Row Space, Column Space, and Nullspace

1 Row space , Column space , and NullspaceLinear AlgebraMATH 2010 Terminology:LetAbe the 2x4 matrixA=[2 3 1 04 5 6 2]Therow vectors ofAare[2 3 1 0][4 5 6 2](the rows ofA) in< vectors ofAare[24],[35],[ 16],[02](the columns ofA) in<2 Definition:LetAbe amxnmatrix (recallmis the number of rows andnis the number of columns ),then Therow spaceofAis the subspace of<nspanned by the row vectors ofA Thecolumn spaceofAis the subspace of<mspanned by the Column vectors ofA. Theorem:If amxnmatrixAis row-equivalent to amxnmatrixB, then the row space ofAis equalto the row space ofB. (NOT true for the Column space ) Theorem:If a matrixAis row-equivalent to a matrixBin row-echelon form, then the nonzero rowvectors ofBform a basis for the row space ofA. Example - Finding a Basis for Row SpaceLetA= 1 1 4 1 20 1 2 1 10 0 0 1 21 1 0 0 22 1 6 0 1 Find a basis for the row space must reduceA: 1 1 4 1 20 1 2 1 10 0 0 1 21 1 0 0 22 1 6 0 1 1 1 4 1 20 1 2 1 10 0 0 1 20 2 4 1 00 1 2 2 3 1 1 4 1 20 1 2 1 10 0 0 1 20 0 0 1 20 0 0 1 2 1 1 4 1 20 1 2 1 10 0 0 1 20 0 0 0 00 0 0 0 0 Thenw1= [1,1,4,1,2]w2= [0,1,2,1,1]w3= [0,0,0,1,2]form a basis for the row space ofA.

2 Example- Finding a basis spanned by a set S:LetS={v1, v2, v3, v4}wherev1= [1, 2,0,3, 4]v2= [3,2,8,1,4]v3= [2,3,7,2,3]v4= [ 1,2,0,4, 3]Find a basis for the subspace of<5spanned we look the matrixA= 1 2 0 3 43 2 8 1 42 3 7 2 3 1 2 0 4 3 then a basis for the row space ofAgives a basis for the subspace of<5spanned byS. 1 2 0 3 43 2 8 1 42 3 7 2 3 1 2 0 4 3 1 2 0 3 40 8 8 8 160 7 7 4 110 0 0 7 7 1 2 0 3 40 1 1 1 20 7 7 4 110 0 0 7 7 1 2 0 3 40 1 1 1 20 0 0 3 30 0 0 7 7 1 2 0 3 40 1 1 1 20 0 0 1 10 0 0 0 0 Thenw1= [1, 2,0,3, 4]w2= [0,1,1, 1,2]w3= [0,0,0,1, 1]form a basis for the subspace spanned byS. The dimension of the row space is 3.

3 Example - Finding a basis for the Column space ofA:There are two ways to find a basis forthe Column the row-echelon form ofA: 1 1 4 1 20 1 2 1 10 0 0 1 21 1 0 0 22 1 6 0 1 1 1 4 1 20 1 2 1 10 0 0 1 20 0 0 0 00 0 0 0 0 The columns from theoriginal matrixwhich have leading ones when reduced form a basis forthe Column space ofA. In the above example, columns 1, 2, and 4 have leading ones. Therefore, columns 1, 2, and 4 of the original matrix form a basis for the Column space , 10012 110 11 11100 form a basis for the Column space ofA. The dimension of the Column space ofAis second way to find a basis for the Column space ofAis to recognize that the Column space ofAis equal to the row space ofAT. Finding a basis for the row space ofATis the same as findinga basis for the Column space 1 0 0 1 21 1 0 1 14 2 0 0 61 1 1 0 02 1 2 2 1 ReducingAT,we get 1 0 0 1 21 1 0 1 14 2 0 0 61 1 1 0 02 1 2 2 1 1 0 0 1 20 1 0 2 10 2 0 4 20 1 1 1 20 1 2 0 3 1 0 0 1 20 1 0 2 10 0 0 0 00 0 1 1 10 0 2 2 2 1 0 0 1 20 1 0 2 10 0 1 1 10 0 0 0 00 0 0 0 0 Thenw1= (1,0,0,1,2)w2= (0,1,0, 2, 1)w3= (0,0,1,1,1)form a basis for the Column space ofA.

