LINEAR ALGEBRA. September 23, 2010

Transcription

1 LINEAR ALGEBRA September 3, 00 Contents 0. LU-decomposition Inverses and Transposes Column Spaces and NullSpaces Ax = 0 and Pivot Variables Solving Ax = b The Four Fundamental Subspaces Orthogonality Projections Gram Schmidt technique Determinants Properties of Determinants Cramers rule Eigenvalues and Eigenvectors Diagonalization of Matrices Symmetric and Positive Semi-Definite Matrices Similar Matrices and Jordan Form Linear Transformation LU-decomposition Aufgabe Solve the nonsingular triangular system u + v + w = b () v + w = b () w = b 3 (3) Show that your solution gives a combination of the columns that equals the column on the right. Lösung zu Aufgabe u = b b, v = b b 3, w = b 3 Aufgabe Explain why the system u + v + w = (4) u + v + 3w = (5) v + w = 0 (6)

2 is singular, by finding a combination of the three equations that adds up to 0 =. What value should replace the last zero on the right side, to allow the equations to have solutions and what is one of the solutions? Lösung zu Aufgabe eq() eq() + eq(3) = 0 = ; ; (3,, 0) is one of the solutions Aufgabe 3 Apply elimination to produce the factors L and U for A =, A = 3 3, A = 4 4 (7) Lösung zu Aufgabe ; ; (8) Aufgabe 4 How could you factor A into a product U L, upper triangular times lower triangular? Would they be the same factors as in A = LU? Lösung zu Aufgabe 4 Gaussian elimination starting from the bottom row (or maybe the first column); no. 0. Inverses and Transposes Aufgabe 5 Which properties of a matrix A are preserved by its inverse (assuming A exists)? () A is triangular () A is symmetric (3) A is tridiagonal (4) all entries are whole numbers (5) all entries are fractions (including whole numbers like 3 ) Lösung zu Aufgabe 5 (), (), (5). Aufgabe 6 (a) How many entries can be chosen independently, in a symmetric matrix of order n? (b) How many entries can be chosen independently, in a skew-symmetric matrix of order n?

3 Lösung zu Aufgabe 6 (a) n(n + )/. (b) (n )n/. Aufgabe 7 If A = L D U and A = L D U, prove that L = L, D = D and U = U. If A is invertible, the factorization is unique. (a) Derive the equation L L D = D U U and explain why one side is lower triangular and the other side is upper triangular. (b) Compare the main diagonals in that equation, and then compare the off-diagonals. Lösung zu Aufgabe 7 (a) The inverse of a lower (upper) triangular matrix is still a lower (upper) triangular matrix. The multiplication of the two lower (upper) triangular matrices gives a lower (upper) triangular matrix. (b) The main diagonals of L L D and D U U are the same as those of D and D respectively. L L D = D U U, so we have D = D. By comparing the off-diagonals of L L D = D U U, they must both be diagonal matrices. L L D = D, D U U = D, D is invertible L L = I, U U = I L = L, U = U. 0.3 Column Spaces and NullSpaces Aufgabe 8 Which of the following subsets of R 3 are actually subspaces? (a) The plane of vectors with first component b = 0. (b) The plane of vectors b with b =. (c) The vectors b with b b = 0 (this is the union of two subspaces, the plane b = 0 and the plane b = 0). (d) The solitary vector b = (0, 0, 0). (e) All combinations of two given vectors x = (,, 0) and y = (, 0, ). (f) The vectors (b, b, b 3 ) that satisfy b 3 b + 3b = 0. Lösung zu Aufgabe 8 (a), (d), (e), (f) Aufgabe 9 Let P be the plane in 3-space with equation x + y + z = 6. What is the equation of the plane P 0 through the origin parallel to P? Are P and P 0 subspaces of R 3? Lösung zu Aufgabe 9 x + y + z = 0; P 0 is a subspace of R 3, P isn t. Aufgabe 0 Which descriptions are correct? The solutions x of [ Ax = x x 0 0 = 0] x 3 (9) form a plane, line, point, subspace, nullspace of A, column space of A. 3

