution set of a hom ogeneoussystem of linear equations. Sure. If the ro ws of N are a bas is for U⊥, then the solutionset of N x = 0 (the null space of N )is(U⊥)⊥= U , So the row space of M is the s olutionset of, for example,x1− 2x2+ x3=0,2x1− 3x2+ x4=0.∗∗∗System s of linear equations againAt the beginning of this course, we were content to ‘solve’ a system of linear equationsAx = b by expressing some of the variables in terms of ‘free’ variables. Here is a moresophisticated and general way to describe the solutions.Theorem 7. The solutions of Ax = b are of the formx = x0+ uwhere x0is any one particular solution (if such exist) and u is any element of the nullspace of A.Proof: If x0is a solution and u in the null space of A,thenx0+u is a solution too, becauseA(x0+ u)=Ax0+ Au = b + 0 = b.Ifx1is a solution, then x0− x1is in the null spaceof A because A(x0− x1)=b − b = 0;andthenx1= x0+ u where u = x0− x1. We may now consider a system Ax = b to be “solved” when we give a p articularsolution and a basis for the null space of