4 Example:LetS={v1, v2, v3, v4}from above wherev1= [1, 2,0,3, 4]v2= [3,2,8,1,4]v3= [2,3,7,2,3]v4= [ 1,2,0,4, 3]Find a basis for the subspace of<5spanned bySthat is a subset of the vectors inS. To do this, weset the columns of a matrixAas the vectorsv1,v2,v3andv4:A= 1 3 2 1 2 2 3 20 8 7 03 1 2 4 4 4 3 3 Then find the Column space ofA: 1 3 2 1 2 2 3 20 8 7 03 1 2 4 4 4 3 3 1 3 2 10 1 7/8 00 0 1 7/30 0 0 00 0 0 0 There are leading ones in columns 1,2, and 3, so columns 1,2, and 3 from the original matrix form abasis for the Column space ofA. This corresponds to the vectorsv1= [1, 2,0,3, 4]v2= [3,2,8,1,4]v3= [2,3,7,2,3] Theorem:IfAis anmxnmatrix, then the row space and Column space ofAhave thesamedimension.

5 Definition:The dimension of the row (or Column ) space of a matrixAis called therankofA; denotedrank(A). Example:LetA= 3 1 22 1 37 1 8 Then 3 1 22 1 37 1 8 1 1/3 2/301 100 0 Therefore, rank(A) = 2 because there are 2 leading ones. Definition:IfAis amxnmatrix, then the set of all solutions of the homogeneous system of linearequationsAx= 0is a subspace of<ncalled thenullspaceofAand denotedN(A).N(A) ={x <n:Ax= 0}The dimension ofN(A) is called thenullityofA. Example: Finding a basis for the Nullspace ofA:LetA= 1 1 4 1 20 1 2 1 10 0 0 1 21 1 0 0 22 1 6 0 1 We need to solve the systemAx= 0: 1 1 4 1 2|00 1 2 1 1|00 0 0 1 2|01 1 0 0 2|02 1 6 0 1|0 1 0 2 0 1|00 1 2 0 1|00 0 0 1 2|00 0 0 0 0|00 0 0 0 0|0 Therefore,x3=sandx5=tare free parameters.

6 The solution to the system is given byx= 2s t 2s+ts 2tt = 2 2100 s+ 110 21 tTherefore, the basis forN(A) is given by 2 2100 , 110 21 We havedim(N(A)) = 2, , nullity(A) = 2. Note that dim(rowspace(A)) = 3 and nullity(A) = +2 = 5 =n(the number of columns ofA). Theorem:IfAis amxnmatrix of rankA(r), then the dimension of the solution space ofAx= 0 isn r, (A) +nullity(A) Example:LetA= 2 4 3 67 14 6 3 2 4 1 22 4 2 2 for row space for Column space ofAthat is a subset of the Column vectors for Nullspace (A) (A) for row space ofA:{[1,2, 1, 1],[0,0,1,4]} for Column space ofAthat is a subset of the Column vectors ofA: 27 22 , 3 61 2 for Nullspace ofA: 2100 , 30 41 (A): (A): 2 Solutions of Systems of Linear Equations:The solutionxtoAx=bcan be written asx=xp+xhwherexpis called a particular solution ofAx=bandxhis called the homogeneous solution ofAx= 0.

7 Example:Consider the systemx1 2x3+x4= 53x1+x2 5x3= 8x1+ 2x2 5x4= 9 Solving the system we have 1 0 2 1|53 1 5 0|81 2 0 5| 9 1 0 2 1|50 1 1 3| 70 2 2 6| 14 1 0 2 1|50 1 1 3| 70 0 0 0|0 Therefore, we get the solutionx= 5 + 2s t 7 s+ 3tst = 5 700 +s 2 110 +t 1301 The particular solution is given byxp= 5 700 and the homogeneous solution is given byxh=s 2 110 +t 1301 Theorem:A system of linear equationsAx=bis consistent (has a solution) if and only ifbis in thecolumn space ofA( ,bcan be written as a linear combination of the columns ofA). Equivalent Statements:IfAis annxnmatrix, then the following are equivalent:(a)Ais invertible(b)Ax= 0 has only the trivial solution(c)The reduced row-echelon form ofAisIn(d)Ais expressible as a product of elementary matrices(e)Ax=bis consistent for everynx1 matrixb(f)Ax=bhas exactly one solution for everynx1 matrixb(g)|A|6= 0(h) = 0 is not an eigenvalue ofA(i)rank(A)=n(j)nrow vectors ofAare linearly independent(k)ncolumn vectors ofAare linearly independent