4 0.4 Ax = 0 and Pivot Variables Aufgabe For the matrix A = [ 0 4 ] (0) determine the echelon form U, the basic variables, the free variables, and the general solution to Ax = 0. Then apply elimination to Ax = b, with components b and b on the right side; find the conditions for Ax = b to be consistent (that is, to have a solution) and find the general solution in the same form as Equation (3). What is the rank of A? Lösung zu Aufgabe U = ; u, w, y are basic variables and v is free; the general solution to Ax = is x = (u 4w, w, y); Ax = bis consistent if b b = 0; the general solution to Ax = b is u x = b 4w w = u w 4 + y b 0 ; r =. y Aufgabe Write the general solution to u [ v = 4 5 4] w () as the sum of a particular solution to Ax = b and the general solution to Ax = 0, as in (3). Lösung zu Aufgabe u v = v 3 v = v w 0 Aufgabe 3 Find the value of c which makes it possible to solve u + v + w = () u + 3v w = 5 (3) 3u + 4v + w = c (4) 0.5 Solving Ax = b Aufgabe 4 Is it true that if v, v, v 3 are linearly independent, that also the vectors w = v +v, w = v + v 3, w 3 = v + v 3 are linearly independent? (Hint: Assume some combination c w + c w + c 3 w 3 = 0, and find which c i are possible.) 4

5 Lösung zu Aufgabe 4 Yes; c (v + v ) + c (v + v 3 ) + c 3 (v + v 3 ) = 0 (c + c )v + (c + c 3 )v + (c + c 3 )v 3 = 0 c + c = 0, c + c 3 = 0, c + c 3 = 0 c = c = c 3 = 0 w, w, w 3 are independent. Aufgabe 5 Find a counterexample to the following statement: If v, v, v 3, v 4 is a basis for the vector space R 4, and if W is a subspace, then some subset of the v s is a basis for W. Lösung zu Aufgabe 5 Let v = (, 0, 0, 0),..., v 4 = (0, 0, 0, ) be the coordinate vectors. If W is the line through (,, 3, 4), none of the v s are in W. Aufgabe 6 Suppose V is known to have dimension k. Prove that (i) any k independent vectors in V form a basis; (ii) any k vectors that span V form a basis. In other words, if the number of vectors is known to be right, either of the two properties of a basis implies the other. Lösung zu Aufgabe 6 (i) If it were not a basis, we could add more independent vectors, which would exceed the given dimension k. (ii) If it were not a basis, we could delete some vectors, leaving less than the given dimension k. Aufgabe 7 Prove that if V and W are three-dimensional subspaces of R 5, then V and W must have a nonzero vector in common. Hint: Start with bases of the two subspaces, making six vectors in all. Lösung zu Aufgabe 7 If v, v, v 3 is a basis for V, and w, w, w 3 is a basis for W, then these six vectors cannot be independent and some combination is zero: c i v i + d i w i = 0, or c i v i = d i w i is a vector in both subspaces. 0.6 The Four Fundamental Subspaces Aufgabe 8 Find the dimension and construct a basis for the four subspaces associated with each of the matrices A = and U = (5)

