Linear Algebra with SageMath∗ Ajit Kumar ICT Mumbai [email protected] July 11, 2019 1 Working with Vectors Vectors in Sage can be defined by vector function and giving the entries in a list. One can also mention the domain from which entries are taken. Suppose we want a vector v 2 Q3, then we define as v=vector(QQ,[1/2,1,3]). Let us explore some inbuilt methods on vectors in SageMath. sage: v1 = vector(QQ,[1,-1,2]) sage: v2 = vector(QQ,[2,3,-1]) sage: 2*v1-3*v2# linear combination (-4, -11, 7) sage: v1.norm() sqrt (6) sage: v1.dot_product(v2)# Gives the dot product of v1 and v2 -3 sage: v1.cross_product(v2)# Gives the dot product of v1 and v2 (-5, 5, 5) Exercise 1.1. Check that for any two vectors u and v in R3, u × v is orthogonal to u and v. Example 1. Find the orthogonal projection of v = (3; 2; 5) onto (1; −2; 1). sage: v = vector([3,2,5]) sage: u = vector([1,-2,1]) sage: p = v.dot_product(u)/(norm(u)^2)*u sage: p (2/3, -4/3, 2/3) Check that u is orthogonal to v − p. sage: u.dot_product(v-p) 0 We can create a SageMath subroutine to find the orthogonal projection of v onto u. sage: def proj(v,u): ....: a = v.dot_product(u)/(norm(u)^2)*u ....: return(a) sage: u1 = vector([3,-2,4]) sage: u2 = vector([2,1,3]) sage: proj(u2,u1) (48/29, -32/29, 64/29) ∗Distributed during \Teachers Enrichment Course in Undergraduate Mathematics Curriculum" at IIT Guwahati during July 1-13, 2019 1 2 System of Linear Equations We can solve a system of linear equations in SageMath in many ways. sage: x,y,z=var('x,y,z')# To declarea,y,z ar variables sage: solve([x-2*y+z==10, 2*y+3*z==-3,9*x+3*y+z==2],[x,y,z]) [ [x == (101/79), y == (-288/79), z == (113/79)] ] We can write the above system as a matrix equation Ax = b and solve. sage: A = matrix(RR,[[1,-2,1],[0,2,3],[9,3,1]]) sage: b = vector(RR,[10,-3,2]) sage: A.solve_right(b) (1.27848101265823, -3.64556962025316, 1.43037974683544) sage: A\b# MatLab like command (1.27848101265823, -3.64556962025316, 1.43037974683544) sage: aug = A.augment(b,subdivide=True) sage: aug [ 1.00000000000000 -2.00000000000000 1.00000000000000| 10.0000000000000] [0.000000000000000 2.00000000000000 3.00000000000000|-3.00000000000000] [ 9.00000000000000 3.00000000000000 1.00000000000000| 2.00000000000000] sage: A1= aug.rref() sage: A1.column(3) (1.27848101265823, -3.64556962025316, 1.43037974683544) 2.1 Cramer's Rule Example 2. Solve the Ax = b using the Cramer's Rule where 0 1 1 2 2 1 0 9 1 B 2 −3 4 −3 C B 1 C A = B C ; b = B C @ 3 7 6 −5 A @ 10 A 8 6 4 −3 5 First of all let us define a Sage subroutine in which we can replace any column of a matrix by appropriate vector. sage: def column_replace(M,col,v): ....: n1=M.nrows() ....: n2=M.ncols() ....: S=matrix(n1,n2) ....: fori in range(n1): ....: forj in range(n2): ....: S[i,j]=M[i,j] ....: fori in range(n1): ....: S[i,col-1]=v[i,0] ....: returnS sage: A=matrix([[1,1,2,2],[2,-3,4,-3],[3,7, 6,-5],[8,6,4,-3]]) sage: b=matrix(4,1,[9,1,10,5]) sage: A1=column_replace(A,1,b) sage: A2=column_replace(A,2,b) 2 sage: A3=column_replace(A,3,b) sage: A4=column_replace(A,4,b) sage: sol = matrix(4,1,[det(A1)/det(A), det(A2)/det(A), det(A3)/det (A), det(A4)/det(A)]) sage: sol [ -140/249] [ 68/83] [1259/498] [ 153/83] 3 Vector Spaces Simplest way to define a vector space in Sage is to begin with the field and use an exponent to indicate the dimension. For example QQ^4 is a vector space of dimension 4 over the field Q. Similarly RR^4, CC^4 are vectors spaces of dimension 4 over R and C. 4 Linear spans and all that Example 3. Find the linear span of the set of vectors f(1; 1; 2; −1); (2; 3; 5; −4); (3; 4; 7; −5)g: sage: V = QQ^4 sage: v1 = vector(QQ, [1,1,2,-1]) sage: v2 = vector(QQ, [2,3,5,-4]) sage: v3 = vector(QQ, [3,4,7,-5]) sage: W = V.span([v1, v2,v3]) sage:W Vector space of degree 4 and dimension 2 over Rational Field Basis matrix: [ 1 0 1 1] [ 0 1 1 -2] Example 4. Check if the following set of vectors in Q4 are linearly dependent? f(1; −1; 2; 4); (−3; 3; 2; 1); (−1; −2; 6; 9)g sage: V = QQ^4 sage: v1 = vector(QQ, [ 