MATH 311Topics in Applied MathematicsLecture 11:Rank and nullity of a matrix.Basis and coordinates.Change of basis.Basis and dimensionDefinition. Let V be a vector space. A linearlyindependent spanning set for V is called a basis.Theorem Any vector space V has a basis. If Vhas a finite basis, then all bases for V are finite andhave the same number of elements.Definition. The dimension of a vector space V ,denoted dim V , is the number of elements in any ofits bases.Row space of a matrixDefinition. The row space of an m×n matrix A isthe subspace of Rnspanned by rows of A.The dimension of the row space is called the rankof the matrix A.Theorem 1 The rank of a matrix A is the maximal numberof linearly independent rows in A.Theorem 2 Elementary row operations do not change therow space of a matrix.Theorem 3 If a matrix A is in row echelon form, then thenonzero rows of A are linearly independent.Corollary The rank of a matrix is equal to the number ofnonzero rows in its row echelon form.Problem. Find the rank of the matrixA =1 1 00 1 12 3 11 1 1.Elementary row operations do not change the rowspace. Let us convert A to row echelon form:1 1 00 1 12 3 11 1 1→1 1 00 1 10 1 11 1 1→1 1 00 1 10 1 10 0 11 1 00 1 10 1 10 0 1→1 1 00 1 10 0 00 0 1→1 1 001 10 010 0 0Vectors (1, 1, 0), (0, 1, 1), and (0, 0, 1) form a basisfor the row space of A. Thus the rank of A is 3.It follows that the row space of A is the entire spaceR3.Problem. Find a basis for the vector space Vspanned by vectors w1= (1, 1, 0), w2= (0, 1, 1),w3= (2, 3, 1), and w4= (1, 1, 1).The vector space V is the row space of a matrix1 1 00 1 12 3 11 1 1.According to the solution of the previous problem,vectors (1, 1, 0), (0, 1, 1), and (0, 0, 1) form a basisfor V .Column space of a matrixDefinition. The column space of an m×n matrixA is the subspace of Rmspanned by columns of A.Theorem 1 The column space of a matrix A coincides withthe row space of the transpose matrix AT.Theorem 2 Elementary column operations do not changethe column space of a matrix.Theorem 3 Elementary row operations do not change thedimension of the column space of a matrix (although they canchange the column space).Theorem 4 For any matrix, the row space and the columnspace have the same dimension.Problem. Find a basis for the column space of thematrixA =1 1 00 1 12 3 11 1 1.The column space of A coincides with the row space of AT.To find a basis, we convert ATto row echelon form:AT=1 0 2 11 1 3 10 1 1 1→1 0 2 10 1 1 00 1 1 1→1 0 2 101 1 00 0 0 1Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis forthe column space of A.Problem. Find a basis for the column space of thematrixA =1 1 00 1 12 3 11 1 1.Alternative solution: We already know from aprevious problem that the rank of A is 3. It followsthat the columns of A are linearly independent.Therefore these columns form a basis for thecolumn space.Nullspace of a matrixLet A = (aij) be an m×n matrix.Definition. The nullspace of the matrix A,denoted N(A), is the set of all n-dimensionalcolumn vectors x such thatAx = 0.a11a12a13. . . a1na21a22a23. . . a2n...............am1am2am3. . . amnx1x2x3...xn=00...0The nullspace N(A) is the solution set of a systemof linear homogeneous equations (with A as thecoefficient matrix).Theorem 1 The nullspace N(A) of an m×nmatrix A is a subspace of the vector space Rn.Definition. The dimension of the nullspace N(A) iscalled the nullity of the matrix A.Theorem 2 The rank of a matrix A plus thenullity of A equals the number of columns of A.Sketch of the proof: The rank of A equals the number ofnonzero rows in the row echelon form, which equals thenumber of leading entries.The nullity of A equals the number of free variables in thecorresponding system, which equals the number of columnswithout leading entries in the row echelon form.Consequently, rank+nullity is the number of all columns in thematrix A.Problem. Find the nullity of the matrixA =1 1 1 12 3 4 5.Elementary row operations do not change the nullspace.Let us convert A to reduced row echelon form:1 1 1 12 3 4 5→1 1 1 10 1 2 3→1 0 −1 −20 1 2 3x1− x3− 2x4= 0x2+ 2x3+ 3x4= 0⇐⇒x1= x3+ 2x4x2= −2x3− 3x4General element of N(A):(x1, x2, x3, x4) = (t + 2s, −2t − 3s, t, s)= t(1, −2, 1, 0) + s(2, −3, 0, 1), t, s ∈ R.Vectors (1, −2, 1, 0) and (2, −3, 0, 1) form a basis for N(A).Thus the nullity of the matrix A is 2.Problem. Find the nullity of the matrixA =1 1 1 12 3 4 5.Alternative solution: Clearly, the rows of A arelinearly independent. Therefore the rank of A is 2.Since(rank of A) + (nullity of A) = 4,it follows that the nullity of A is 2.Basis and coordinatesIf {v1, v2, . . . , vn} is a basis for a vector space V ,then any vector v ∈ V has a unique representationv = x1v1+ x2v2+ · · · + xnvn,where xi∈ R. The coefficients x1, x2, . . . , xnarecalled the coordinates of v with respect to theordered basis v1, v2, . . . , vn.The mappingvector v 7→ its coordinates (x1, x2, . . . , xn)is a one-to-one correspondence between V and Rn.This correspondence respects linear operations in Vand in Rn.Examples. • Coordinates of a vectorv = (x1, x2, . . . , xn) ∈ Rnrelative to the standardbasis e1= (1, 0, . . . , 0, 0), e2= (0, 1, . . . , 0, 0),. . . ,en= (0, 0, . . . , 0, 1) are (x1, x2, . . . , xn).• Coordinates of a matrixa bc d∈ M2,2(R)relative to the basis1 00 0,0 10 0,0 01 0,0 00 1are (a, b, c, d).• Coordinates of a p olynomialp(x) = a0+ a1x + · · · + an−1xn−1∈ Pnrelative tothe basis 1, x, x2, . . . , xn−1are (a0, a1, . . . , an−1).Vectors u1=(2, 1) and u2=(3, 1) form a basis for R2.Problem 1. Find coordinates of the vectorv = (7, 4) with respect to the basis u1, u2.The desired coordinates x, y satisfyv = xu1+yu2⇐⇒2x + 3y = 7x + y = 4⇐⇒x = 5y = −1Problem 2. Find the vector w whose coordinateswith respect to the basis u1, u2are (7, 4).w = 7u1+ 4u2= 7(2, 1) + 4(3, 1) = (26, 11)Change of coordinatesGiven a vector v ∈ R2, let (x, y ) be its standardcoordinates, i.e., coordinates with respect to thestandard basis e1= (1, 0),