6 Lösung zu Aufgabe 8 R(A) : r =, (, ); N(A) : n r = 3, (, 0, 0, 0), (0, 4,, 0), (0, 0, 0, ); R(A T ) : r =, (0,, 4, 0); N(A T ) : m r =, (, ); R(U) : (, 0); N(U) : (, 0, 0, 0), (0, 4,, 0), (0, 0, 0, ); R(U T ) : (0,, 4, 0); N(U T ) : (0, ). Aufgabe 9 If the product of two matrices is the zero matrix, AB = 0, show that the column space of B is contained in the nullspace of A. (Also the row space of A is the left nullspace of B, since each row of A multiplies B to give a zero row.) Lösung zu Aufgabe 9 AB = 0 A(b,..., b n ) = 0 Ab = 0,..., Ab n = 0 b N(A),..., b n N(A) R(B) is contained in N(A). Aufgabe 0 Explain why Ax = b is solvable if and only if rank A = rank A, where A is formed from A by adding b as an extra column. Hint: The rank is the dimension of the column space; when does adding an extra column leave the dimension unchanged? Lösung zu Aufgabe 0 Ax = b is solvable b R(A) R(A) = R(A ) rank A = rank A. Aufgabe does Suppose A is an m by n matrix of rank r. Under what conditions on those numbers (a) A have a two-sided inverse: AA = A A = I? (b) Ax = b have infinitely many solutions for every b? Lösung zu Aufgabe (a) m = n = r (b) n > m = r Aufgabe If Ax = 0 has a nonzero solution, show that A T y = f fails to be solvable for some right sides f. Construct an example of A and f. Lösung zu Aufgabe Ax = 0 has a nonzero solution r < n R(A T ) smaller than R n A T y = f is not solvable for some f. 0.7 Orthogonality Aufgabe 3 In R 3 find all vectors that are orthogonal to (,, ) and (, -, 0). Produce from these vectors a mutually orthogonal system of unit vectors (an orthogonal system) in R 3. 6

7 Lösung zu Aufgabe 3 All multiples of (,, ); (/ 3, / 3, / 3), (/, /, 0), (/ 6, / 6, / 6). Aufgabe 4 Show that x y is orthogonal to x + y if and only if x = y. Lösung zu Aufgabe 4 (x y) T (x + y) = 0 x T x + x T y y T x y T y = 0 x T x = y T y x = y. Aufgabe 5 Let P be the plane (not a subspace) in 3-space with equation x + y z = 6. Find the equation of a plane P parallel to P but going through the origin. Find also a vector perpendicular to those planes. What matrix has the plane P as its nullspace, and what matrix hast P as its row space? 0.8 Projections Aufgabe 6 Suppose A is the 4 4 identity matrix with its last column removed. A is 4 3. Project b = (,, 3, 4) onto the column space of A. What shape is the projection matrix P and what is P? Lösung zu Aufgabe Gram Schmidt technique Aufgabe 7 If u is a unit vector, show that Q = I uu T is an orthogonal matrix. (It is a reflection, also known as a Householder transformation.) Compute Q when u T = [ ]. Lösung zu Aufgabe 7 (I uu T ) T (I uu T ) = I 4uu T + 4uu T uu T = I; Q = Aufgabe 8 Show, by forming b T b directly, that Pythagoras law holds for any combination b = x q x n q n of orthonormal vectors: b = x x n. In matrix terms b = Qx, so this again proves that lengths are preserved: Qx = x. Lösung zu Aufgabe 8 (x q x n q n ) T (x q x n q n ) = x x n b = b T b = x x n Aufgabe 9 Apply the Gram-Schmidt process to a = 0 0, b = 0, c =. (6) 7

8 and write the result in the form A = QR. Lösung zu Aufgabe = 0 0, Determinants Aufgabe 30 How are det(a), det( A), and det(a ) related to det A, when A is n by n? Lösung zu Aufgabe 30 n det(a); ( ) n det(a); (det(a)) Aufgabe 3 (a) a rank one matrix Find the determinants of: A = 4 [ ] (7) (b) the upper triangular matrix U = (8) (c) the lower triangular matrix U T ; (d) the inverse matrix U ; (e) the reverse-triangular matrix that results from row exchanges, M = (9) Lösung zu Aufgabe 3 (a) 0 (b) 6 (c) 6 (d) /6 (e) 6 Aufgabe 3 If every row of A adds to zero prove that det A = 0. If every row adds to prove that det(a I) = 0. Show by example that this does not imply det A =. Lösung zu Aufgabe 3 Adding every column of A to the first column makes it a zero column, so det ] A = 0. If every row of A adds to, every row of A I adds to 0 det(a I) = 0; A =, det(a I) = 0, but det A = 0. 8 [

