LINEAR ALGEBRA. September 23, 2010


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1 LINEAR ALGEBRA September 3, 00 Contents 0. LUdecomposition 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 SemiDefinite Matrices Similar Matrices and Jordan Form Linear Transformation LUdecomposition 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 skewsymmetric 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 offdiagonals. 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 offdiagonals 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 3space 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 threedimensional 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 twosided 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 3space 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 GramSchmidt 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 reversetriangular 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 offdiagonal 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 rankone 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 SemiDefinite 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 crossdiagonal (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 crossdiagonal 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
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