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Math 304–504 Linear Algebra Lecture 17: Change of coordinates (continued). Rank and nullity of a matrix Basis and coordinates If {v1 , v2, . . . , vn } is a basis for a vector space V , then any vector v ∈ V has a unique representation v = x 1 v1 + x 2 v2 + · · · + x n vn , where xi ∈ R. The coefficients x1, x2, . . . , xn are called the coordinates of v with respect to the ordered basis v1 , v2, . . . , vn . The mapping vector v 7→ its coordinates (x1, x2, . . . , xn ) is a one-to-one correspondence between V and Rn . This correspondence respects linear operations in V and in Rn . Change of coordinates Let v1, v2, . . . , vn and u1 , u2, . . . , un be two bases for a vector space V . Take any vector x ∈ V . Let (x1, x2, . . . , xn ) be the coordinates of x with respect to the basis v1, v2, . . . , vn and (x1′ , x2′ , . . . , xn′ ) be the coordinates of x with respect to u1, u2 , . . . , un . Then ′ x1 u11 . . . u1n x1 . . . .. = .. . . . .. ... , xn′ un1 . . . unn xn where U = (uij ) is the transition matrix from the basis v1, . . . , vn to u1 , . . . , un . Columns of U are coordinates of the vectors v1 , . . . , vn with respect to the basis u1 , . . . , un . Example. Vectors u1 = (1, 1, 0), u2 = (0, 1, 1), and u3 = (1, 1, 1) form a basis for R3 . The transition matrix from the basis u1 , u2, u3 to the standard basis e1 , e2, e3 is 1 0 1 1 1 1 . 0 1 1 The transition matrix from e1, e2 , e3 to u1, u2 , u3 is −1 1 0 1 0 1 −1 1 1 1 = −1 1 0. 0 1 1 1 −1 1 Problem. Find the transition matrix from the basis v1 = (1, 2, 3), v2 = (1, 0, 1), v3 = (1, 2, 1) to the basis u1 = (1, 1, 0), u2 = (0, 1, 1), u3 = (1, 1, 1). To change coordinates from v1, v2, v3 to u1, u2 , u3, we first change coordinates from v1, v2, v3 to the standard basis e1 , e2, e3, and then from e1, e2, e3 to u1 , u2, u3. Transition matrix from v1, v2, v3 to e1 , e2, e3: 1 1 1 U1 = 2 0 2 . 3 1 1 Transition matrix from e1 , e2, e3 to u1 , u2, u3: −1 0 1 −1 1 0 1 U2 = 1 1 1 = −1 1 0. 0 1 1 1 −1 1 Transition matrix from v1, v2, v3 to u1 , u2, u3: 1 1 1 −1 −1 1 0 1 −1 U2U1 = −1 1 0 2 0 2 = 1 −1 1. 1 −1 1 3 1 1 2 2 0 Nullspace Let A = (aij ) be an m×n matrix. Definition. The nullspace of the matrix A, denoted N(A), is the set of all n-dimensional column vectors x such that Ax = 0. x 1 0 a11 a12 a13 . . . a1n x a 2 0 21 a22 a23 . . . a2n x .. 3 = .. .. .. .. .. . . .. . . . . . 0 am1 am2 am3 . . . amn xn The nullspace N(A) is the solution set of a system of linear homogeneous equations (with A as the coefficient matrix). Let A = (aij ) be an m×n matrix. Theorem The nullspace N(A) is a subspace of the vector space Rn . Proof: We have to show that N(A) is nonempty, closed under addition, and closed under scaling. First of all, A0 = 0 =⇒ 0 ∈ N(A) =⇒ N(A) is not empty. Secondly, if x, y ∈ N(A), i.e., if Ax = Ay = 0, then A(x + y) = Ax + Ay = 0 + 0 = 0 =⇒ x+y ∈ N(A). Thirdly, if x ∈ N(A), i.e., if Ax = 0, then for any r ∈ R one has A(r x) = r (Ax) = r 0 = 0 =⇒ r x ∈ N(A). Definition. The dimension of the nullspace N(A) is called the nullity of the matrix A. Problem. Find the nullity of the matrix 1 1 1 1 A= . 2 3 4 5 Elementary row operations do not change the nullspace. Let us convert A to reduced row echelon form: 1 1 1 1 1 1 1 1 1 0 −1 −2 → → 2 3 4 5 0 1 2 3 0 1 2 3 x1 = x3 + 2x4 x1 − x3 − 2x4 = 0 ⇐⇒ x2 = −2x3 − 3x4 x2 + 2x3 + 3x4 = 0 General 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) forms a basis for N(A). Thus the nullity of the matrix A is 2. Row space Definition. The row space of an m×n matrix A is the subspace of Rn spanned by rows of A. The dimension of the row space is called the rank of the matrix A. Theorem 1 Elementary row operations do not change the row space of a matrix. Theorem 2 If a matrix A is in row echelon form, then the nonzero rows of A are linearly independent. Corollary The rank of a matrix is equal to the number of nonzero rows in its row echelon form. Theorem 3 The rank of a matrix A plus the nullity of A equals the number of columns of A. Problem. Find the rank of the matrix −1 0 −1 2 A = 2 0 2 0. 1 0 1 −1 Elementary row operations do not change the row space. Let us convert A to row echelon form: −1 0 −1 2 −1 0 −1 2 2 0 2 0 → 0 0 0 4 1 0 1 −1 1 0 1 −1 −1 0 −1 2 −1 0 −1 2 → 0 0 0 4 → 0 0 0 0 0 0 0 1 0 0 0 1 −1 0 −1 2 1 0 1 −2 → 0 0 0 1 → 0 0 0 1 0 0 0 0 0 0 0 0 Vectors (1, 0, 1, −2) and (0, 0, 0, 1) form a basis for the row space of A. Thus the rank of A is 2. Remark. The rank of A equals the number of nonzero rows in the row echelon form, which equals the number of leading entries. The nullity of A equals the number of free variables in the corresponding system, which equals the number of columns without leading entries. Consequently, rank+nullity is the number of all columns in the matrix A. Theorem 1 Elementary row operations do not change the row space of a matrix. Proof: Suppose that A and B are m×n matrices such that B is obtained from A by an elementary row operation. Let a1 , . . . , am be the rows of A and b1 , . . . , bm be the rows of B. We have to show that Span(a1 , . . . , am ) = Span(b1 , . . . , bm ). Observe that any row bi of B belongs to Span(a1 , . . . , am ). Indeed, either bi = aj for some 1 ≤ j ≤ m, or bi = r ai for some scalar r 6= 0, or bi = ai +r aj for some j 6= i and r ∈ R. It follows that Span(b1 , . . . , bm ) ⊂ Span(a1 , . . . , am ). Now the matrix A can also be obtained from B by an elementary row operation. By the above, Span(a1 , . . . , am ) ⊂ Span(b1 , . . . , bm ). Problem. Find the nullity of the matrix 1 1 1 1 A= . 2 3 4 5 Alternative solution: Clearly, the rows of A are linearly 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.