9 0. Properties of Determinants Aufgabe 33 diagonals: Suppose A n is the n by n tridiagonal matrix with s everywhere on the three A = [ ], A = 0, A 3 =,... (0) 0 Let D n be the determinant of A n ; we want to find it. (a) Expand in cofactors along the first row of A n to show that D n = D n D n. (b) Starting from D = and D = 0 find D 3, D 4,..., D 8. By noticing how these numbers cycle around (with what period?) find D 000. Lösung zu Aufgabe 33 (b) 6; D 000 = D = D 4 = Aufgabe 34 Explain why a 5 by 5 matrix with a 3 by 3 zero submatrix is sure to be a singular (regardless of the 6 nonzeros marked by x s): x x x x x x x x x x the determinant of A = x x is zero. () x x x x Lösung zu Aufgabe 34 In formula (6), a α...a 5v is sure to be zero for all possible (α,..., v). A + = 4. Or by 3.6.4, rank Aufgabe 35 If A is m by n and B is n by m, show that ( ) 0 A I 0 det = = det AB. Hint: Postmultiply by. B I B I () Do an example with m < n and an example with m > n. Why does the second example have det AB = 0? Lösung zu Aufgabe 35 0 A I 0 AB A I 0 0 A AB A =, det = det = det = B I B I 0 I B I B I 0 I det(ab); e.g.a = [ ] [ [ 0 A, B =, det = 5 = det(ab); A =, B = ] B I ] [ ], 0 A det = 0 = det(ab), because AB is a matrix with rank(ab) rank(a) n < m. B I 9

10 0. Cramers rule Aufgabe 36 The determinant is a linear function of the column. It is zero if two columns are equal. When b = Ax = x a + x a + x 3 a 3 goes into the first column of A, then the determinant of this matrix B is b a a 3 = x a + x a + x 3 a 3 a a 3 = x a a a 3 = x deta (a) What formula for x comes from left side = right side? (b) What steps lead to the middle equation? Lösung zu Aufgabe 36 (a) x = det([b a a 3 ])/deta, if deta 0 (b) The determinant is linear in its first column so x a a a 3 + x a a a 3 + x 3 a a a 3. The last two determinants are zero because of repeated columns, leaving x a a a 3 which is x deta. 0.3 Eigenvalues and Eigenvectors Aufgabe 37 Suppose that λ is an eigenvalue of A, and x is its eigenvector: Ax = λx. (a) Show that this same x is an eigenvector of B = A 7I, and find the eigenvalue. (b) Assuming λ 0, show that x is also an eigenvector of A and find the eigenvalue. Lösung zu Aufgabe 37 Ax = λx (A 7I)x = (λ 7)x; Ax = λx x = λa x A x = (/λ)x. Aufgabe 38 Show that the determinant equals the product of the eigenvalues by imagining that the characteristic polynomial is factored into det(a λi) = (λ λ)(λ λ) (λ n λ) (3) and making a clever choice of λ. Lösung zu Aufgabe 38 Choose λ = 0 Aufgabe 39 Show that the trace equals the sum of the eigenvalues, in two steps. First, find the coefficient of ( λ) n on the right side of (5). Next, look for all the terms in a λ a a n a a λ a n det(a λi) = det (4)... a n a n a nn λ which involve ( λ) n. Explain why they all come from the product down the main diagonal, and find the coefficient of ( λ) n on the left side of (5). Compare. 0

