MATH 304Linear AlgebraLecture 18:Rank and nullity of a matrix.Basis and coordinates.Change of coordinates.Rank of a matrixDefinition. The row space of an m×n matrix A isthe subspace of Rnspanned by rows of A. Thecolumn space of A is a subspace of Rmspanned bycolumns of A.The row space and the column space of A have thesame dimension, which is called the rank of A.Theorem 1 Elementary row operations do not change therow space of a matrix.Theorem 2 If a matrix A is in row echelon form, then t henonzero rows of A form a basis for the row space.Theorem 3 The rank of a matrix is equal to the number ofnonzero rows in its row echelon form.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. . . amnx1x2x3...xn=00...0The nullspace N(A) is the solution set of a systemof linear homogeneous equations (with A as thecoefficient matrix).Let A be an m×n matrix. Then the nu llspaceN(A) is the solution set of a system of linearhomogeneous equations in n variables.Theorem The nullspace N(A) is a subspace of thevector space Rn.Definition. The dimension of the nullspace N(A) iscalled the nullity of the matrix 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 3x1− 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.rank + nullityTheorem The rank of a matrix A plu s the nullityof A equals the number of columns in 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.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 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 (called thedimension of V ).Example. Vectors e1= (1, 0, 0, . . . , 0, 0),e2= (0, 1, 0, . . . , 0, 0),. . . , en= (0, 0, 0, . . . , 0, 1)form a basis for Rn(called standard) since(x1, x2, . . . , xn) = x1e1+ x2e2+ · · · + xnen.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 matrixa bc d∈ M2,2(R)relative to the basis1 00 0,0 01 0,0 10 0,0 00 1are (a, c, b, d).• Coordinates of a polynomialp(x) = a0+ a1x + · · · + an−1xn−1∈ Pnrelative tothe basis 1, x, x2, . . . , xn−1are (a0, a1, . . . , an−1).Vectors u1=(3, 1) and u2=(2, 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⇐⇒3x + 2y = 7x + y = 4⇐⇒x = −1y = 5Problem 2. Find the vector w whose coordinateswith respect to the basis u1, u2are (7, 4).w = 7u1+ 4u2= 7(3, 1) + 4(2, 1) = (29, 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), e2= (0, 1), and let(x′, y′) be its coordinates with respect to the basisu1= (3, 1), u2= (2, 1).Problem. Find a relation between (x, y ) and (x′, y′).By definition, v = xe1+ ye2= x′u1+ y′u2.In standard coordinates,xy= x′31+ y′21=3 21 1x′y′=⇒x′y′=3 21 1−1xy=1 −2−1 3xyChange of coordinates in RnThe usual (standard) coordinates of a vectorv = (x1, x2, . . . , xn) ∈ Rnare coordinates relative to thestandard basis e1= (1, 0, . . . , 0, 0), e2= (0, 1, . . . , 0, 0),. . . ,en= (0, 0, . . . , 0, 1).Let u1, u2, . . . , unbe another basis for Rnand (x′1, x′2, . . . , x′n)be the coordinates of the same vector v with respect to thisbasis.Problem 1. Given the standard coordinates(x1, x2, . . . , xn), find the nonstandard coordinates(x′1, x′2, . . . , x′n).Problem 2. Given the nonstandard coordinates(x′1, x′2, . . . , x′n), find the standard coordinates(x1, x2, . . . , xn).It turns out thatx1x2...xn=u11u12. . . u1nu21u22. . . u2n............un1un2. . . unnx′1x′2...x′n.The matrix U = (uij) does not depend on the vector v.Columns of U are coordinates of vectorsu1, u2, . . . , unwith respect to the standard basis.U is called the transition matrix from the basisu1, u2, . . . , unto the standard basis e1, e2, . . . , en.This solves Problem 2. To solve Problem 1, we haveto use the inverse matrix U−1, which is thetransition matrix from e1, . . . , ento u1, . . . , un.Problem. Find coordinates of the vectorx = (1, 2, 3) with respect to the basisu1= (1, 1, 0), u2= (0, 1, 1), u3= (1, 1, 1).The nonstandard coordinates (x′, y′, z′) of x satisfyx′y′z′= U123,where U is the transition matrix from the standard basise1, e2, e3to the basis u1, u2, u3.The transition matrix from u1, u2, u3to e1, e2, e3isU0=
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