MATH 23: ASSIGNMENT 4 SOLUTIONS More on the LU factorization Q.: pg 96, q 24. Find the P t LU factorization of the matrix 2 A = 3 2 2 A.. By interchanging row and row 4 we get a matrix that may be easily reduced to upper-triangular form using only row operations of the form R j kr i. To find U we apply R 2 + R, R 3 R 2, R 4 + R 3 giving U = 2 2 2 By examining the multiplier of each row operation we find that L is of the form L = Finally by applying the row interchange R R 4 to the identity matrix we find the required permutation matrix and so P t is P t = Thus the matrix A may be factorized as A = P t LU Q.2: pg 96, q 28. Solve the system Ax = b using the given factorization A = P t LU. As P P t = I, the linear system may be rewritten as LUx = P b. Use this fact to solve the linear system: 8 3 5 4 2 6 A = P t LU = 4 2 =, b = 4. 4 3 2 2 4
2 MATH 23: ASSIGNMENT 4 SOLUTIONS A.2. To start we compute (P t ) t = P, and apply it to both sides of the equation, P A = P b, as we have the P t LU factorization for A we compute the effect of P on b on the right hand side 4 4 P b = 6 = 6. 4 4 As P A = LU we may now apply the LU factorization approach, by setting Ux = y, we solve the problem Ly = P b, which has three linear equations: y = 4, y + y 2 = 6, 2y y 2 + y 3 = 4. Solving these equations we find a unique solution 4 y = 2. Taking the components of y we now solve the system Ux = y with the three corresponding linear equations 4x + x 2 + 2x 3 = 4, x + x 2 = 2, 2x 3 =, for which back substitution yields the solution 4 x = 2. There is a typo in the textbook here, if you compute the P t LU factorization one gets the matrix P A where A is given in the question. Subspaces, Basis, Dimension and Rank [ x Q.3: pg 25 q 2,3,4. Let S be the collection of vectors in R y] 2 that satisfy the given conditions. Show that each of these S form a subspace of R 2 or give a counterexample to show it doesn t form a subspace. x, y y = 2x xy A.3. This is not a subspace as u = [ ] belongs in the subspace but u = [ ] does not belong in the set. [ Any member of this set may be written as bfx = x for arbitrary x R. 2] As this describes a line which is a subspace we conclude that it is in fact a subspace.
MATH 23: ASSIGNMENT 4 SOLUTIONS 3 [ ] [ ] This set does not satisfy the addition property since u = and v = [ ] both belong to this set, however their sum u + v = does not belong to the set. Therefore it is not a subspace. Q.4: pg 25 q 2. Determine whether b is in col(a) and whether w is in row(a), where A = 3, b = 2, w = [, 3, 3]. 3 5 A.4. Forming the augmented matrix, row reduction gives the following reduced row echelon form 3 2 3 2 2 2 3 5 this is a consistent system and so b belongs to col(a). To see if w belongs to row(a) we form the analogue of the augmented matrices for row vectors: 3 3 5 3 3 2 As we have reduced to the row belong the line to the zero vector we have shown that it may be expressed as some linear combination of the row vectors w = 3 (first row of A) - 2 (second row of A), thus it must lie in the row space of A. Q5: pg 26 q 2. Give bases for row(a), col(a) and null(a) where A is the matrix 2 4 2 2 2 3 2 4 4 A.5. The reduced row echelon form, R, of the matrix A is 2 2 R = 7 3 2 Noting the linear dependence of the column vectors r 2 = 2r, r 4 = r + 3r 3 and r 5 = 2 r 3 + 7 2 r 3 we find that the column space consists of the first and third column vectors 2 Basis for col(a) =,
4 MATH 23: ASSIGNMENT 4 SOLUTIONS By examination we see that the linearly independent row vectors of R are the first and second, hence the basis for row(a) is Basis for row(a) = {[, 2,,, 2 ], [,,, 3, 7 2 ]} Finally taking R and forming the augmented matrix [R ] we find that a particular solution to Ax = satisfies x = 2x 2 x 4 2 x 5, x 3 = 3x 4 7 2 x 5 And so the basis for the null space of A is then 2 Basis for null(a) =, 3, 2 7 2 Q.6: pg 27 q 52. Show that w is in Span(B) and find the coordinate vector [w] B, B = 3, 5, w = 3. 4 6 4 A.6. Forming the augmented matrix, we apply row reduction 3 5 7 3 4 4 6 4 from the reduced row echelon form we find the coordinates relative to the basis [ ] 7 [w] B = 4 Linear Transformations Q.7: pg 23 q 4,6. Show that the given transformation is a linear transformation using the definition of a linear transformation. [ ] x T = x + 2y x y 3x 7y x x T y = x + y z x + y + z
MATH 23: ASSIGNMENT 4 SOLUTIONS 5 A.7. Expressing this in vector form we find a matrix [ ] 2 2 x x T = x + y = y y 3 7 3 7 z As this is a matrix transformation we conclude it is linear. Expressing this in vector form x x T y = x + y + z = y. z z Again, we conclude that is this a linear transformation as we may write it as a matrix transformation. Q.8: pg 27 q 9. In two dimensions, the three types of elementary matrices give rise to five 2 2 matrices of the form: [ ] [ ] [ ] [ ] [ ] k k, ; ;,. k k Each of these elementary matrices correspond to a linear transformation from R 2 to R 2. Draw pictures to illustrate the effect of each one on the unique square with vertices at (, ), (, ), (, ), (, ). (Hint: Think about vectors in standard position for each of these points.) Q.9 pg q 55. If A is an invertible 2 2 matrix, what does the Fundamental Theorem of Invertible Matrices say about the corresponding linear transformation T A in light of the previous question Q.8? A.9. The Fundamental Theorem of Invertible Matrices tells us that if T A is invertible it gives rise to an invertible matrix that is built out of elementary matrices. Thus any invertible transformation is built out composing elementary invertible transformations T Ei i =,..., 5 Q. pg q. Using the definition, prove that T : R n R m is a linear transformation if and only if T (c v + c 2 v 2 ) = c T (v ) + c 2 T (v 2 ) where c, c 2 are scalars and v, v 2 belong to R n A.. To regain the original two properties consider the case when c = c 2 = and c = c 2 = c respectively we find T (v +v 2 ) = T (v )+T (v 2 ) and T (cv 2 ) = ct (v 2 ) Alternatively, we take the two properties and prove the statement. By the first property the left hand side becomes T (c v + c 2 v 2 ) = T (c v ) + T (c 2 v 2 ). Applying the second property twice we find the right hand side becomes T (c v + c 2 v 2 ) = T (c v ) + T (c 2 v 2 ). = c T (v ) + c 2 T (v 2 )
6 MATH 23: ASSIGNMENT 4 SOLUTIONS References [] D. Poole, Linear Algebra: A modern introduction - 3rd Edition, Brooks/Cole (22).