11 Lösung zu Aufgabe 39 The coefficient is λ λ n. In det(λ λi), a term which includes an off-diagonal a ij excludes both a ii λ and a jj λ. Therefore such a term doesn t involve ( λ) n. The coefficient of ( λ) n must come from the main diagonal and it is a +...+a nn = λ +...+λ n. 0.4 Diagonalization of Matrices Aufgabe 40 Factor the following matrices into SΛS : A = and A =. (5) 0 0 Lösung zu Aufgabe 40 [ 0 0 ; Aufgabe 4 ] Suppose A = uv T is a column times a row (a rank-one matrix). (a) By multiplying A times u show that u is an eigenvector. What is λ? (b) What are the other eigenvalues (and why)? (c) Compute trace(a) = v T u in two ways, from the sum on the diagonal and the sum of λ s. Lösung zu Aufgabe 4 (a) Au = uv T u = (v T u)u λ = v T u. (b) All other eigenvalues are zero because dim N(A) = n. Aufgabe 4 If A is diagonalizable, show that the determinant of A = SΛS is the product of the eigenvalues. Lösung zu Aufgabe 4 det A = det(sλs ) = det S det Λ det S = det Λ = λ...λ n. 0.5 Symmetric and Positive Semi-Definite Matrices Aufgabe 43 If A = QΛQ T is symmetric positive definite, then R = Q ΛQ T is its symmetric positive definite square root. Why does R have real eigenvalues? Compute R and verify R = A for Lösung zu Aufgabe 43 Because Λ > 0. R = A = [ and A = ] ; R = 0 6. (6) Aufgabe 44 If A is symmetric positive definite and C is nonsingular, prove that B = C T AC is also symmetric positive definite.

12 Lösung zu Aufgabe 44 If x T Ax > 0 for all x 0, then x T C T ACx = (Cx) T A(Cx) > 0 (C is nonsingular so Cx 0). Aufgabe 45 If A is positive definite and a is increased, prove from cofactors that the determinant is increased. Show by example that this can fail if A is indefinite. Lösung zu Aufgabe 45 det A = a A +... If A is positive definite, then A > 0. As a is increased, a A is increased while the others don t change det A is increased. 0.6 Similar Matrices and Jordan Form Aufgabe 46 Find the Jordan forms (in three steps!) of 0 A = and B = (7) Lösung zu Aufgabe J =, J = Aufgabe 47 Show that each Jordan block J i is similar to its transpose, J T i = P J i P, using the permutation matrix P with ones along the cross-diagonal (lower left to upper right). Deduce that every matrix is similar to its transpose. Lösung zu Aufgabe 47 Every matrix A is similar to a Jordan matrix J = M AM, and by part (a), J = P J T P. (Here P is formed block by block from the cross-diagonal permutations used on each block J i.) Therefore A is similar to A T : M AM = J = P J T P = P M T A T (M T ) P, or A = (MP M T )A T (MP M T ). Aufgabe 48 Find by inspection the Jordan forms of 3 A = and B =. (8) Lösung zu Aufgabe J = , J = Linear Transformation Aufgabe 49 Suppose a linear T transforms (, ) to (, ) and (, 0) to (0, 0). Find T (v):

13 (a) v = (, ) (b) v = (3, ) (c) v = (, ) (d) v = (a, b) Lösung zu Aufgabe 49 Write v as a combination c(, ) + d(, 0). Then T (v) = c(, ) + d(0, 0).T (v) = (4, 4); (, ); (, ); if v = (a, b) = b(, ) + a b (, 0) then T (v) = b(, ) + (0, 0). Aufgabe 50 Suppose T is reflection across the 45Â line, and S is reflection across the y axis. If v = (, ) then T (v) = (, ). Find S(T (v)) and T (S(v)). This shows that generally ST T S. Lösung zu Aufgabe 50 S takes (x, y) to ( x, y). S(T (v)) = (, ). S(v) = (, ) and T (S(v)) = (, ). Aufgabe 5 Suppose we have two bases v,..., v n and w,..., w n for R n. If a vector has coefficients b i in one basis and c i in the other basis, what is the change of basis matrix in b = Mc? Start from b v b n v n = V b = c w c n w n = W c. (9) Your answer represents T (v) = v with input basis of v s and output basis of w s. different bases, the matrix is not I. Because of Lösung zu Aufgabe 5 If V b = W c then b = V W c. The change of basis matrix is V W. 3