1, -1, 2, 4]) sage: v2 = vector(QQ, [-3, 3, 2, 1]) sage: v3 = vector(QQ, [-1, -2, 6, 9]) sage: L = [v1, v2, v3] sage: V.linear_dependence(L) == [] True Example 5. Show that the following set of vectors in Q4 is a basis of Q4. B = f(2; −1; 3; 1); (3; 3; 4; 0); (1; −2; 0; −1); (4; 5; 5; −1)g Hence find the coordinates of w = (2; 3; 5; 9) with respect to B. sage: V=QQ^4 sage: v1 = vector(QQ, [2, -1, 3, 1]) 3 sage: v2 = vector(QQ, [3, 3, 4, 0]) sage: v3 = vector(QQ, [1, -2, 0, -1]) sage: v4 = vector(QQ, [4, 5, 5, -1]) sage: L = [v1,v2,v3,v4] sage: V.linear_dependence(L) == [] True sage: w = vector(QQ,[2,3,5,9]) sage: W = V.subspace_with_basis(L) sage: s = W.coordinates(w) sage: s [-19/3, 26, 2/3, -16] sage: sum([s[i]*L[i] fori in range(4)])==w True We can obtain the coordinates of w with respect the above basis using RREF as follows: sage: M = column_matrix([v1,v2,v3,v4,w]) sage: M1 = M.rref() sage: M1.column(4) (-19/3, 26, 2/3, -16) Example 6. Complete the following linearly independent set of vectors f(2; 0; 2; 0); (1; 0; 2; 0)g in Q4 to a basis of Q4. sage: v1 = vector(QQ, [2,0,2,0]) sage: v2 = vector(QQ, [1,0,3,0]) sage: M = column_matrix([v1,v2]) sage: I4 = identity_matrix(4) sage: M1=M.augment(I4) sage: M1.rref() [ 1 0 3/4 0-1/4 0] [ 0 1-1/2 0 1/2 0] [ 0 0 0 1 0 0] [ 0 0 0 0 0 1] sage: [M1.column(p) forp in M1.pivots()] [(2, 0, 2, 0), (1, 0, 3, 0), (0, 1, 0, 0), (0, 0, 0, 1)] 5 Matrix Space The four fundamental subspaces associated to an m × n matrix A are: • The column space, col(A), spanned by all of the columns of A. This is a subspace of Rm. • The row space, row(A), spanned by all of the rows of A. This is a subspace of Rn. This also happens to be the column space of AT . • The nullspace (or kernel),null(A), consisting of all those vectors x for which Ax = 0. This is a subspace of Rn. • The left nullspace, which is just the nullspace of AT . This is a subspace of Rm. 0 1 3 −2 0 2 0 1 B 2 6 −5 −2 4 −3 C Example 7. Consider a matrix B = B C. Find the four fundamental @ 0 0 5 10 0 15 A 2 6 0 8 4 18 subspaces associated to B. 4 sage: B = matrix([[1, 3,-2 ,0 ,2 , 0],[2,6 ,-5,-2,4,-3],[0,0,5,10,0,15],[2,6,0,8,4,18]]) sage: B.column_space() Free module of degree 4 and rank 3 over Integer Ring Echelon basis matrix: [ 1 0 10 4] [ 0 1 -5 2] [ 0 0 0 6] sage: B.row_space() Free module of degree 6 and rank 3 over Integer Ring Echelon basis matrix: [1 3 0 4 2 0] [0 0 1 2 0 3] [0 0 0 0 0 6] sage: B.right_kernel()# Null space ofB Free module of degree 6 and rank 3 over Integer Ring Echelon basis matrix: [ 1 1 0 0 -2 0] [ 0 2 0 0 -3 0] [ 0 0 2 -1 2 0] sage: B.left_kernel() Free module of degree 4 and rank 1 over Integer Ring Echelon basis matrix: [10 -5 -1 0] sage: (B.T).right_kernel() Free module of degree 4 and rank 1 over Integer Ring Echelon basis matrix: [10 -5 -1 0] Exercise 5.1. Find the four fundamental subspaces associated to B using rref. 6 Gram-Schmidt Process Example 8. Apply the Gram-Schmidt orthogonalization process find an orthogonal basis from a basis Show that the following set of vectors in Q4 is a basis of Q4. L = f(2; −1; 3; 1); (3; 3; 4; 0); (1; −2; 0; −1); (4; 5; 5; −1)g of R4. sage: v1 = vector([2, -1, 3, 1]) sage: v2 = vector([3, 3, 4, 0]) sage: v3 = vector([1, -2, 0, -1]) sage: v4 = vector([4, 5, 5, -1]) sage: L = [v1,v2,v3,v4] We shall use the orthogonal function we have defined earlier. sage: u1 = v1/norm(v1).n() sage: q2 = v2 - proj(v2,u1) sage: u2 = q2/norm(q2).n() sage: q3 = v3 - proj(v3,u1)-proj(v3,u2) sage: u3 = q3/norm(q3).n() sage: q4 = v4 - proj(v4,u1)-proj(v4,u2)-proj(v4,u3) sage: u4 = q4/norm(q4).n() sage: G = [u1,u2,u3,u4] 5 sage:G [(0.516397779494322, -0.258198889747161, 0.774596669241483, 0.258198889747161), (0.229415733870562, 0.917662935482247, 0.229415733870562, -0.229415733870562), (0.489142426032054, -0.286738663536032, -0.151802821872017, -0.809615049984090), (-0.664410597026756, -0.0949157995752659, 0.569494797451495, -0.474578997876241)] Next we shall write a SageMath subroutine to find an orthonormal basis.